CALCULUS 3 DR JASON SAMUELS LECTURE NOTES Table of Contents: 4 Sequences and Series 4 Sequences 8 Series 8 Intro 9 Tests for Convergence 9 n-th term

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1 1 CALCULUS 3 DR JASON SAMUELS LECTURE NOTES Table of Contents: 4 Sequences and Series 4 Sequences 8 Series 8 Intro 9 Tests for Convergence 9 n-th term test 10 Integral test 12 Direct Comparison test (DCT) 12 Limit comparison test (LCT) 13 Alternating Series Test (AST) 14 Ratio Test 14 Root test 14 Summary 15 Power Series 16 Power Series as Functions 16 find the power series of a function using algebra & calculus 18 Finding the Power Series of a Function: Taylor and Maclaurin Series 21 Taylor and Maclaurin Series, math applications 22 Other coordinate representations 22 Parametric equations 23 Calculus with Parametric equations 23 derivative 25 integral 25 area 25 arc length 26 surface area 26 Polar equations 28 Calculus with polar equations 28 derivative 29 integrals 29 area 29 arc length 30 surface area 31 Three Dimensions and Vectors 31 a coordinate system for 3 dimensions 32 distance 33 what is a vector 33 vector arithmetic- scalar multiplication and vector addition 34 vector length 35 vector direction (the unit vector) 35 dot product 36 Application in math: vector components (or projections) 36 Application in physics: force & work 37 cross product 38 math applications: area and volume

2 39 Application in Physics: force & torque 40 lines 41 planes 42 math applications with lines & planes 44 Vector Functions and Calculus 44 Curves in space 45 Derivatives 47 Integrals 47 Arc Length 48 Unit Tangent and Unit Normal Vectors 48 Curvature 49 Surfaces & Functions of several variables 49 Surfaces 51 Functions of several variables 51 Level Curves 54 Derivatives for functions of several variables 54 A review of derivatives for y=f(x) 54 Intro to Partial Derivatives (numerically, graphically, algebraically) 57 Higher partial derivatives 58 Summary of intro to partial derivatives 58 Tangent Plane 59 Linear Approximation 59 Differentials 61 Chain Rule 61 Implicit Differentiation 63 The Directional Derivative and the Gradient 64 The Connection Between the Gradient and Level Curves 66 Application in Math: Maximum & Minimum 68 Solving Optimization Problems with Lagrange Multipliers 70 Taylor Polynomials for Functions of Several Variables 71 Integration with Functions of Several Variables 71 Intro 72 Double Integral as a Volume 76 Double Integral as an Area 77 Applications of Multiple Integrals 77 Average value 78 Surface area 79 Density, mass, and center of mass 81 Probability 82 Double integral with polar coordinates 83 Triple integrals 84 Vector Calculus 84 Intro to vector fields 86 Vector field operators swirl, divergence and curl 89 Line integrals 89 For scalar functions and for vector field functions 92 Application in Physics: Density and Mass 93 Fundamental Theorem of Line Integrals (FTLI) 94 Application in Physics: Conservation of Energy 2

3 95 Green s Theorem 97 Parametric Surfaces 97 Tangent Plane for Parametric Surfaces 97 Surface Integrals 98 Application in Math: Surface Area for Parametric Surfaces 99 Application in Physics: density, total mass and center of mass 100 Surface Integral for Vector Fields 100 Application: Electrodynamics 100 Application: Thermodynamics 101 Limits for Multivariable Calculus 101 Numerical 102 Graphical 104 Algebraic 104 Limit definition of the derivative 105 Limit definition of the integral Appendix 3

4 4 Sequences and Series Sequence: a list of numbers Series: a list of numbers that are added together Term: each number (or expression) in the sequence or series Sequences Ex) 2,3,4,5,6,7,8,9 that s a sequence, just not a very interesting one What is the fifth term? Ex) 2,3,5,7,11,13, What s the next term? Ex) 1,1,2,3,5,8,13,21, What s the next term? Ex) 1,4,9,16, What is the next term? Notation: to represent the whole sequence, we write {a n }. In this example, a n =n 2 To represent one term, we write a 3. In this example, a 3 = 9. It is a tradition to use {a n } and {b n } or {a i } and {b i } for sequences (and series). That subscript is called the index. Ex) a n = (.3) n write the first five terms (starting at n=1) Note that some sequences (and series) allow you to write down a formula for a term (like the fourth example), some let you write down a formula but using the other terms (like the third example), and some do not permit any formula (like the second example). In this course we will study the easiest situation, the first one. In this class we usually will talk about infinite sequences and series. If the list was finite, you would just write the terms down (for a sequence) then add them up to find the sum (for a series). This is more difficult when there are an infinite number of terms. We will study the techniques of how to analyze that situation. Limits, or describing the target value of the sequence Ex) for the sequence {1, 1.9, 1.99, 1.999, , , } is there a number the terms are heading toward? If so, what is it? Notation: for the sequence, {a n } = {2 }, lim n a n = 2 It is tradition to refer to the limit as L. so in this example, L=2. Note: In this formula, n starts at 0. If we want to make n start at 1, we could write {2 we only care about what happens at infinity, starting at n=0 or n=1 does not change the answer. In problems about series, when we are adding terms, it does change the answer. } Since We can graph a sequence by putting the index values on the horizontal axis. An illustration of different ways sequences can converge to a limit:

5 5 We can use the laws of limits to make some rules to rewrite limits on complicated sequence formulas into easier formulas. Note that the formulas basically say that you can split up addition, subtraction, multiplication, division, exponents, and (continuous) functions into separate limits to calculate them individually Ex) lim + = lim + lim = cos lim + lim = cos(0) +0 = 1 One useful technique for finding the limit of a sequence is called the Squeeze Theorem Note how b n is caught between a n and c n, and gets squeezed to the value L. Ex) find the limit of a n = ( )

6 6 Do all sequences have a limit? Of course not. Ex) find the limit of a n = (-1) n The sequence is: 1, -1, 1, -1, 1, -1, Is there a single target value? No. The limit does not exist. We say the sequence diverges. A graph of this diverging sequence: Determine if the sequence converges or diverges. If it converges, state the limit. Ex) a n = e 1/n Ex) a n = 3 n Some tricky limits: Ex) find the limit of a n =! [note that for the limit of a sequence we assume that n if nothing is wri en] How do we find the limit? If you take the limit of numerator and denominator separately, you get We try to use L Hopitals Rule, but we cannot because we don t know the derivative of n! So let s explore the sequence by writing out some terms: a 1 = a 2 = a 3 = in general, a n = a n = so therefore a n < so 0 < a n < also lim n (0) = 0 and lim n = 0, so lim n a n = 0 ex) find lim! Let lim!= lim ln!= ln lim ln(!)= ln Now, note that!> + 1 ( 1) > (this is true for all even n, and that is sufficient because n! is an increasing expression) So, lim ln(!)> lim ln = lim ln Some other terminology and definitions A sequence is increasing if it is always getting bigger: a n+1 > a n A sequence is decreasing if it is always getting smaller: a n+1 < a n A sequence is bounded above if there is a value it never goes above: a n < M for all n (M for max)

7 A sequence is bounded below if there is a value it never goes below: a n > m for all n (m for min) A sequence is bounded if it is bounded above and below (Note that if M is a boundary above, so is M+1 or M+5. It is a different question to find the best boundary.) Note: one technical point. A sequence is a list of values, so it is discrete. Our familiar functions take on continuous values. Does this make a difference when taking a limit? The following picture makes clear that there is no difference. 7 This can be stated as a formal theorem: There is a formal definition for limits of sequences. It is not necessary for our purposes. It is presented here, with a brief explanation, as advanced material. The intuitive idea behind the limit of a sequence is, you can make the terms as close as you want to the limit value by going as far along in the sequence as you need. Ex) For a n = 2 (starting at n=0), suppose we want to get within.01 of the target value. You need to go to n=2, then that term and every term after is within.01 of the target value. Suppose we want to get within of the target value. Then a 5 and every term after is close enough. That within amount is called the error E. Thus, we can make our error as small as possible. In formal notation, this is: lim n a n = L means: for every error E, there is some place in the sequence N so that if n>n then a n -L <E

8 8 Series What is ? Clearly the sum is getting bigger and bigger, going toward infinity Notation: sum S = 2 diverges. What is ? After 1 term, sum = 1 After 2 terms, sum =1 After 3 terms, sum = 1 After 4 terms, sum = 1 Can you make a guess what the entire sum equals? Notation: sum S = =? We will prove this later. For now, here is a picture that supports this result: For a series, when we take the sum of the first k terms, that is called a partial sum. S 1 = 1 S 2 = 1 + S 3 = S 4 = Notation: We write it S k. In this example, S 4 = 1. Note that the sum of the series can be determined by looking at the list of partial sums, which is a sequence, and taking the limit. Notation: S = & S k = & S = lim Note: It is handy to call the series S and the partial sum S k so you don t have to write the summation notation repeatedly when solving a problem. Geometric Series Let s prove that earlier result about the sum S = Multiply both sides by ½ to get S = Subtract equations to get S = 1 So S = 2 This particular technique works on a special type of series called a geometric series. Notation: a geometric series S can be written S = a+ar+ar 2 +ar 3 + =. a is the first term and r is the ratio. Given one term, you multiply by r to get the next term. In our example, the first term is a=1 and the ratio is r = What is the general formula for the sum of a geometric series? Let s look at the partial sum, then use our technique.

9 9 S k = a + ar + ar 2 + ar ar k-1 + ar k rs k = ar + ar 2 + ar 3 + ar ar k + ar k+1 S k - rs k = a - ar k+1 (1-r)S k = a(1-r k+1 ) S k = ( ) S = lim = lim ( ) =? The result depends on r k+1 which depends on r. If r is large, r k+1 will become infinity, so the sum diverges. If r is small, r k+1 will become 0, and the sum is S = To be precise, for -1<r<1 (or r <1), the sum converges. Otherwise the sum diverges. Ex) find S = 2 Ex) find S = 3 Telescoping Series Ex) find S = ( ) To solve this requires some algebraic manipulation. S = Write out the terms of the partial sum: S k = Terms cancel! S k = 1 - Now find the sum S = lim = lim 1 = 1 This is called a telescoping series because terms cancel, knocking out the middle like a collapsing telescope. This type of series appears when we can factor the bottom and split it into fractions. Ex) calculate S = Tests for Series Convergence For most series, we will only have the tools in this course to determine whether or not the series converges, not the actual sum. N-th term test Ex) does the series converge? [discuss] This sum cannot converge, because it is always getting bigger by at least 1.

10 10 (Note: do not confuse this with the sequence, which converges to 1.) The n-th term test: For the series, if lim 0, then the series diverges. Ex) does converge? Note that the converse is not true. If lim prove using the next test. = 0 we do not know that the series converges, as we will What are the indications to use the N-th term test? - You should always use the N-th term test first (unless another test is clearly indicated) The Integral Test Consider the sum S = The graph of the sequence {a n } looks like this: We can rewrite the sum as S = 1, and depict 1 as the width of the rectangle, then the sum is an area. Now the graph of the sum, compared with its function graphed continuously, looks like this:

11 Notice that the sum is an approximation of the area under the curve, the actual area (plus the first sum term) is bigger. So + >. But now, let s draw the rectangles on the other side: 11 Now the integral is less than the whole sum: < Put it together and you have: < < +. Therefore the integral is a good approximation for the sum. In general, we have that ( ) < < ( ) +. Further, if the integral converges, then the sum converges, and if the integral diverges, then the sum diverges. Note that this only works if the terms (or function) are positive and decreasing. If not, then sometimes the rectangles are above the function, sometimes below, so the inequalities aren t true and you cannot make a generalization. Ex) for which values of p does the series converge: This is called a p-series, and will be very useful going forward, since we will often compare other series to it to determine if they converge or diverge. Ex) use the integral test to check if the series converges: What are the indications to use the integral test? - When the formula is a function we know how to integrate What are the requirements to use the integral test? - The terms (or function) must be positive and decreasing

12 Estimating a Series using a partial sum, and estimating the error with an integral (bonus material) One way to estimate the sum of an infinite series S is to add the first k terms and then give up. That is easy, but is it useful? Yes if we can show that the error (or remainder) from doing that isn t too big. If the infinite series is S = a 1 +a 2 +a 3 + and the partial sum is S k = a 1 + +a k, then S k leaves out the remainder R k = a k+1 + a k+2 + =. If {a n } is a positive, decreasing sequence, we can estimate the error (the sum R k ) using an integral from our formula earlier: Ex) for = ( ) < < ( ) a) Estimate the sum with S 6 b) Estimate the maximum error for S 6 c) How many terms are needed to ensure that the error is less than.001? 12 The Direct Comparison Test (DCT) We know that S = 2 converges, because it s a geometric series with r <1. But what about T= That s not a geometric series. But how does it compare to our known geometric series? It s smaller, in that every term is smaller (because the denominator is bigger). S converges, and T is smaller so T converges. That s the basic idea. Here is the formal statement: For two series S= a n and T= b n with all positive terms If S converges and b n a n, then T converges If S diverges and b n a n, then T diverges Ex) show that / diverges Note that this test does not work the other way. If a series T has terms bigger than the terms of a series S that converges, it could converge or diverge. For example, if S converges to 3 and T is bigger, it could be 5 or. What are the indications to use the Direct Comparison Test? - When the given series closely resembles a known series Limit Comparison Test (LCT) Let s start with a simple idea, and extend it to a harder one. If you double every term of a series, then the sum doubles. And whether the first series converges or diverges, the second series does the same. What if you approximately double each term? Then the series should approximately double. Again, whether the first series converges or diverges, the second series does the same. Now let s make that idea rigorous. Suppose Σ a n and Σ b n are infinite series with positive terms. If lim n =c, where c is a finite number (not 0), then either both series converge or both diverge. A proof of the LCT is in the appendix. Ex) determine whether the sum converges or diverges: S =

13 13 ex) determine whether the series converges or diverges: S = what are the indications to use the LCT? - When the given series, in the limit, resembles a known series What are the requirements to use the LCT? - Both series must have all positive terms Alternating Series Test (AST) An alternating series is a series whose terms alternate between positive and negative. Ex) Notation: S = b 0 b 1 + b 2 b 3 + b 4 - where b n > 0 OR S = ( 1) (so a n = (-1) n b n ) If it starts with a negative term, that could be written S = ( 1) This type of series converges or diverges under very straightforward conditions. AST: for S = ( 1), b n positive, if lim n b n = 0 and b n decreasing, then the series converges. If not then it diverges. The reason for this can be explained with a picture: For an alternating series, what if you approximate the sum S with the partial sum S k, what is the error? From the picture, you see that it is at most the next term, b k+1. Note that this method to estimate the error only works for alternating series (think of the picture). Ex) for the series S = ( 1) (a) prove that it converges (b) find S 4 and the maximum error of that estimate (c) find the sum with error less than.001 What are the indications to use the Alternating Series Test? - When you have an alternating series you must use the AST, and at no other times Before the next test, we need to define a few terms. We saw that diverges, but ( 1) converges. When a series converges as it is, but not when you take the absolute value of the terms, it is conditionally convergent. If it converges even when you take the absolute value, it is absolutely convergent. Note that if a series is absolutely convergent then it is convergent. That is because taking the absolute value of all terms can only make the sum of a series bigger.

14 14 Ratio Test For a series =, if lim < 1, then the series converges absolutely, so it converges. If the limit >1, the series diverges. If the limit = 1, there is no conclusion. Ex) use the ratio test to prove that converges Ex) use the ratio test to prove that! diverges Ex) what does the ratio test tell us about the two series and The proof of the Ratio Test uses geometric series, and appears in the appendix. What are the indications to use the Ratio Test? - This is one of the most powerful tests, it is used in many different problems - When you see a factorial or a variable exponent The Root Test For a series =, if lim < 1 then the series converges. If the limit > 1, the series diverges. If the limit = 1, there is no conclusion. Ex) determine if the series S converges, S = What are the indications to use the root test? - When the entire term is raised to a power involving n. A Summary of the Tests for Series Convergence What are the indications to use the N-th term test? - You should always use the N-th term test first (unless another test is clearly indicated) What are the indications to use the integral test? - When the formula is a function we know how to integrate What are the requirements to use the integral test? - The terms (or function) must be positive and decreasing

15 15 What are the indications to use the Direct Comparison Test (DCT)? - When the given series closely resembles a known series what are the indications to use the Limit Comparison Test (LCT)? - When the given series, in the limit, resembles a known series What are the indications to use the Alternating Series Test (AST)? - When you have an alternating series you must use the AST, and at no other times What are the indications to use the Ratio Test (RaT)? - This is one of the most powerful tests, it is used in many different problems - When you see a factorial or a variable exponent What are the indications to use the Root Test (RoT)? - When the entire term is raised to a power involving n. For the following series, which test would you use to make a conclusion about convergence? (Note the abbreviated sum notation. We assume the sum is for n going to, which is all we need to solve it.) Ex) ( ) Ex) ( 1) Ex) ( ) Ex) Ex) 1.1( 0.4) Ex) ln When you solve test-for-convergence problems, make sure you (a) state which test you are using (b) show that you have satisfied the test conditions (c) state your conclusion about your original series Power Series Next we will look at a very important type of series and see how it is a powerful tool to do advanced mathematics. A power series has the form S = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + =. It may help to think of it as an infinite polynomial. Again we want to know if a given series converges. With power series, we always use the ratio test (because it always works). Ex) test the convergence of S = Ratio test: lim ( ) = lim ( ) = We want this to be L < 1. So x < 3 (or -3<x<3) But recall that there is no conclusion for L=1. So when x =3, we don t know if it converges or diverges, so we have to check.

16 16 x=3: S = = by the n-th term test, lim a n 0 so the series diverges x=-3: S = ( ) = ( ) = ( ) by the n-th term test, lim an 0 so the series diverges Therefore, the power series converges for -3<x<3. This is called the interval of convergence. The center of the interval a is at x=0. The interval extends by 3 in each direction, so 3 is the radius of convergence R. Ex) find the interval of convergence for S =!( 2) RaT: This is always infinity and the series always diverges unless x=2. So the interval of convergence is x=2 (a single point). The radius of convergence is R=0. Ex) find the interval of convergence for S = RaT:! So the series converges for all real values of x. The interval of convergence is (-, ) OR - <x< and the radius of convergence is R=. Power Series are Functions This is true in three different ways. First, consider f(x) = Then f(0.5) = which is some value Second, a power series is a function of x. For example, f(x) = is a geometric series with first term 1 and ratio x. Thus, f(.5) = domain of the function.. = 2. Note that this series has an interval of convergence -1<x<1, so that is the Third, as a result, it is now clear that f(x) = = as long as -1<x<1. So we see that a typical function can be written as a power series! There are many different ways to rewrite a function as a power series. We will look at three. Developing the power series for function through algebra We know that: So what about: Ex) what is the power series for? By a simple substitution, Ex) what is the power series for? By multiplication,

17 17 Note: it is not at all obvious that this is always allowed, since we are working with an expression with an infinite number of terms. However, theorems from higher mathematics prove that this works. Those proofs are beyond the scope of this course. Developing the power series for a function through calculus Ex) what is the power series for ( )? By taking the derivative, we get: Several amazing things happened in that calculation which deserve explanation. First of all, are we allowed to take the derivative of an infinite number of terms? Yes (the theorem is given below). Second, how do we know that, after taking the derivative, the function and the power series are still equal? That s another theorem. Part of that theorem says that the equality is only true for appropriate values of x. For example, at x=5, the original equality does not hold, so it still fails after you take the derivative. (The interval of convergence of the new series is almost the same as the interval for the original series. It has the same center and radius. However, the new series may diverge at the endpoints.) One tricky detail is the way the sum boundary changed in the last step from n=0 to n=1. Why did it change? Because when the derivative is taken, the n=0 term is a constant and so it disappears. It is quicker to do this calculation using the summation notation, because there is less to write down. However, that makes it easy to miss small details like this one. Now, lets do the same thing with integration. Ex) find the power series for ln 1-x Since we want to have the function ln 1-x, plug in x=0 to solve for C: ln(1)=c so C=0 (Note that specifying the result after integrating is like specifying an initial condition, so you can solve for C) Note that the center and radius of convergence are same as the original series, a=0, R=1. Theorem for the derivative and integral of power series (the proof is beyond the scope of this course) For the power series f(x),

18 18 with radius of convergence R around center of convergence a, the derivative and integral are Both the integrated and differentiated series have the same center and radius of convergence. Developing the power series of any function: Taylor and Maclaurin series Previously, we have approximated a function with a polynomial specifically, a linear function. f(x) L(x) = f(a) + f (a) (x-a) Can we make a better approximation? Well, since this approximation has the same value and first derivative (or slope) at x=a, we could find an expression with the same second derivative. And third derivative The proper way to do this is with a polynomial. You will see why this works as we develop the formula. Also, one advantage of having a polynomial is that it is much easier to calculate with than many other types of functions. (For example, we have no good way to calculate sin(1.2). We will see several applications of the benefits of these polynomials.) We will see that the approximation becomes exact with an infinite number of terms; that infinite polynomial is a power series. Here is the notation for our goals: For some function f(x), we will start with the approximation: f(x) c 0 + c 1 (x-a) + c 2 (x-a) 2 + c 3 (x-a) c k (x-a) k = ( ) Eventually, we want the equality: f(x) = c 0 + c 1 (x-a) + c 2 (x-a) 2 + c 3 (x-a) 3 + = ( ) [The expression simplifies if we use a=0, since you get c n x n. We will often do that, but not always.] All we need to know is what those coefficients c n are. For the first step, we will very cleverly plug in x=a, and almost every term cancels, leaving f(a) = c 0. So we can solve for the constant term. Now, we will one by one put each coefficient in the constant position. How can we do that? By taking the derivative. f (x) = c 1 + 2c 2 (x-a) + 3c 3 (x-a) 2 + Plug in x=a to get f (a) = c 1 f (x) = 2c 2 + 6c 3 (x-a) + So f (a) = 2c 2 f (x) = 6c 3 + So f (a) = 6c 3 Before we write the general formula, note that we are accomplishing two goals. One, assuming we want an expression with all the same derivatives as the function, we found out what the expression must be. Two, assuming we want a power series equal to the function, we found out what the coefficients must be. In general, the n-th derivative reveals that f (n) (a) = n!c n and therefore: = ( ) ( ) Therefore, f(x) = ( ) ( )! ( ) = ( )+ ( ) ( )+ ( )! ( ) + ( )! ( ) + This special power series is called the Taylor Series of the function. In the special case where a=0, this is called the Maclaurin Series. If the expression stops at the k-th term (which includes x k ), that is called the k-th Taylor Polynomial, T k. For example, note that T 1 is just the linear approximation for the function: T 1 (x)=f(a)+f (a) (x-a)!

19 19 Ex) find the Maclaurin series for f(x) = e x. Also find the interval of convergence. Ex) find the Taylor series for f(x) = e x at a=3 Ex) find the Maclaurin series for f(x) = sin x Ex) find the Maclaurin series for ( )= 1 + Ex) find the Maclaurin series for ( )= 1 + The hard way: start from step one and calculate derivatives, etc The easy way: take the Maclaurin series for ( )= 1 + and replace x with x 2

20 20 You can also integrate and differentiate. When you integrate remember to solve for C Ex) find the series for cos(x) by integrating the series for sin(x) sin(x) =! +! -sin(x) = +!! + cos(x) = -sin(x)dx = +!! + = +!! + Then plug in x=0 cos(0)=c C=1 So, when you know what the function should be after you integrate, don t forget to solve for C In general, using a=0 makes the expression simplify a lot, so the Maclaurin series is easier to calculate and more commonly used. A table of the most common Maclaurin series (with the radius of convergence): An example of finding a Taylor Series using clever shortcuts Ex) find the Taylor Series for sin 2 x sin 2 x = ½ (1-cos(2x)) = ½ (1 1 ( )! + ( )! ( )! + ) = ½ ( )! ( )! + ( )! = +!!! Or, using summation notation, ( ) ( )! The Taylor Remainder Theorem [bonus material] In most functions encountered in this course, a function is equal to its Maclaurin or Taylor series; this is the situation we will focus on. However, this is not automatically the case. You can determine if they are equal by the Taylor Remainder Theorem. We will state it, but not cover it in depth.

21 Let T n be the n-th Taylor polynomial, the Taylor series cut off after the term with (x-a) n. When approximating f(x) with T n (x), let R n (x) be the remainder (or error): f(x) = T n (x) + R n (x). then R n = ( ) ( ) ( ) for some value c with a<c<x ( )! Also, if f (n+1) (x) <M, then R n (x) < ( )! and lim n R n (x) = 0 and lim n T n (x) = f(x) Roughly speaking, this means that it s a fact that x-a is finite and n! goes to infinity, so as long as all derivatives are finite, then the error term goes to 0. A proof is in the appendix. Applications: Some uses of Taylor (and Maclaurin) series Finding approximate values Ex) Approximate (without a calculator) to within Finding limits Ex) find lim x 0 ( ) What if we used l hopitals rule? Finding integrals Ex) find (leave answer as a series) Finding sums Ex) find [hint: S=! ] An amazing application of Taylor series Question: what is the value of e iπ? (recall that i= 1 ) Fact: e ix = cos(x) + i sin(x) We can prove this using Taylor Series: e ix = ( )! + ( )! + ( )! + ( )! + ( )! + ( )! + ( )! +

22 22 Other coordinate representations: parametric equations and polar coordinates What is the equation of a circle (radius=1, center is the origin)? x 2 + y 2 = 1 It s a simple equation for a simple graph, and yet it is not a function y=f(x). In this section, we will explore alternate ways to represent a graph which can be very useful. For example: We will actually learn the equation of this graph. Parametric Equations If we tried to describe to mathematically describe this graph, we could not write a function equation y=f(x). How could we describe it? In words, it looks like a path. Therefore, one way to describe it would be with a formula that gives our coordinates as we walk along the path. We could keep track of our journey using time. So at every time t, we need to describe our location. Our position has two components, x and y. therefore, x is a function of t and y is a function of t. Notation: this function relationship can be written explicitly as x = f(t), y = g(t), or it can be written implicitly as x = x(t), y = y(t). Either way, we assume that x & y are functions of t. (Note: Thinking of time as the parameter is the most common approach. Another common approach is the use of distance as the parameter. The important aspect is that a single variable changes to track our changing position on the path.) Ex) what is the graph of x=t 2 & y=t+1? Ex) what is the graph of x = 4t 2 & y = 2t+1?

23 Important lesson: different equations can represent the same graph. What changes is the starting position (when t=0) or the velocity at which the graph is traveled relative to the parameter t. Each set of equations is called a parameterization of the graph, and changing from one to another is called a reparameterization. In this example, the graph was reparameterized by replacing t with 2t: for x, (2t) 2 = 4t 2 ; for y, (2t)+1 Ex) What is the graph of x = cos(t) & y = sin(t)? 23 A typical function y=f(x) can be graphed as parametric equations: x = t and y = f(t) Also, cases in which x is a function of y can be represented as parametric equations: y=t & x=g(t) Parametric equations are one way to generalize the idea of a function. Sometimes, we can rewrite the parametric equations with only x & y. This is called eliminating the parameter. Ex) eliminate the parameter for: x=t+3, y=t 3 t=x-3 so y=(x-3) 3 Ex) eliminate the parameter for: x=3+2e t, y=5e 2t =e t now we use the fact that e 2t =(e t ) 2 so y=5 Calculus with Parametric Equations Previously, you learned calculus for expressions in two variables, typically x & y. Everything you learned can be applied to parametric equations. Calculating Derivatives We want to find the derivative, or the slope of the curve. This is possible because, at any point, when we zoom in we see a straight line, and the slope of that line is the derivative. However, we cannot calculate it in the usual way because we start, not with y=f(x), but with x=x(t) & y=y(t). We do know that on any segment of the graph we can treat y as a function of x, which itself is a function of t. y = y(x) y is a function of x y(t) = y( x(t) ) both are functions of t y = [y(x(t))] take the derivative, and your variable is t = this equation comes from the chain rule. It also makes sense if you cancel dx So: = (if = 0, this may be undefined)

24 24 To find the second derivative, take the derivative again in this way. (Note that ) = = A proof of the derivative formula using limits is in the appendix. Ex) for the curve x=t 2 & y=t 3-3t a) Find the slope of the tangent when t=2 b) Find the slope of the tangent at (1,-2) c) Confirm that there are two tangents at the point (3,0), then find the slope of each one. d) Find the points where the tangent is horizontal and where it is vertical e) Calculate the second derivative Ex) the graph x = 2t π sin(t) & y = 2 π cos(t) is called a prolate cycloid. find the slope of the tangent at the point (0, 2-π) First, note that we want y=2-π, so cos(t)=1, so probably t=0. Confirm that this gives x=0.

25 25 Slope = = Calculating Area We have previously seen that area can be calculated with an integral A = y dx When working with parametric equations, the variable is t, not x. So let s adjust the equation. Observe that x is a function of t, x=x(t), so dx=. Or, quite simply, uncancel dt. A = Note that your boundaries of integration are now t-values. They will correspond with your starting and ending x-values. Ex) x=4cos(t), y=3sin(t) defines an ellipse. Find the area enclosed by the ellipse. for the boundaries of integration, x goes from -4 to 4 as t goes from π to 0 Calculating Arc Length Recall the integration formula for arc length, L = 1 + When we proved it, at a certain step we multiplied by dx in the numerator and denominator. This time, since t is our variable, we will multiply by dt. In the notation, represent arc length using s. (ds) 2 = (dx) 2 + (dy) 2 (ds) 2 = ( ) ( ) ( ) ds = ( ) ( ) ( ) ds = + ds = + s = + ( ) ( ) where the boundaries of integration are t-values [Recall that you can convert to classical function notation y=f(x) by letting x=t. Applying that to this formula, you get the formula for arc length that you already knew. Once again, the parametric notation x=x(t) & y=y(t) is a generalization of the function notation y=f(x).] ex) find the circumference of a circle with radius r=3

26 26 Calculating Surface Area Previously, we saw that the formula for the surface area from revolving y=f(x) around the x-axis was SA = 2πhL = For parametric curves, let s just plug in the new formula for L: SA = 2πhL = 2 + Ex) find the surface area of a sphere with radius r=3 Finally, recall that graph from the beginning of the unit: Its equation is: x=sin(t)+.5cos(5t)+.25sin(13t), y= cos(t)+.5sin(5t)+.25cos(13t) use a grapher to see more Polar Coordinates [omit] In many cases, the most natural way to describe a location or a graph in the plane is by using (x,y)- coordinates. However, this is not always the case. For example, a natural way to describe a circle (centered at the origin) is to give the radius, say r=4. Notice how that completely describes the graph. Similarly, a straight line (through the origin) could easily be described by the angle it makes with the x-axis, say =. Again, notice how this completely describes the line. We will now develop the coordinate system that relies on r and θ, and find it to be useful in a variety of situations. Any point on the plane can be defined by (r, θ), where r is the distance from the origin and θ is the angle formed with the positive x-axis by the line through the origin and the point. Ex) plot the points: A (3, ) B (1, ) C (-2,π)

27 27 Note that the easiest way to locate the point usually is to start at the origin, find the direction (angle), and then move by the amount of the radius. Note that if r=0, the point is the origin, and it does not matter what the angle is. It is possible to convert between rectangular and polar coordinates, using the definition of trig functions. sin θ = cos θ = solve for y to get y=r sin θ solve for x to get x = r cos θ to solve for r, find an equation that involves only r,x,y: x 2 + y 2 = r 2 = + to solve for θ, find an equation that involves only θ,x,y: tan θ = θ=arctan Note that tan is a periodic function and the arctan function only provides values for θ so that θ. You must be careful to choose θ so it is in the correct quadrant, when this matters. So to be precise, θ=arctan +kπ, for some integer k. Given (x,y): (r, θ) = +, arctan + kπ Given (r, θ): (x,y) = (r cos θ, r sin θ) Ex) for r = 2cos θ, (a) sketch the graph (b) find an equation in x&y Here are some interesting examples of graphs of polar equations

28 28 Ex) sketch r=2 cos 3θ Calculus in Polar Coordinates Again, everything you learned about calculus can be applied using polar coordinates. Calculating the derivative We want to find the slope of the tangent line. We borrow from the approach of parametric equations. Instead of introducing the variable t, we use the variable θ. (On a segment of the graph, r is a function of θ.)

29 29 = It is possible (but not necessary) to get a formula in r & θ. Assume that r = r(θ), and we get: ( ) = ( ) Ex) for r=1-cos θ, (a) sketch the graph (b) find the horizontal and vertical tangents Note: at θ=0, both and are 0. By examining the graph, we can conclude that this point is a cusp, the derivative is undefined, and no tangent line exists. (This could also be determined by calculating handed derivatives.) Calculating the integral Calculating Area with Polar Coordinates If we are using polar coordinates, we need a new integral formula for area; x y is an area, but r θ is not. To develop it, we follow a familiar approach, using a sum of small areas. In polar coordinates, what is the basic unit for which we know the area? A circle, or a slice of pie called a sector. The area of a circle is πr 2 and the area of a sector is πr 2, or θr 2. Assume r is a function of θ, r=r(θ). In a small slice, the width is Δθ and the height is r. The area of one small slice is ΔA r 2 Δθ. The total area is the sum of all of those slices: Approximate Area: A = ΔA r 2 Δθ Exact Area: A = da = r 2 dθ Ex) find the area enclosed by a limacon, r=2+cosθ Arc Length with Polar Coordinates We start with the classic arc length formula and use the same technique from parametric equations to convert to polar coordinates. Instead of t, we want θ to be our variable

30 30 L = + L = + θ We could stop there, or we can rewrite it completely in r & θ. = θ r sinθ + sinθ + r cosθ θ Under the radical: = + θ Ex) for a circle radius r=3, find the circumference Ex) find the arc length of the cardioid, r=2-2cos θ, 0< θ<2π [set up, do not solve] Calculate the Surface Area with Polar Coordinates Previously, we saw that the formula for the surface area from revolving y=f(x) around the x-axis was SA = 2πhL = For polar coordinates, the horizontal axis is θ=0, and plug in the new formula for L: SA = 2 + θ OR 2 sinθ + θ To rotate around the vertical axis, θ=, x replaces y as the height: SA = 2 + θ OR 2 cosθ + θ Ex) find the surface area when revolving r=cos θ around the vertical axis this is called the pinched torus

31 31 Three Dimensions and Vectors Space The Final Frontier. We live in three dimensions (spatially). We will now learn how to do mathematics in three dimensions. A Coordinate system for 3 dimensions First we need a coordinate system. In addition to the x-axis and y-axis, we introduce the z-axis. If the x & y axes are where they usually are, then the positive z-axis comes straight at you from your sheet of paper. It is impossible to actually draw a 3-dimensional object on a 2-dimensional piece of paper. We have to choose a perspective, and there are several typical ways to do this. In the first one, the x-axis is in the usual place. However, the axes are bunched together, so graphs tend to get squished. In the second one, the axes are spread out, but y is the horizontal axis, which is different from what we are used to. We will make drawings using the second option because they tend to be clearer. The third option is from the professional mathematics software, Mathematica. We will use that sometimes as well. (The software allows you to rotate your graph in any direction, so the orientation changes.) In all representations, z is on the vertical axis. (Often we will think of z as a function of x & y, z=f(x,y), so our function values are still represented by height.) Ex) graph all the points which satisfy z=0 Ex) graph y=2 This is just a copy of the standard xy plane (notice that from this angle its hard to see precisely where the plane lies)

32 32 Ex) graph x=1 Distance in 3 dimensions The distance between 2 points A (x 1, y 1, z 1 ) and C (x 2, y 2, z 2 ) is Dist(A,C) = ( ) + ( ) + ( ) That formula should seem pretty familiar and reasonable as an extension of the formula from 2 dimensions. Also, here is a picture and an explanation: Consider the point B (x 2, y 2, z 1 ). A & B are in the same plane, the distance between them is Dist(A,B) = ( ) + ( ) B & C are on a vertical line, so dist(b,c) = z 2 z 1 ABC is a right triangle, AC is the hypotenuse, so Dist(A,C) = ( (, )) + ( (, )) = ( ) + ( ) + ( ) Ex) find the distance between (2,3,-1) & (4,-3,0) Note: you can extend this formula beyond 3 dimensions, to any dimension.

33 33 What is a Vector A vector is a object with a magnitude and a direction. For example, speed is not a vector since it is always positive and does not describe the direction. Speed is a magnitude. However, velocity is a vector. In the 1- dimensional case, it has a magnitude, and the sign indicates forward or backward. Note: in the discussion of vectors, sometimes the definition or an example will be given for 2 dimensions, sometimes for 3 dimensions. You should assume that any number of dimensions is possible, and the definition is what seems natural. (We already saw this for the distance formula.) In cases where an idea does not apply for all dimensions, this will be emphasized. A vector is determined by a starting point and an ending point. However, when just a vector is given, the starting point is not known. (We usually draw it at the origin for simplicity.) Notation: to emphasize the endpoints: To emphasize the vector as one object: or v Ex) what is the vector with start point (4,7) and endpoint (5,3)? Picture: calculation: The vector from (x 1 y 1 ) to (x 2 y 2 ) is (x 2 x 1, y 2 y 1 ). Ex) draw the vector (-2,3)? Each number is called a component. Note: one option to describe a vector is to state the magnitude, then give the direction as an angle. This indicates a choice to use polar coordinates (or the equivalent in higher dimensions) instead of xyz coordinates Another notation for vectors: sometimes vectors are written as columns. = This notation is useful to compare the same component of different vectors. Scalar multiplication of vectors We can multiply a vector by a constant, called a scalar. This will scale the vector in size (and reverse it if the number is negative). Ex) for = (1,2) 3 = (3,6) 0.5 = (.5,1) -2 = (-2,-4) This also works for vectors in 3 dimensions (or any dimension) In general: = c(v 1, v 2, v 3 ) = (cv 1, cv 2, cv 3 ) Ex) for = (-1,4,2) 3 = (-3,12,6) Vector Addition and Subtraction It can be helpful to think of a vector as a change (or displacement). Imagine making a trip starting at point A, taking the move indicated by (to point B), then making the move indicated by v (to point C). Each part of the trip ( and v ) is a vector, and the entire trip (w ) is a vector. + For two vectors, = (u 1, u 2 ) & = (v 1, v 2 ) + = (u 1 +v 1, u 2 +v 2 ) OR + = +

34 34 Ex) find (3,-2) + (1,4) and draw a picture to represent the calculation This works for 3-dimensional vectors as well. Define two vectors in 3 dimensions, = (u 1, u 2, u 3 ) & = (v 1, v 2, v 3 ) + = (u 1 +v 1, u 2 +v 2, u 3 +v 3 ) Ex) find (3,-2,7) + (-2,4,-4) A parallelogram helps to visualize addition and subtraction of vectors: In the first parallelogram, starting at the dot, tracing then yields +. Tracing then yields +. Observing the diagram should confirm that these are the same vectors. (You can also prove this algebraically using the definition.) In the second parallelogram, tracing then yields. Alternately, the vector connecting the ends of u & v also yields. This is clear from the parallelogram, or from the fact that + ( ) =. We can emphasize the x,y,z components of a vector using vector addition. = (1,0,0) the unit vector in the x-direction [in 2-dimensions it would just be (1,0)] = (0,1,0) the unit vector in the y-direction [in 2-dimensions it would just be (0,1)] = (0,0,1) the unit vector in the z-direction In 2 dimensions: : = (u 1, u 2 ) = u 1 + u 2 OR u 1 i + u 2 j In 3 dimensions: = (v 1, v 2, v 3 ) = v 1 + v 2 + v 3 OR v 1 i + v 2 j + v 3 k,, are called the standard basis vectors. Ex) rewrite = (3,-4,1) using the standard basis vectors = Note that sometimes there is no distinction between the notation for a point and for a vector. We can think of a vector as a point. Or, we can think of a point as a vector. (Its starting point would be the origin.) This equivalence can be useful, but it can also be confusing. We will keep this in mind as we proceed. So far, all vector properties discussed also apply to vectors in 3 dimensions. Of course, on a sheet of paper, it is easier to draw vectors in 2 dimensions, so often the examples and explanations are presented like that. Length of a vector The length (or magnitude or norm) of a vector can be calculated by taking the origin as the starting point, the coordinates as the endpoint, and using the distance formula. Notation: the magnitude of = (v 1, v 2, v 3 ) is or = ( ) + ( ) + ( ) Ex) find (5,-2,4) Note that, for a vector and a number c, c = c As an obvious example, (1,0) = 1 and 3(1,0) =3

35 35 Direction of a vector: the Unit Vector We have a way to represent the magnitude of a vector. How can we represent the direction? One way is with a vector that has a magnitude of 1. This is called a unit vector. To turn a vector into a unit vector, divide by the magnitude. The unit vector in the direction of is: = or Ex) for =(8,6), find the direction (as a unit vector) and the magnitude Why does a unit vector represent direction? Note that, in polar coordinates, we could give the direction with an angle. If you think back to some activities from trigonometry, an angle could also be specified with a point on the unit circle; a unit vector does exactly that. Also note that for a unit vector and a positive number c, c = c = c. This is why we say a unit vector only indicates a direction and does not contribute to the magnitude. The Dot Product So far we added vectors. Is it possible to multiply vectors? What would it mean? To explain the dot product of two vectors, first we show the formula, then we discuss what it means. In 2 dimensions, for = (u 1, u 2 ) & = (v 1, v 2 ), we define the dot product = u 1 v 1 + u 2 v 2 In 3 dimensions it works as you expect. For = (u 1, u 2, u 3 ) & = (v 1, v 2, v 3 ), = u 1 v 1 + u 2 v 2 + u 3 v 3 Using column vector notation: = + + Note that the result of a dot product of two vectors is a real number (or scalar) and not a vector. Ex) find (1,4,1) (2,0,-3) The dot product has properties like multiplication with real numbers. If a,b,c are vectors, k is a scalar: a a = a 2 a b = b a a (b+c) = a b + a c (k a) b = k(a b) = a (kb) It is possible to prove these by writing out the components. For example, = a 1 a 1 + a 2 a 2 + a 3 a 3 = ( ) + ( ) + ( ) = The dot product has a nice property: = cos θ where θ is the angle between the vectors This can be proven using the Law of Cosines with the vectors,, and.

36 36 Ex) find the angle between (3,0) and (0,5) cosθ = cos θ = 0, so θ = arcos(0) =. Since the vectors lie on the x-axis and y-axis, that is what we expected. Note that these two vectors are perpendicular. So we can observe that two vectors are perpendicular (or orthogonal or normal) precisely when their dot product is 0. Notation: precisely when = 0 Two vectors are parallel if they point along the same line (same or opposite direction). This means that one must be a multiple of the other. and are parallel precisely when = Also, note that same direction means that the angle between them is 0. Therefore: and are parallel precisely when = Ex) are (3,1) & (-6,-2) parallel, perpendicular or neither? Ex) are (3,5,1) & (0,-1,5) parallel, perpendicular or neither? Dot product math application: Splitting vectors into components (or projections) Often we are interested in breaking down a vector into components. When we write down vectors in the usual way, we are giving the x-, y- and z-components. Ex) find the x-component of (4,3), also find its magnitude But we might want the component in a different direction. In the picture are two vectors, and. has been split into two components, one parallel to, called one perpendicular to called. Note that + = To find the component (or projection) in the direction, we find its magnitude and direction. Note that it is one leg of a right triangle, so its length is cosθ. It is parallel to so it is a multiple of. We want the vector to contribute direction, but not magnitude, so we divide by the magnitude of. Length = = v cos θ = direction (unit vector) = Component of in direction: vector = cosθ = Ex) find the component of (-1,4) in the (2,3) direction. = Application in Physics of dot product and vector components: Force and Work In Physics, recall that we can measure the amount of energy expended, called work, by applying a force over a certain distance. In the basic case, in which quantities are scalar and constant, W = F x, where W is work (in Joules), F is force (in Newtons) and x is distance (in meters). When you drag a bag of laundry along the

37 ground, the only part of your pulling force doing work is the component in the direction the bag is moving, which is parallel to the ground. Notation: W is work (in Joules), is force (in Newtons), is displacement (distance with a direction) (in meters) W = cos θ = Note that we can calculate the work with either the vectors or the magnitudes and angle. Ex) if your force vector is (9,7) and your displacement is the vector (30,35), find the work done. W = = (9,7) (30,35) = 515 J Note that the magnitude of the force doing work is cos θ and the magnitude of the distance is. Ex) suppose you are pulling a bag of laundry along with a force of 15N so that your arm makes a 40 o angle with the ground. (a) Find the magnitude of your pulling force which is doing work. (b) Find the work done to move the bag 20 meters. 37 Vector Cross Product Sometimes, if we have a vector, it will be useful to construct a perpendicular vector. How can this be done in 2 dimensions? One vector by itself determines a perpendicular vector, as can be seen in the picture. Notation: for a vector, a perpendicular vector is Here is the formula: if = (a,b) then = (b,-a) Here is the proof: (a,b) (b,-a) = ab-ab = 0 Note that you can multiply by a constant since that does not change the direction. Ex) find a vector perpendicular to (3,-4) If you are familiar with linear algebra, you can define the formula with a determinant: = How can this be done in 3 dimensions? We will now define an operation between two vectors which produces a third vector that is perpendicular to both vectors. (Note: It help to see the reason that, in 3 dimensions, 2 vectors are required to specify a perpendicular vector. The object perpendicular to 1 vector is a plane and we will discuss that situation later. For now, note that this means in three dimensions, vectors in infinitely many directions are perpendicular to one vector.) The cross product of two vectors is: = = (read it as u cross v ) Here is a device to help remember the formula. In the x-component of the result, the input vectors x- components are omitted. The first component of the result starts with the 2 nd component of times the 3 rd component of. Note that the indices cycle: 1,2,3,1,2,3. So the second component starts with the 3 rd component of and the 1 st component of. The proof that the result of the cross-product is perpendicular to each original vector is accomplished by showing that each dot product is 0. If you know linear algebra, you can use determinants to define the cross-product:

38 38 The cross product has another striking property. x = sinθ Proof: Note that the cross-product of two vectors is a vector, and we have described its magnitude and direction. Ex) for the vectors (-1,3,2) and (-4,-2,5), (a) find a vector perpendicular to both (b) find the sine of the angle between them. Here is a way to visualize the magnitude of the cross-product. The length of the parallelogram is, the height is sinθ, so the area is sinθ, which is X Ex) find the area of the triangle with vertices P (4,2,7) & Q (-3,5,1) & R (2,0,2) Here are some properties of the cross-product. Each can be proven by writing each vector in terms of its components and using the definitions of cross and dot.

39 39 If a, b, c, are vectors and k is a scalar, 1) a X b = -b X a 2) (ka) X b = k(a X b) = a X (kb) 3) a X (b + c) = a X b + a X c 4) (a + b) X c = a X c + b X c 5) a (b X c) = (a X b) c 6) a X (b X c) = (a c)b (a b)c The expression in equation 5 is related to an amazing property. Imagine the shape determined by the vectors a,b,c as in the picture below. It is called a parallelepiped, which looks like a leaning cardboard box. Its height is a cosθ, where θ is the angle between a and (b X c). The area of the base is b X c. Therefore the volume is a cosθ b X c = a (b X c) Ex) find the volume of the parallelepiped with vector sides (5,3,1) & (3,2,3) & (4,2,0) at the same corner Application: Force and Torque Suppose you are trying to turn a bolt with a wrench. For simplicity, whre the bolt is, Call that the origin. Where the wrench lies, call that the x-axis. With your hand, you push the wrench, call that the y-direction. To describe the impact on the bolt, we need to know two things: which way does it turn (or, what is the axis of rotation)? What is the force of the turning? [discuss] Torque measures the tendency of a body to rotate around the origin. is the point where the force is applied, is the force vector, and is the torque vector. Then: = X The axis of rotation is the direction of the torque vector. The magnitude of the torque (the intensity of rotation) is the magnitude of the torque vector. Ex) a wrench is being used to turn a bolt. The bolt is at the origin, and the handle of the 0.3m wrench is on the x-axis. You apply 20N of force in the downward (negative z) direction. Find the torque vector and its magnitude. = X = (.3,0,0) X (0,0,-20) = (0, 6, 0) = X = r F sinθ = (.3)(20)(1) = 6 OR = = 6 Note that we can calculate the magnitude of the torque with either the vectors or the magnitudes and angle. Note that the torque axis is the y-axis, both in the calculation and the picture.

40 40 Lines In 2 dimensions a line is determined by a point and a slope (or direction). Let s rewrite this using parametric equations, so both x and y change as t changes. Suppose the line goes through (a,b) with slope m. The initial value for x is a, and let s say that changes in x are the same as changes in the parameter t, so x=a+t. For y, the initial value is b, and as x (or t) changes by 1, y changes by m. so y=b+mt. The line is described by the system of parametric equations: x = a+t & y=b+mt (Notice that if you eliminate the parameter t, you get y-b = m(x-a), the point-slope form of a linear equation.) Now let s turn that into vector notation. = + + To find the general vector formula, split it into constant and variable parts = + 1 Using the notation: is the position vector, is the known point, is the known direction of the line = + Note: you can multiply by any constant and it will still represent the same direction. As t changes, changes, and that is the point on the line. It is the vector version of the equation of a line. But it is extremely powerful, because it now can be used in any dimension. Let s see where & & & t appear on the graph of a line. Ex) find the equation of the line through (0,4) parallel to (1,2). Write the answer (a) as a vector equation (b) as a system of parametric equations (c) what is the slope (d) what is the standard equation y=f(x)? (Note that we think of (p,q) as a point when we want a position and as a vector when we want a direction.) Ex) find the equation of the line through (-2,4,5) parallel to (3,4,2). Write the answer (a) as a vector equation (b) as a system of parametric equations Here is a question to think about that we will ask now then answer later in the course, and it is connected to one of the key ideas of the course: what is the slope of this line? Alternate ways to express the equation of a line: Symmetric equations: in each equation, solve for t and set the expressions equal to each other

41 41 Ex) write the symmetric equations for the previous example, the line through (-2,4,5) parallel to (3,4,2). Find the equation of a line: Given two points Solve this by using two points to find a direction vector. Ex) find the equation of the line through (2,-3,1) and (5,2,0) Planes Previously, we saw that, to determine a line, we need a point and a vector (or direction), or two points. What determines a plane? [discuss] A plane is determined by either (a) 3 points (b) a point and 2 vectors in the plane (c) a point and a vector perpendicular to the plane. [(c) only works in 3 dimensions. The reasons are explored in linear algebra.] Next, we will use (c) to write down the equation. Assume that the known point is P (x 0, y 0, z 0 ) and the perpendicular (also called normal) vector is = (a,b,c). For any point in the plane Q (x,y,z), the vector is perpendicular to so their dot product is 0. = 0 (x-x 0, y-y 0, z-z 0 ) (a, b, c) = 0 a(x- x 0 ) + b(y- y 0 ) + c(z- z 0 ) = 0 If you combine all the constants, the equation looks like: ax + by + cz + d = 0 Or, to write the equation using the given information: = Exploring the graph using 3D graphing technology can help understand the equation. Ex) find the equation of a plane through the point (3,-2,4) with normal vector (2,5,-2) Using 3 points to find the equation of a plane A pair of points determines a vector, let s get two of those. Those two vectors are in our plane, and we need a vector perpendicular to them so take the cross product. Use that normal vector and any one of the points to make the equation Ex) find the equation of the plane through the points P (2,5,1) & Q (0,3,-2) & R (6,1,2)

42 As an alternate to generating a normal vector, one could write the equation of a plane using parametric equations with two vectors in the plane and two parameters. Starting from the point P, we can get to any point Q in the plane by moving a certain amount in the a direction and another amount in the b direction. In this example, the equation is (x,y,z) = (0,3,-2) + (2,2,3)r + (6,-2,4)s OR (0+2r+6s,3+2r-2s,-2+3r+4s). In general, for a plane through point P with direction vectors a and b, Q = P +ra + sb. In vector notation, r = r 0 + ra + sb. As a system of equations, (x,y,z) = (x 0 +ua 1 +vb 1,y 0 +ua 2 +vb 2,z 0 +ua 3 +vb 3 ). This idea is covered more thoroughly in Linear Algebra. The proof that the last two methods are equivalent involves lots of algebra; it is in the appendix. 42 Math Applications with Lines and Planes The easiest way to describe the direction of a plane is using its normal vector, since we already know how to describe the direction of a vector. Therefore, Two planes are parallel if their normal vectors are parallel ( = ) Two planes are perpendicular if their normal vectors are perpendicular ( = 0 ) In general, the angle between two planes is the angle between their normal vectors. Ex) find the angle between these two planes: 4x+3y-z-5=0 & -2x+5y+z+2=0 the line of intersection between two planes There are two ways to find it: Method 1: With standard algebra set one variable equal to the parameter t, then solve for the other variables in terms of t. Ex) find the line of intersection between 4x+3y-z-5=0 & -2x+5y+z+2=0 L: x = t & y = & z = OR (,, )= 0,, + 1,, Method 2: With vector operations For the vector equation of a line, we need a point and a direction vector. To find a vector: the line of intersection is in both planes, therefore it is perpendicular to both normal vectors, and therefore it is in the direction of their cross product. To find a point: take the two plane equations and set one variable equal to 0, then solve the two equations in two variables. (This works because any point will do.) Ex) find the line of intersection between 4x+3y-z-5=0 & -2x+5y+z+2=0 The normal vectors are (4,3,-1) & (-2,5,1) direction vector = (4,3,-1) X (-2,5,1) = (3+5,2-4,20+6) = (8,-2,26)

43 43 The distance between a point and a plane The distance between a point P 1 and a plane ax+by+cz+d=0 with normal vector is the magnitude of the component of in the direction, for some point P 0 in the plane. Let s work out what the formula is: Dist= cosθ = Dist(point, plane) = for the point (x 1 y 1 z 1 ) & the plane ax+by+cz+d=0 Ex) find the distance between the point (5,4,1) and the plane 3x-y+2z=6 The distance between a point and a line To find the distance between a point P and a line L, take any point Q on the line. The distance is the component of perpendicular to the line. dist = sinθ = = dist(point P,line L) = where is the line s direction vector and Q is any point on the line ex) find the distance between the point P (8,2,-5) and the line (x,y,z)=(6,2,1)+(3,-3,4)t Note that, when finding the distance from a point, when the other object is a line, the vector you use in the calculation is parallel to the line, so finding the component of the vector requires sinθ, which involves the cross product. If the other object is a plane, the vector you use in the calculation is perpendicular, so finding the component of the vector requires cosθ, which involves the dot product.

44 44 Vector Functions Curves in space Previously we used parametric equations to describe a curve in 2 dimensions. We can do exactly the same in 3 dimensions. So far we have done this for the simplest graph, a line. Ex) (t) = (3+t, 1+2t, 3-4t ) OR = Ex) (t) = ( sin(t), cos(t), t ) These expressions are simple enough that we can connect them to the graph. Note that if this was just a graph in x & y, it would be a circle with radius 1. However, as we pass around the circle, the height z is increasing. That makes a corkscrew, which is called a helix. (If you look from overhead, all you see is a circle.) Note that we can draw the graph over a certain range of t- values. Here, -5<t<5. Since the domain of each component is - <t<, we could choose to graph that parameter domain. Then the graph would be long. We are starting to see that the path in the graph can be anything you can imagine. In general, it is very difficult to plot a vector function (or space curve) by hand. 3 Ex) = (3-t, t 2, cos(t) ) OR cos( ) What does the curve look like from different perspectives? You can solve it on paper or see it with a 3D grapher (or your mind). From straight ahead, you see x on the horizontal axis and z on the vertical axis (it s the xz-plane), and you 3 cannot distinguish y-values. The graph is = or z = cos(x+3), so it s a cosine curve. cos( ) Ex) what graph is seen looking at the xy-plane? At the yz-plane?

45 45 Calculus with Vector Functions Derivative of vector functions The derivative of a vector function is exactly what you think it is take the derivative of each component. Note that the variable is the parameter. Ex) = (3-t, t 2, cos(t) ) Find the derivative = (-1, 2t, -sin(t) ) Notation: the derivative of a vector function (t) is written or or or or or r (t) The most typical application of curves in space is where they represent the movement of some object. At time t, the object is at a given point on the curve. Therefore, if represents the position vector, then the derivative represents the velocity vector. And the second derivative represents the acceleration vector. What about speed? Speed is the magnitude of the velocity, = Ex) for the position vector (t) = (2t-1, e 3t, 4t-t 4 ), find the velocity and acceleration vectors as well as their magnitudes. Let s describe the tangent line. In calculus in 2 dimensions, the direction of the derivative is tangent to the curve. This is also true in 3 dimensions. At a given point on the curve, the velocity vector is tangent to the curve. (To see that it is tangent, we draw it starting at the point on the curve, not starting at the origin.) Ex) 2 dimensions: for (x,y) = (3t, t 2 ) (a) draw the curve and a tangent vector at t=2 (b) find the equation of the tangent line at t=2 Ex) 3 dimensions: for (x,y,z) = (cos(t), sin(t), t) (a) draw the curve and a tangent vector at t = equation of the tangent line at t = (b) find the We can explore the tangent line at other points using a 3D grapher.

46 46 Here are some properties of the derivative. They can be proven by writing down the vectors by components. Ex) prove that if a vector (t) (a position vector) has constant magnitude, it is perpendicular to its derivative (a velocity vector) (t) = c (t) 2 = c 2 = c 2 Take the derivative 2 = 0 Therefore and are perpendicular. The derivative of a vector-valued function can be defined, and its properties proven, with limits. Definition: Theorem: Proof:

47 47 Integration of vector functions The integral of a vector function is exactly what you think it is take the integral of each component. Note that the variable is the parameter. Ex) take the integral of (t) = (6t, cos(t), e t ) Notation: the integral of (t) is ( ) OR ( ) The Fundamental Theorem of Calculus applies to vector functions as well. It looks almost the same as before: ( ) = ( ) ( ) Ex) a particle has acceleration vector function (t) = (2,3t,e t ) At t=0, its velocity is (1,0,0) and its position is (2,3,1). Find the position vector function (t) Arc Length for Vector Functions In two dimensions, the formula for arc length is L = + In three dimensions, the formula for arc length is L= + + Since we know that ( )=,, and = = ( ) + ( ) + ( ) we can rewrite the formula as: L = OR dt Recall that is the velocity, so is the speed, so the formula says that total distance traveled is the integral of speed with respect to time. That makes sense! Ex) find the arc length of the helix = (sin t,cos t, t) for 0<t<2π (one revolution)

48 48 The Unit Tangent and Unit Normal Vectors [omit] Notation: a tangent vector is (t) This vector is tangent to the graph. It captures the direction of change, as well as the magnitude of change. The unit tangent vector has the same direction as the tangent vector, but has length 1 = ( ) ( ) There are an infinite number of vectors which are perpendicular to the curve. However, since =1, therefore =0, therefore and are orthogonal. Therefore is orthogonal to the curve. We want to make it a unit vector, so divide by the magnitude. The principal unit normal vector (or unit normal) is = Ex) for (t) = (3t 2,2t,5-t 2 ) find the unit tangent and unit normal Curvature [omit] If you travel at a constant speed in a straight line, no matter how fast that speed is, you feel no forces or acceleration. However, if you travel at a constant speed on a curve, you feel acceleration. The acceleration tends to go away from the direction in which your path is bending. The more tightly curved the path is, the more acceleration you feel. The measure of this is called the curvature. The curvature is the magnitude of the change in relative to how far you have traveled on the curve = An equivalent formula for curvature is: = ( ) ( ) In the special case of 3 dimensions, this reduces to a nice formula: = (The proof of this special case is in the Appendix.) Ex) prove that the curvature of a circle in the plane with radius a is 1 / a Ex) find the curvature κ(t) of the curve = (sin t, cos t, 1) Ex) find the curvature κ(t) of the curve = (ln t,1-2t,2e 3t ) then find the curvature at t=1

49 49 Surfaces & Functions of Several Variables The next step is to look at surfaces in 3 dimensions which are more complicated than planes. Surfaces of translation We can take a known graph from 2 dimensions, then just extend it into 3 dimensions. Ex) graph y=x 2 At every point, y=x 2. This is true when z=1, z=2, z=5.7, etc. One way to visualize the 3 dimensional graph is to take a slice (or crosssection) at each z-value. In this example, every cross-section looks the same, it is y=x 2. Typical 3 dimensional graphs are more complicated than that. Ex) graph x 2 + y 2 = 4 Ex) graph z=x 2 + y 2 This graph is a rotation of a parabola. It is called a paraboloid. How can we connect the equation to the graph? When z is constant, we are in a plane parallel to the xy-plane. When z=0, both x & y must be 0. For z=constant, it produces the equation of a circle. Therefore, horizontal crosssections are circles, with a larger radius for larger z-values. Visualize each cross-section by intersecting a plane with the surface. Let s consider vertical cross-sections for constant y. This means we are in a plane parallel to the xzplane. When y=0, it produces a standard parabola, z=x 2. In fact for any y=constant, it produces a parabola (with z as a function of x). Also, x=constant produces cross-sections of parabolas (with z as a function of y). (In the graph, these lines are already drawn.) A cross-section with x=constant is called an x-trace. It lives in a plane parallel to the yz-plane. Similary, a cross-section with y constant is called a y-trace, and it lives in a plane parallel to the xz-plane. Finally, a cross-section with z=constant is a z-trace, and it lives in a plane parallel to the xy-plane. Each trace is a slice of the surface. Stack them together to build the

50 50 whole surface. In this example, x-traces and y-traces are parabolas, and z-traces are circles. Ex) graph + + = 1 This graph is the 3 dimensional version of an ellipse. It is called an ellipsoid. What shape are the traces? From the graph: From the equation: These are examples of quadric surfaces. These are surfaces whose equations can be made from x,y,z,x 2,y 2,z 2 and constants. Here are some more examples.

51 51 Functions of several variables formulas, graphs, and more Functions as formulas Previously we saw that a function y=f(x) can be represented in 2 dimensions with xy-axes. Plugging in a value for x yields a function value y. We will now look at functions z=f(x,y). These can be represented in 3 dimensions with xyz-axes. Plugging in values for x & y yields a function value z. Ex) for z= + find f(4,5) and find the domain of the function z Functions as graphs Ex) what shape is depicted in this graph? What is the function? It looks like a hemisphere. We can find the equation of this graph. The radius is r=3 and the center is the origin. In three dimensions, x 2 +y 2 +z 2 =3 2. We want to write it as a function so, solving for z, we get z = 9 (Take the positive root to get the top half of the sphere). What is the domain of the function? Notice that the domain is a 2-dimensional region in the xy-plane, over (or under) which the graph of the surface exists. Every point (x,y) in the domain has a function value z represented by the height. Level curves (or contour lines) Before there was mathematical software which could render beautiful images indicating depth, people needed ways to visualize 3-dimensional objects in a 2-dimensional image on paper. The most common technique was developed by map makers who wanted to represent mountains and valleys. They drew a curve to show every point which was at the same height. The mountain in the picture is represented by the map on the right, called a contour map. The lines in a contour map are called level curves (or contour lines). Moving along a level curve means moving along the same height (or function value). Crossing from level curve to level curve means moving to a different height. Level curves are just z-traces. When z is a function, they have a special status.

52 52 This technique is used to measure things other than actual height. Your function can represent any real wolrd quantity; for example, it could measure temperature. (These contour map lines are called isothermals, iso means same, therm means temperature.) The variables x,y,z can represent any quantities in any real-world application. What does the contour map look like for a given surface? Ex) z = 64

53 53 This can also be solved algebraically. For the level curve at z=c, we have c= 64 c 2 = 64 x 2 y 2 x 2 + y 2 = 64 c 2 So the level curve is a circle centered at the origin, radius r=64-c 2. Except for the simplest cases, to draw level curves by hand given the formula of a surface is very hard. Algebraically, the problem is possible, but does not give simple objects. Analysis of the graph is more direct. Ex) for =, What are the x-traces? What are the y-traces? What are the level curves? In general, to graph functions and contour maps, 3 dimensional graphing software is useful, almost necessary.

54 54 Derivatives for functions of several variables A review of derivatives for y=f(x) Previously, we saw derivatives for one-variable functions y=f(x) as a rate which represented the change in y over the change in x. This could be presented graphically by taking the graph of the function (which was a curvy line) and zooming in on a point (a,b) of the graph to reveal a straight line called the tangent line; the slope of that line is the derivative at that point. Numerically, we made an estimate by taking a point and a nearby point, then calculating change in the function divided by change in x. Algebraically, we calculated that rate as the change in x goes to 0. Numerically: x y=x Grahpically: Algebraically: ( )= ( ) ( ) Intro to Partial Derivatives (numerically, graphically, algebraically) We will be able to understand the derivative for the surface z=f(x,y) in a similar way. We will answer two questions. First, with the extra variables, how does it make sense to talk about derivatives in 3 dimensions? Second, once we believe that it is possible to have a derivative, how can we say what it is? Consider the plane z=1-x+2y Numerically: Here is a table of values for this function x: y: What does it mean to talk about the rate of change of the function? [discuss] How does the function change as x changes? How does the function change as y changes? Graphically: This surface was rotated to view the xz-plane. The red line represents all values on the surface for which y=2. Since y is constant, z is a function of x. What is the slope of that line (estimate visually)?

55 55 This surface was rotated to view the yz-plane. The green line represents all values on the surface for which x=3. Since x is constant, z is a function of y. What is the slope of that line (estimate visually)? That is what a derivative (or slope) could mean for z=f(x,y). Numerically, find the change in z over the change in x as y is constant (or the change in z over the change in y with x constant). Graphically, you draw a straight line and the slope of that line is your derivative. Notation: the derivative of f with x as the variable (and other variables constant) is This is called the partial derivative of f with respect to x. That curvy d indicates that there are other variables we hold constant. Other notations: f f x D x f There are many ways to say it. I suggest partial dee zee dee ex. Sometimes mathematicians do not say partial when it is clear from the context. Ex) for f(x,y)=1-x+2y = =

56 But how can we talk about slopes of straight lines when the surface is curvy? Consider the following example: 56 Ex) What is the rate of change of the function f(x,y) =4 at the point (1,0.5,0.57)? Numerically: x: y: What does it mean to talk about the rate of change at the point (1,0.5,0.57)? [discuss] How does the function value change as x changes? If x increases by 1, f(x,y) changes by So = We get a better approximation of the rate at that point if we change x by 0.5, now f changes by So = In each case, the approximation of the rate at that point is better if Δx (or Δy) is closer to 0. Next, at the same point, how does this function value change as y changes? If y increases by 1, f(x,y) changes by So = We get a better approximation of the rate at that point if we change y by 0.5; and now f changes by So = Graphically: Ex) f(x,y) =4 This surface was rotated to view the xz-plane. The red line represents all values on the surface for which y=0.5 (so z is a function of x). What is the slope of that line at x=1, indicated by the green line (estimate visually)? This surface was rotated to view the yz-plane. The green line represents all values on the surface for which x=1 (so z is a function of y). What is the slope of that line at y=0.5, indicated by the red line (estimate visually)?

57 57 The final step graphically is that, when we zoom in on a point, the surface appears as a (nearly) flat plane containing our (nearly) straight lines. The plane is called the tangent plane. The lines are called tangent lines. To explore this idea, use a 3D grapher. Symbolically: Here s the key idea: when we determine the rate of change as the change in the function over the change in a particular variable, the other variables are constant. For example, to find calculate the derivative from the formula, we use the same idea. Ex) for f(x,y) = 0.1x xy y 2 find the derivative = (0.1x xy y 2 ) = 0.2x + 0.2y y is treated as a constant, so in the second term, x is the variable and the coefficient is 0.2y = and, y does not change. When we Note that you can no longer use f to indicate a derivative because it is necessary to specify the variable. Formal definition of the partial derivative: = (, ) (, ) Notice how, comparing the two terms in the numerator, the x-value is changing but the y-value is constant Higher partial derivatives: The notation for the second derivative of the function f with respect to x is It is possible to take several derivatives with respect to different variables Take the derivative of f with respect to x and then with respect to y. the notation is: Ex) for f(x,y) = x 5 + y 4 + x 2 y 2 find: (a) f x (b) f y (c) f xx (d) f yy (e) f xy (f) f yx

58 Is there a difference between f xy and f yx? In other words, if you take the derivative twice, once with respect to x and once with respect to y, does it matter which one you do first? If the derivatives of f(x,y) are continuous, then f xy = f yx This is called Clairaut s Theorem. Summary of the introduction to partial derivatives 58 Numerically, we can calculate by keeping the other variable, y, constant. Geometrically, suppose we want to know the derivative at the point (a,b,c). We can zoom in at that point. What we will see is a flat plane called the tangent plane. Note that, in any direction (for example the x-direction), the plane has a slope. You can visualize this by intersecting the tangent plane with a vertical plane having a fixed y- value, which produces a line in an xz-plane. The slope of that line will be the derivative of the function in that direction - in this case showing how z changes as a function of x (You can do the same by taking a plane with a fixed x-value, which intersects the tangent plane in a line in a yz-plane.) Note that we could zoom in (to see the tangent plane) then make a trace (which is a straight line), or make a trace (which is a function) and then zoom in (to see a straight line). Algebraically, consider how to calculate the derivative for functions of 2 variables. If a derivative should be a rate of change, it should be the ratio between the change in the function over the change in a variable. However, there is a choice of variables. The solution to this difficulty is to consider each variable separately. Therefore, we can have two different derivatives. (In fact, we can have many, as we will see later.) The Tangent Plane Previously, for y=f(x) at the point (x 0,y 0 ), we used derivatives to find the equation of the tangent line: y-y 0 = f (x 0 ) (x-x 0 ) Now, for z=f(x,y) at the point (x 0,y 0,z 0 ), we will do the equivalent thing, which is to use derivatives to find the equation of the tangent plane. Tangent Plane Equation Any plane passing through the point P=(x 0, y 0, z 0 ) has an equation A(x-x 0 ) + B(y-y 0 ) + C(z-z 0 ) = 0 for some A,B,C. Let s move the terms with z to one side to get C(z-z 0 ) = -A(x-x 0 ) - B(y-y 0 ) Divide by C to get z-z 0 = -A/C (x-x 0 ) B/C (y-y 0 ) Rename the constants, to get z-z 0 = a(x-x 0 ) + b(y-y 0 ). Now consider the traces at this point. First consider the y-trace. Geometrically, in the surface, you get f as a function of x and in the tangent plane you get the tangent line to that function, so it has slope. Algebraically, when y = y 0, we get z-z 0 = a(x-x 0 ). Therefore, a =. It works out similarly for the x-trace. Geometrically, in the surface, you get f as a function of y and in the tangent plane you get the tangent line to that function, so it has slope. Algebraically, when x = x 0, we get z-z 0 = b(y-y 0 ). Therefore, b =. Therefore we know the equation of the tangent plane: z-z 0 = f x (x 0,y 0 ) (x-x 0 ) + f y (x 0,y 0 ) (y-y 0 ).

59 59 Notation: recall that f x (x 0,y 0 ) is, the partial derivative of f with respect to x, evaluated at the point (x 0,y 0 ). Note: we will assume that these derivatives exist, and so the tangent plane exists. To conceive of what the surface would look like so that the partial derivatives did not exists, think of the 1-variable case in 2- dimensions, then extend it to 2-variables in 3-dimensions. Ex) find the equation of the tangent plane for the paraboloid z = 1 -.1x 2 -.4y 2 at the point (1,1,0.5) Ex) Find the equation of the tangent plane for z = 2x 2 y + x 4 y 2 at (2,3,168) Linear approximation using the tangent plane For a function of 1 variable, we saw that the tangent line is an approximation to the graph. Now, for a function of 2 variables, the tangent plane is an approximation to the graph. The equation of the tangent plane for z=f(x,y) at the point (a,b,c) can be written as a linear function: f(x,y) L(x,y) = f(a,b) + f x (a,b) (x-a) + f y (a,b) (y-b) This looks a lot like the equation of the tangent line (or linear approximation) in 2 dimensions: f(x) L(x) = f(a) + f (a) (x-a) It has been updated because now the function changes, not from one variable changing, but from two variables changing. The equation can be described as: (new z-value) (initial z-value) + (change in z due to change in x) + (change in z due to change in y) z z 0 + Δ x z + Δ y z. Those changes are calculated with linear approximations. Ex) for z=f(x,y)=x 3 + 2y + 3, (a) find f(3,2) (b) find f(3.2, 2.1) using the linear approximation (c) find f(3.2,2.1) exactly Ex)for z=f(x,y)=2x xy, (a) find f(4,.5) (b) find f(4.1,.55) using the linear approximation (c) find f(4.1,.55) exactly

60 Differentials These equations for the linear approximation can be written with differentials Equations for in 2 dimensions in 3 dimensions Linear Approximations Without differentials y b = f (a) (x-a) z c = f x (a,b) (x-a) + f y (a,b) (y-b) Written as an Δy ΔL(x) = f (x) Δx approximation Δz ΔL(x,y) = f x Δx + f y Δy OR Δz Δx + Δy Written exactly dy = f (x) dx OR dy = dx dz = f x dx + f y dy OR dz = dx + dy In words the change in the function is its rate of change times the change in the variable the change in the function is its rate of change in the x-direction times the change in x plus its rate of change in the y-direction times the change in y Notation: ΔL = Δ x L + Δ y L where Δ x L = f x Δx, similar for y. Also, dz = d x z + d y z where d x z = dx, similar for y ex) for z=ln(x 2 +2y) (a) find dz. Now, at the point (1,0,0), (b) find L (c) approximate f(1.1,0.2) (d) find f(1.1,0.2) exactly. Now, for the two points, find (e) ΔL between the two points (f) Δ x L (the amount of ΔL from the change in x) and Δ y L (the amount from the change in y) 60 A Geometric explanation of Δf, ΔL and dz (same as df) The surface z=f(x,y) is outlined in blue and the tangent plane at the point (a,b,c) is outlined in purple. Four points are marked off, the ones with (x,y)-values (a,b) & (a+δx,b) & (a,b+δy) & (a+δx,b+δy). They are represented, first, in dark green on the xy-plane below, second, in dark blue on the surface, and third, in dark purple on the tangent plane. The edges of this piece of the tangent plane are T x, the tangent line in the x-direction (when you hold y constant and z is a function of x) and T y. To find the new z=l(a+δx,b+δy), climb along T x (z changes as x changes while y is constant), then climb along the line parallel to T y (z changes as y changes, x constant). Δz (or Δf) represents the change between old and new point of the function, and ΔL represents the change between old and new point of the linear approximation (the tangent plane). dz is an infinitesimal like dx and dy. The formula dz = p(x,y)dx+q(x,y)dy involves functions. That formula is valid everywhere (at every (x,y) point in the domain). If we evaluate the functions at a point (a,b), we get m dx + n dy where m & n are constants (the slopes in the x- and y-directions at that point). Now the expression is valid only at the point (a,b). [dx, dy, dz are best conceived of as tiny changes in their variable, and their ratios represent rates of change. The full theory behind differentials and infinitesimals is quite advanced and beyond the scope of this course.] The tangent plane L(x,y), or the change in the value ΔL, is a linear approximation of the function at the point (a,b). It is exact at (a,b), but an approximation at any other point. The closer the point is to (a,b), the flatter the surface is for that range of values, and the better the approximation is.

61 f(x,y), or the change in the value Δz (or Δf), is the actual function value. Those calculations represent points on the actual surface. Note: in higher mathematics, the coordinates are not named x,y, but rather x 1, x 2, In that case, if z=f(x 1, x 2 ), then =. This formula can be extended to any number of variables. The Chain Rule Suppose that z=f(x,y), but in addition, x & y are both functions of another variable, t; that is, x=x(t) and y=y(t). The derivative can be understood in two different ways. One, simply plug in the expressions for x & y so z=z(t), then find z (t). Two, use the chain rule for functions of several variables. This formula comes immediately from the expression for the differential dz. Since dz = dx + dy we have that: = + 61 ex) for z=2xy+cos(y) where x=6t 2 and y=ln(t), find at t=1 method 1 (using t only): z = 2(6t 2 )(ln t) + cos(ln t) = 24t(ln t) + 12t 2 - sin(ln t) = 12 method 2 (using the chain rule): = + = (2y)(12t) + (2x-sin y) = 2 ln(t) 12t + (2(6t 2 )-sin(ln(t)) = 12 Note: instead of the last two steps, you could use the fact that if t=1, then x=6 and y=0, and plug in: (2y)(12t) + (2x-sin y) = 2(0)12(1) + (2(6) - sin(0)) = 12 Ex) for z = x 3 y + 3x 2 y 1.7 where x=sin(2t) and y=cos(t), find at t=0 Ex) Application from physics: In an electrical circuit, Ohm s Law relates voltage V (the potential ability to move charge, measured in volts), current I (the speed at which charge is moving, measured in amps) and resistance R (the wire s tendancy to resist charge moving through it, measured in ohms): V=IR. If the current I=4 amps, the rate of change of the current is -1 amp/min, The resistance R = 300 ohms, the rate of change of the resistance is -5 ohms/min, find (a) the voltage V right now (b) the rate of change of the voltage (in volts/min).

62 An alternate proof of the chain rule uses the idea of linear approximation. f(x,y) r(x-a) + s(y-b) + c Where at (x,y)=(a,b) we have f x (a,b)=r and f y (a,b)=s and f(a,b)=c Suppose x=x(t) and y=y(t). now we have the linear approximations x mt+a y nt+b Where at t=0 we have x(0)=a and =m and y(0)=b and =n Note that all approximations produce the exact values and first derivatives at t=0, or equivalently (x,y)=(a,b). F(x,y) r(mt+a-a) + s(nt+b-b) + c rmt + snt + c rm + sn + (, ) (, ) (, ) (, ) This approximation is exact at the point (a,b), so: = + (, ) (, ) (, ) (, ) (, ) Every expression is evaluated at the same point (a,b), that point can be any point. As a result we have: The Chain Rule: = + For a discussion of the chain rule with more variables, see the appendix. Implicit Differentiation (using partial derivatives) Ex) find given y 2 + 3x 4 y 3 tan(y) + x = 0 The old way: 2y + 12x 3 y 3 + 3x 4 (3y 2 ) sec 2 y + 1 = 0 2y + 3x 4 (3y 2 ) sec 2 y = x 3 y 3 = ( ) New method: Call the left side F(x,y) F x = 12x 3 y F y = 2y + 9x 4 y 2 sec 2 y = = That s pretty quick and straightforward. Why does it work? F(x,y)=0 F(x,y) = (0) + = 0 = 62 Ex) find given e xy cos(y) = 4x 2

63 63 Note the position of the variables in the fraction on the left side and the right side of the formula they switch! This extends to several variables: For F(x,y,z)=0, = and similarly for the other partial derivatives. Ex) e x z 2 y 3 = 2xy find and The Directional Derivative and the Gradient We have seen how to find, the derivative of f(x,y) in the x-direction, and, the derivative of f(x,y) in the y-direction. Now we will find the derivative of f(x,y) in any direction. Directional Derivative understood geometrically Previously, we described the partial derivatives and. They came from drawing a trace in the x-direction or the y-direction, then finding the slope of the trace at that point. To help draw the trace, we could intersect a plane. To help find the slope of the trace, we could zoom in to see straight lines in the tangent plane and visually approximate the slope. However, we could draw a trace in any direction. The slope of the trace at that point is the derivative of the function in that direction. To better understand this, let s illustrate this on the graph, then develop the formula. Assume we have a function of two variables f(x,y) and a point in the domain (in 2 dimensions) p=(a,b). The surface z=f(x,y) has a tangent plane at the point (in 3 dimensions) q=(a,b,c). To indicate a direction from the point, there are 4 equivalent ways: (a) slice the surface with a vertical half-plane (b) starting at (a,b) in the domain, draw a vector v=(r,s) (c) starting at (a,b,c) on the surface, draw a line in the tangent plane (d) starting at (a,b,c) on the surface, draw a trace on the surface. We are most interested in the line in the tangent plane. Its direction matches v, so call it T v. For that line, its run is measured in the v-direction, its rise (or fall) is measured in the z-direction, and that slope = is the derivative of f(x,y) in the v-direction.

64 Directional derivative - Developing the formula Lets determine the rate of change of f(x,y) at a point p = (a,b) in the direction indicated by a vector v = (r,s). Thus v moves r units in the x-direction and s units in the y-direction. But we know the rate of change in the x- direction, it is, and the rate of change in the y-direction is. So moving r units in the x-direction produces r units of change, and moving s units in the y-direction produces s units of change. So the function changes by r + s However, if v is not a unit vector, v 1, then the function changes r + s per v units. Therefore, its rate of change in the v direction is: If the vector v is a unit vector, then the rate of change of f in the directon of v is: r f x + s f y Notation: the derivative of the function f in the direction of the vector v is D v f Ex) for z=3x 3 + 5y 2 and v=(-2,5), find D v f at the point (2,3) [or ] 64 We will introduce some notation for a very important concept, which will make that expression very simple. For a function f(x,y), its gradient vector is, The notation is f or, read as grad f or del f. Always remember that f is a vector. This is the first of many times we will use the gradient vector. The numerator can now be rewritten as the vector dot product f v. So the formula is: D v f = f If v is a unit vector, call it u, this formula simply becomes D u f = f u Ex) for f(x,y) = find (a) the gradient vector f (b) the derivative in the direction of v = (3,4) (c) D v f at (3,2) Finding the maximum and minimum directional derivative D u f = f u = f u cosθ = f cosθ where θ is the angle between f & u. Therefore, the directional derivative D u f is largest when θ=0, meaning f & u are in the same direction. Thus, the direction of the greatest rate of change is the direction of the gradient vector f, and the magnitude of the greatest rate of change is f. Also, D u f is minimal when θ=π, meaning f & u are in opposite directions, and then the minimal value of D u f is - f. In addition, note that when the gradient vector and the direction vector are perpendicular, θ = and the directional derivative is 0. or ex) for f(x,y) = x ln(y) at the point (2,1) (a) what is the unit vector in the direction of the greatest rate of change? (b) What is the greatest rate of change? (c) What is the directional derivative in the direction the gradient direction? from

65 65 The Connection Between the Gradient and Level Curves We know that z=f(x,y) generates a surface in 3 dimensions. Also, f(x,y)=c for different values of c are level curves which are graphs in 2 dimensions. Note that in the direction along (or parallel to) a level curve, the function is not changing. For the unit vector u along (tangent to) a level curve of f(x,y), D u f = 0 & D u f = f u f u = 0 f & u are perpendicular Thus, the gradient vector (and the direction of greatest rate of change) is perpendicular to the level curves Ex) for the surface f(x,y) = y e x (a) graph the level curve at f=0 (b) find the gradient f (c) find each normal vector on this level curve for x=0,1,2 (d) bonus: find a parallel (tangent) vector to the level curve at x=1 For a 3-dimensional surface defined by g(x,y,z)=0, its gradient is a 3-dimensional vector, and it will be perpendicular to the surface. (The reason is that the equation is a level curve for w=g(x,y,z). Note that this formula has 4 variables so it exists in 4 dimensions! We will not attempt to visualize that.) Ex) for the surface xyz=12, find the equation of the normal line at the point (2,-2,-3) First, let f(x,y,z) = xyz-12 f = (f x, f y, f z ) = (yz, xz, xy) f(2,-2,-3) = ((-2)(-3), (2)(-3), (2)(-2)) = (6, -6, -4) Thus the normal line goes through the point (2,-2,-3) with direction (6,-6,-4) The equation is (x,y,z) = (2,-2,-3) + t(6,-6,-4) OR = =

66 66 (Note: Since we have a point and a normal vector, we could now easily write down the equation of the tangent plane; this is an alternate method to do that.) Ex) find a vector perpendicular to the surface 0 = 4-x 2 -y 2 -z 2 at the point (0.5,0.5, 3.5) Application: Maximum & Minimum For functions of 1 variable, to find the maxima and minima, we saw the need to identify critical points (f (x) = 0 or undefined). Now let s investigate the process for functions of 2 variables. How can we identify all extreme points (local max & min)? In terms of calculations, all derivatives are 0. In terms of the geometry, the tangent plane is completely horizontal. Any such points are critical points of the function. But once we have the critical points, how can we tell which are maxima and which are minima? Also, just like in 1-variable calculus, there may be some points like that which are neither a max nor a min. And in 3 dimensions there are new ways for this to happen. What would that look like? How do we classify critical points as max, min, or neither? This graph shows a saddle point. At (0,0,0), = 0 and = 0, but the point is neither a local max nor local min To find critical points: = 0 and = 0 To classify critical points as local max, local min, neither, use the concavity (or second derivative) test: Calculate the discriminant d=(f xx )(f yy ) (f xy ) 2

67 67 If d<0, then the point is a saddle point, neither max nor min If d>0, then it is an extreme point and: if f xx >0 then the graph is concave up and the point is a relative min if f xx <0 then the graph is concave down and the point is a relative max If d=0, then there is no conclusion Ex) find the critical points of f(x,y) = x 4 + y 4 4xy +1 and classify each as a max, min, or saddle Absolute Maximum & Minimum Recall from 2-dimensional calculus, to find the absolute max & min for a function on a closed interval, you had to check the critical points and the boundary values. In 3-dimensional calculus, you do the same thing. Note that, for the curve y=f(x), the boundary is two points. For a surface z=f(x,y), the boundary is a curve! (The mathematical theorem says that the absolute maximum and minimum of a function on a closed domain each occur either at a critical point or a boundary point.) Ex) find the absolute max & min for f(x,y)=x 2-2xy+2y on the region defined by 0 x 3, 0 y 2. f x = 2x-2y = 0 x=y classify as local max/min/neither: f y = -2x+2 = 0 x=1 y=1 f xx = 2 f yy = 0 f xy = -2 So the critical point is (1,1) d(1,1) = (2)(0) (-2) 2 = -4 saddle point Now, check the boundary values. We need to identify which points on the boundary are possible extrema. But the boundary is a curve in 3 dimensions, so how do we do that? We can first break it into 4 straight lines. On the boundary x=0, the function is f(0,y) = 2y Find boundary c.p.: (2y) = 2 = 0 never, so no c.p., only need endpoints include (0,0) and (0,2) On the boundary x=3, the function is f(3,y) = 9-6y+2y = 9-4y Find boundary c.p.: (9-4y) = -4 = 0 never, so no c.p., only need endpoints include (3,0) & (3,2) On the boundary y=0, the function is f(x,0) = x 2 Find boundary c.p.: (x 2 ) = 2x = 0 x=0 is the critical point, also need endpoints include (0,0) & (3,0) On the boundary y=2, the function is f(x,2) = x 2-4x+4 Find boundary c.p.: (x 2-4x+4) = 2x-4 = 0 x=2, also need endpoints include (0,2) & (2,2) & (3,2) Note that we have three types of points to check: interior critical points, boundary critical points, and corners.

68 68 Check values: (x,y) (0,0) (0,2) (3,0) (3,2) (2,2) (1,1) f(x,y) Max point: (3,0,9) Min point: (0,0,0) & (2,2,0) confirm this solution by looking at the graph (A proof of the formula for the second derivative test is in the appendix.) Note: in this problem, we assume that every part of the boundary is a straight line x=constant or y=constant. But what if it was something different, like a circle? Or any curve? That problem can be solved with the techniques of the next section. Solving Optimization Problems with Lagrange Multipliers It is a typical problem that you want to optimize something (such as maximizing profit) subject to some constraint (such as costs). In single variable calculus, this optimization problem starts with a function f(x,y) to optimize, and a constraint g(x,y)=k. It is completed by solving for one variable in the constraint, substituting to eliminate one variable in the function, then taking the derivative and setting it to 0. This problem can be solved in general for several variables by the technique of Lagrange Multipliers. Geometric Discussion Start with a contour map of the function to be optimized, f(x,y). Also draw the constraint g(x,y)=k. At the point where the constraint overlaps with the level curve with the largest value, they meet at only that one point, you can see they are tangent. (If the constraint crossed the level curve, then it would be headed for higher level curves.) The gradient vector of the constraint is perpendicular to the constraint graph, and the gradient vector of the level curve is perpendicular to the level curve. Therefore, the two gradient vectors are parallel to (or in the same direction as) each other. In other words, f = λ g for some constant λ. That creates one equation for each component. Note that the gradient is a vector, so the number of equations is the same as the number of dimensions. λ is another variable, and the constraint is another equation. So there are n+1 variables and equations. While it is easiest to draw the picture of this discussion for functions of two variables, this is true for any number of variables. Notation: that constant λ is called a Lagrange multiplier. Most textbooks use λ, lambda, a lower case Greek L. Use any letter you like. (Sometimes I like to use c.) Algebraic Method Ex) minimize the surface area of a cylinder with volume 100cm 3 Set up: Function: surface area S = 2πr 2 + 2πrh Constraint: V = πr 2 h = 100 Old method eliminate one variable from the function and set the derivative to 0: h = r -2 S = 2πr 2 + 2πr r -2 = 2πr r -1 S = 4πr 200r -2 = 0 r 3 =

69 69 r = 2.52cm h 5.03cm Surface area cm 2 New Method Lagrange Multipliers: = h, = 2 h 2 first component equation: S r = c V r 4πr + 2πh = c(2πrh) second component equation: S h = c V h 2πr = cπr 2 solve for c to get: c = Constraint: πr 2 h = 100 solve for h to get: h = r -2 Plug those expressions for c & h into S r to get: 4πr + 2π r -2 = 2πr r -2 r 3 = r = 2.52cm then h 5.03cm and surface area cm 2 With 2 variables, both methods work with about the same level of difficulty. What about 3 variables? Ex) A rectangular box without a lid will be made from 12m 2 of cardboard. Find the maximum volume. Setup: Function: volume V = xyz Constraint: surface area S = xy + 2xz + 2yz = 12 Old method eliminate a variable from the function and set the derivatives to 0: = = = = = ( ) ( ) ( ) ( ) ( ) ( ) = 0 = 0 Let s stop right there, because solving those two equations in two variables will be an algebraic mess. Lagrange multiplier method: 2 + =, = V x = c S x yz = c(2z+y) V y = c S y xz = c(2z+x) V z = c S z xy = c(2x+2y) Constraint S: 2xz+2yz+xy=12 There are several ways to solve this system of equations. The typical first step is to eliminate c the Lagrange multiplier. Solving the V x equation: c = Plug into the V y equation: xz = Plug into the V z equation: xy = ( ) 2xz+xy = 2xz+2yz x = 2z ( ) 2xz+xy = 2yz+xy x = y Plug into S to get: 4z 2 + 4z 2 + 4z 2 = 12 12z 2 = 12 z=±1 (we need a length so take the positive value) z=1, x=2, y=2 and the maximum volume is V = (2)(2)(1) = 4m 3. Ex) Math application: minimize the distance between the point (-3,0) and a point on the parabola y=x 2

70 70 Function: distance = ( + 3) + ( ) A helpful trick: minimizing distance is the same as minimizing (distance) 2, and the math is much easier f(x,y) = (x+3) 2 +y 2 Constraint: y=x 2 g(x,y) = y-x 2 = 0 f = c g f x = c(g x ) 2x+6 = c(-2x) f y = c(g y ) 2y = c(1) Also, y=x 2 So 2x+6=(2y)(-2x) 2x+6=2(x 2 )(-2x) 4x 3 + 2x + 6 = 0 (x+1)(4x 2-4x+6) = 0 x=-1 so y=1 and the distance is ( 1 + 3) + (1) = 5 At this point, it is useful to revisit some aspects of the gradient. Previously, we saw that, for a function defining a graph, the gradient vector is tangent to the graph. Further, it is perpendicular to the level curve of the graph. Yet, in Lagrange multiplier problems, the gradient is perpendicular to the constraint. Why is that? This is because, in that problem context, the constraint equation is a level curve, not a function. For example, z=x+y is written as a function. It has a total of 3 variables, so it will be graphed in 3 dimensions, and the gradient will be tangent to the graph. On the other hand, 0=x+y is written as a level curve. It has a total of 2 variables, so it is graphed in 2 dimensions, and the gradient will be perpendicular to the graph. Similarly, w=3x-2y+z is written as a function. It has a total of 4 variables, so it will be graphed in 4 dimensions, and the gradient will be tangent to the graph. On the other hand, 0=3x-2y+z is written as a level curve. It has a total of 3 variables, so it is graphed in 3 dimensions, and the gradient will be perpendicular to the graph. The first and last case can be confused for each other. Consider the equation 0=x 2 +4y-z and its graph. It contains 3 variables and can be graphed in 3 dimensions. Is the gradient tangent or perpendicular to the graph? It depends on which gradient one takes. If the equation is treated as z=x 2 +4y, a function of 2 variables, its gradient has 2 components, and the gradient is tangent. On the other hand, if it is treated as the level curve of w=x 2 +4y-z, which is a function of 3 variables, the gradient will have 3 components and be perpendicular to the original graph. The graph in 3 dimensions is really a level curve of a graph in 4 dimensions. To understand the properties of the gradient, its important to keep in mind whether your equation is a function or a level curve. Extra Credit material: Taylor Polynomials for Functions of Several Variables For 1-variable functions, we have seen the linear approximation of f(x) at x=a, which is the equation of the tangent line at that point, L(x)=f(a)+f (a) (x-a). Then we extended that to the Taylor Polynomial approximation, P n (x) = f(a) + f (a) (x-a) +! f (a) (x-a) 2 +! f (a) (x-a) ! f(n) (a) (x-a) n For 2-variable functions, we have seen the linear approximation of f(x,y) at (x,y)=(a,b), which is the equation of the tangent plane at that point, L(x,y)=f(a,b)+f x (a,b) (x-a) + f y (a,b) (y-b). Now, we expand the approximation to higher powers. P 2 (x,y) = f(a,b)+f x (a,b) (x-a) + f y (a,b) (y-b) +! f xx (a,b) (x-a) 2 + f xy (a,b) (x-a) (y-b) +! f yy (a,b) (y-b) 2 Note that the expression is more complicated because of the presence of mixed partial derivatives. You can check that every second partial derivative of f(x,y) agrees with P 2 (x,y) at the point (a,b). Ex) for f(x,y) =, find P 1 (which is the same as L) and P 2.

71 71 Integration with Functions of Several Variables The simplest interpretation of the integral is that it is the opposite of the derivative. In 1-variable calculus: Ex) f (x) = 2x, find f(x) f(x) = f (x) dx = 2x dx = x 2 + C if you know an initial condition, you can solve for the constant suppose that f(0)=1 f(0) = 0 2 +C=1 C=1 f(x) = x 2 +1 Now let s apply this to 2-variable calculus. Ex) f xy (x,y) = x 2 y-4xy 3 find f(x,y) [assume all constants of integration are 0] First, undo the derivative with respect to y f x (x,y) = f xy dy = x 2 y-4xy 3 dy = ½ x 2 y 2 xy 4 Now undo the derivaqtive with respect to x f(x,y) = ½ x 2 y 2 xy 4 dx = 1 / 6 x 3 y 2 ½ x 2 y 4 [Note: dealing with constants of integration is trickier in multivariable calculus than in single-variable calculus. We will return to it in chapter X.] Since we will integrate twice, we can write that in the formula initially. We can also include boundary values. Ex) calculate 6 Interpret it like this: 6 first do the inner integral, where x is the variable and it has boundary values 1 & 3 = [3 ] = (27 ) (3 ) = 24 = [8 ] = 448 So what we found is that f(x,y) = f xy (x,y) dy dx. Since we can take derivatives in either order and the result is the same (as long as the derivatives are continuous), we can also integrate in either order and the result is the same. We could also calculate these double integrals with boundary values. Note that when we integrate with respect to one variable, we assume all other variables are constant just as with partial derivatives. Now let s apply this to functions of 3 variables. Ex) if f xyz = x 2 y + e x cos(z) + yz 3/2, find f(x,y,z) [assume all constants of integration are 0] f xy = f xyz dz = x 2 y + e x cos(z) + yz 3/2 dz = x 2 yz + e x sin(z) + yz 5/2 f x = f xy dy = x 2 yz + e x sin(z) + yz 5/2 dy = x 2 y 2 z + e x ysin(z) + y 2 z 5/2 f = f x dx = x 2 y 2 z + e x ysin(z) + y 2 z 5/2 dx = x 3 y 2 z + e x ysin(z) + xy 2 z 5/2

72 72 Since we will integratre thrice, we can write that in the formula initially. We can also include boundary values. Ex) calculate + 1 That was a brief introduction of the multiple integral with several variables as an extension of the single integral with one variable. Now let s look at possible meanings and uses. Math Application: Double Integral as a 3-Dimensional Volume Let s review using integration to find area for functions of 1 variable. To find the area under y=f(x) over an interval, the interval is broken up into short intervals with length Δx. We take a point x i on each interval and make a rectangle with height f(x i ). The approximate area over each interval is f(x i ) Δx, and the approximate total area is f(x i ) Δx. The exact total area is found by taking Δx 0. Thus, area = ( ) For a function of 2 variables in 3 dimensions, the space below the graph is a volume. It seems that we should be able to find it using the integral. There are two ways to think about this. The first way: For 2-variable functions, there is a large region R which is broken up into small regions, each with area ΔA. Each one is the base of a rectangular solid, (x i, y j ) is a point in the base and the height is f(x i, y j ). The volume over that small region is f(x i, y j ) ΔA, the total volume is i j f(x i, y j ) ΔA, and the exact volume comes from

73 taking ΔA 0. Just as in the 1-variable case, that limit is hard to calculate directly. We define the volume as a double integral: V = R f(x,y) da The second way: 73 Think of the volume as a stack of slices. Let s say each slice is at a different y-value. For each slice, the area is A(y) and the width is Δy, so the approximate volume of each slice is A(y) Δy, and the approximate total volume is the sum A(y) Δy Next, any single slice area A(y) for y=constant can be split into rectangles, each at a different x-value. For each one, the height is f(x,y) and the width is Δx, so the approximate area of one rectangle is f(x,y) Δx, and the approximate total area is the sum f(x,y) Δx. Therefore the approximate volume is A(y) Δy = ( f(x,y) Δx) Δy. This can be made exact by converting to an integral. So the volume is: V = A(y) dy = [ f(x,y)dx ] dy As a tradition, we do not write the brackets, but we calculate the inner integral first. Also, note that we could have sliced the other way, which would switch the order of x & y. Therefore, the volume is: V = f(x,y) dx dy OR f(x,y) dy dx What are the boundaries of integration? They are determined by the boundaries on the values for x & y, which are the boundaries of the region in the xy-plane. If the region is a rectangle, with a x b and c y d, then the volume is: c d a b f(x,y) dx dy Note: be careful to match the correct boundaries with the correct variable Ex) find the volume under f(x,y) = 2xy over the region enclosed by x=1, x=3, y=2, y=5

74 What if the region is not a rectangle? In that case, the boundaries of the inner integral change depending on the outer variable. Ex) find the volume under f(x,y)= 4x 2 + y 3 + 4xy 2 and over the region R enclosed by y=x 2, y=0, x=2 74 Note that you can NOT say the boundaries are 0 x 2 & 0 y 4, because that would be the whole rectangle. First step: draw the region R in the xy-plane and make slices. There are two ways to draw the slices. In the first picture with vertical slices, notice that the slices range from x=0 to x=2. Further, on each slice, x is constant and y ranges in value from 0 to x 2. Thus the boundaries are 0 x 2 & 0 y x 2 and the volume is V = Note that x is fixed first, so the outer integral is dx. Note that y has a variable boundary so the inner integral is dy. Next, consider the second picture with horizontal slices, where the slices range from y=0 to y=4. Further, on each slice, y is constant and x ranges from y 1/2 to 2. Thus the boundaries are 0 y 4 and y 1/2 x 2 and the volume is V = / Here x has a variable boundary and the inner integral is dx. Vertical slice choice: V = = [4 + + ] = = + + = Horizontal slice choice: V = / = / = / + / + 2 = + / / = When we calculate volume using double integrals of f(x,y), it often helps to have two pictures in mind: the graph of the solid in 3 dimensions, and the graph of the region R in 2 dimensions. The fact that you can switch the order of integration is a deep mathematical theorem.

75 Fubini s Theorem: if f(x,y) is a continuous function on the rectangle R = [a,b] x [c,d], then: R f(x,y) da = c d a b f(x,y) dx dy = a b c d f(x,y) dy dx The theorem can be extended to more general regions, such as those defined by differentiable functions (like the problem above). The proof of these theorems is difficult and beyond the scope of this course. Ex) find R x+2y da, where R is the region bounded by y=2x 2 & y=x 2 +1 First step: 75 Note that it only makes sense to make vertical slices, not horizontal slices Ex) find the volume under z=xy over the region enclosed by y=x+1 & y 2 =2x+10 First step: Note that one way of drawing slices will make a much easier calculation than the other. With vertical slices, sometimes the bottom function is the straight line, sometimes it s the negative branch of the parabola this will require a set-up with two integrals, which we would hope to avoid. With horizontal slices, the larger function is always the straight line and the smaller function is always the parabola. We previously encountered this idea of needing to choose a slice direction when calculating areas enclosed by curves (as well as volumes of solids of revolution). Find x-boundary values (not constant): x=y-1 & x= y 2-5 Find y boundary values (constant) with points of intersection: y-1= y 2-5 y 2-2y-8=0 (y-4)(y+2)=0 y=-2,4 V = R f(x,y) da = = = ( 1) ( 5) = + +

76 76 = + 3 = + = (4) + (4) (4) (4) ( 2) + ( 2) ( 2) ( 2) = 0 How can the volume be 0? We are calculating signed volume, just like signed area from 1-variable calculus. The volume counts positively for positive z-values, and negatively for negative z-values, so the net volume might be 0, as it is here. If you want to count all volume positively, take the absolute value of the function. This is typically very difficult to calculate by hand, but can be done numerically with technology. Ex) find R sin(y 2 ) da, where R is the region bounded by y=x & y=1 & x=0 First, sketch R and make slices With the first slice choice, R sin(y 2 ) da = 0 1 x 1 sin(y 2 ) dy dx =?? That inside integral is impossible. So try the other slice choice, which changes the order of integration. R sin(y 2 ) da = y sin(y 2 ) dx dy Area as a double integral Note that, in single variable calculus, an integral with no function for height would give the length of the interval of x-values: a b dx = b a In its approximate form, it is the sum of lengths Δx. In the same way, double integrals can be used to find area, by including only width and length (with no function for height). For a region R, to find its area, break it up into small rectangles. Each rectangle has height Δy, width Δx, and area ΔA. The total area of R is: Approximately, Area ΔA = Δx Δy (The sum is a double sum, once over the widths and once over the lengths.) Exactly, Area = R da = R dx dy What are the boundaries of integration? They are determined by the boundaries on the values for x & y, which are the boundaries of the region in the xy-plane. Ex) find the area of the rectangle with -2 x 3 & 0 y 4 Area = R da = = [ ] = 4 = [4 ] = 20 As an easier way to find the area, since this is just the area under the function y=f(x) for y=4, we could have calculated Area= 4 = [4 ] = 20 An even easier way to find the area, since the region is a rectangle, is (4)(5) = 20. The double integral method to find the area is useful when the region is not a rectangle, or not defined by a function. Ex) find the area enclosed by x+9=2y 2 and x=y 2 and above y=0

77 77 Find the intersections: x+9=2x -> 9=x -> y=3 also y=0 Choose to slice horizontally, which is dy. On each slice, 2y x y 2 Area = R da = = [ ] = 9 = 9 = 18 ex) find the area enclosed by the ellipse + = 1 Area = R da Here is another way to see why both a single integral and a double integral can calculate a 2-dimensional area. If the region can be sliced vertically, then the height is y=f(x) (or the difference between two functions), and the area reduces to ( ) ( ) Area = = [ ] = ( ) Similarly, if the region can be sliced horizontally, then the height (measured horizontally) is x=f(y) (or the difference between two functions), and the area reduces to Area = ( ) = [ ] ( ) More Applications of Multiple Integrals Math Appliction: Average Value = ( ) Previously, we saw that the average value of f(x) over the interval [a,b] is f avg = ( ) The average value for multivariable functions works in a similar way.

78 78 The average value of f(x,y) on the region R is f avg = ( ) (, ) Ex) find the average value of the function f(x,y) = 4x+2x 2 y 3 on the region R: 0 x 3, 1 y 5 f avg = 1 / x+2x 2 y 3 dy dx = 4 + = = [ ] = 2880 = 240 One way to interpret the answer is that the volume under the function and over this rectangular region is the same as a box over the rectangle with constant height 240. Math Application: Surface Area For a function f(x,y) over a region R, the surface area is S = ( ) + ( ) + 1 Ex) find the surface area of f(x,y) = cos(x)+sin(y) over the region R: 0<x<4, 1<y<4 [set up, do not solve] f x = -sin(x) f y = cos(y) S = ( sin ) + (cos ) + 1 = One way to see why this is true is to recall that the area of a parallelogram with sides a and b is Area = a X b. At each point, the graph is approximated by a tangent plane. In the domain, each rectangle of area ΔA has sides Δx and Δy. Over that rectangle is a piece of the surface with area ΔS and a parallelogram in the tangent plane. One parallelogram side has constant x and one side has constant y. For constant y, z is a function of x, the tangent line has slope f x. As a vector, this is (1,0,f x ) Δx. Similarly, for constant x, z is a function of y, the tangent line has slope f y. As a vector, this is (0,1,f y ) Δy. So the area of this parallelogram is: (1,0,f x )Δx X (0,1,f y )Δy = (-f x, -f y, 1) Δx Δy= ( ) + ( ) + 1 Δx Δy. So, the total surface area is: Approximately: surface area = ΔS ( ) + ( ) + 1 Δx Δy Exactly: surface area = R ds = ( ) + ( ) + 1 or ( ) + ( ) + 1 Physics Application: Density, Mass, and Center of Mass Center of mass, introduction The center of mass is the balancing point. (Think of people sitting on a seesaw.) To find this point, we must consider mass, but we also must consider position. (Think of a lighter person balancing a heavier person on the seesaw by sitting further out.) A system of two weights is in balance when: m 1 d 1 = m 2 d 2 where m 1 m 2 are masses of each object and d 1 d 2 are the distances from the balance point. If we refer to position instead of distance, since one mass is on the positive side and one is on the negative side, this is the same as: m 1 d 1 = -m 2 d 2 or m 1 d 1 + m 2 d 2 = 0 If the object is sitting at x 1, the position can be written as x 1 - and we get: m 1 (x 1 - ) + m 2 (x 2 - ) = 0 solve for m 1 x 1 m 1 = m 2 m 2 x 2 m 1 x 1 + m 2 x 2 = m 1 + m 2 =

79 79 What if you have many objects? It is easy to generalize the formula. = Notation and names: is called the center of mass M = Σ m i x i is called the moment of the system m = Σ m i is called the mass of the system (total mass) center of mass = note: the moment and the center of mass need to be calculated for each variable Ex) where is the center of mass of a standard 1-meter stick? At the half-meter mark. This is just obvious, by symmetry. Ex) if a 3kg mass sits at 4cm and a 6kg mass sits at 10cm, where is the center of mass? mass m = 3+6 = 9kg Moment M = (3)(4)+(6)(10) = 72 kg cm Center of mass = = 8cm Center of mass, one variable calculus Where is the center of mass of a meter stick that gets denser the closer you get to the 1m end? Ex) find the center of mass for a 1m stick which has density 0.1x kg/m at the x-distance To find, we need find the moment and the mass. mass = ( h) = (density)(length) In a little piece of the stick, Δm = r(x) Δx, where r(x) = density To find the total mass, add up all the little masses: approximately, m r(x) Δx exactly, m = r(x) dx For our stick, mass = x dx = 0.05x = 0.05 kg moment = (position)(mass) =(position)(density)(length) The total moment is Approximately: M Σ x r(x) x Exactly: M = x r(x) dx For our stick, moment M = x(0.1x)dx = x 2 dx = 0.033x = kg m Center of mass = =. = 0.66 m. To summarize the general formulas: mass m = r(x) dx Moment M = x r(x) dx center of mass = r(x)=density, x=position Center of mass, several variable calculus The size of a 2-dimensional object is measured as area instead of length. Then we have: mass = ( ) = (density)(area) Now, density is a function of two variables, r(x,y)

80 80 mass m = R r(x,y) da x-moment M x = R x r(x,y) da y-moment M y = R y r(x,y) da x-center of mass = y-center of mass = ex) find the total mass and center of mass of a sheet with density r(x,y)=x+y 2 0<y<3 m = R r(x,y) da = + = [0.5 + ] = = 2 + = 24 kg M x = R x r(x,y) da = + = + = + 2 kg/m 2 over the region R: 0<x<2, = + = 26 kg m M y = R y r(x,y) da = + = [0.5 + ] = = + = 49.5 kg m x-center of mass = y-center of mass = m 2.06 m the center of mass is (1.08, 2.06) Math Application: Probability A brief introduction The probability of an event is its frequency out of the total number of possibilities, P = Ex) if you roll a regular die, what is P(3)? 1/6 Ex) what is P(roll larger than a 4)? 2/6 Suppose you have the following probability distribution for family size: # children Probability 0.11

81 Ex) find P(2 children) Ex) find P(from 1 to 3 children) Ex) find the average number of children Probability with calculus, one variable The previous method assumes you can list all of the outcomes with their probabilities. But what if there is a continuum of possibilities? (For example, the temperature can be any real number.) Notice that P(a x b) = P(a) + + P(b) = a x b P(x) If there is a continuum of possibilities, the probability is: Approximately: P(a x b) P(a) + + P(b) = P(x) Exactly: P(a x b) = a b p(x) dx where p(x) is a function called the probability density function. Note that the total probability is always 1, so the integral over all possibilities is 1: ( ) = 1 If there is a continuum of possibilities, the mean is: Approximately: μ x 1 P(x 1 ) + + x n P(x n ) = x P(x) where this includes all possible x-values Exactly: μ = ( ) Note that when we calculuate the mean, typically we divide by the total number of values. Here, however, that total number is represented by the total probability, which is 1, so to divide by 1 is redundant. Ex) for the probability density function p(x)=0.5x for 0<x<2 (p=0 elsewhere), (a) confirm that this is a probability density function (b) find the probability P(x>1) (c) find the mean (a) ( ) = 0.5 = [0.25 ] = 0.25(2) (0) 2 = 1 (b) P(x>1) = x dx = [0.25x 2 ] 1 2 = 0.75 (c) Mean μ = x p(x) dx = 0 2 x(0.5x)dx = x 2 dx = [ x 3 ] 0 2 = 1.33 Probability with calculus, several variables Now, we consider the probability of two events. In general, the probability that the point (x,y) is within the region R can be written R p(x,y) da, where p(x,y) is called the joint probability density function. Note that, since the total probability is 1, S p(x,y) da=1, where S is the entire plane, representing all possibilities. The probability that x is between a & b and y is between c & d is: P(a<x<b, c<y<d) = a b c d p(x,y) dy dx. The mean (or expected value) for x is μ x = S x p(x,y) da. The mean for y is μ y = S y p(x,y) da. Ex) When you go to a movie, you have to wait on the ticket line and then wait for concessions. At one cinema, we have a model for total wait times For x=minutes waiting for tickets and y=minutes waiting for concessions, p(x,y) = / / for x,y>0. Set up (do not solve) (a) the probability that both wait times are less than 5 minutes (b) the probability that the total wait time is less than 20 minutes (c) find the mean ticket wait time (a) P(x<5,y<5)= / /

82 82 (b) P(x+y<20) = / / (c) mean μ x = / / Double integrals with polar coordinates Previously, we found that Area = R da = R dx dy in rectangular coordinates, and Area = r 2 dθ in polar coordinates. We can turn this into a double integral. Using formulas, note that r 2 = r dr. So the area expression can be written as a double integral, Area = R r dr dθ. Using a graph, note that in polar coordinates, a tiny sector from small changes in r & θ is nearly a rectangle, with side lengths dr and r dθ, so da = r dr dθ. Ex) find the area enclosed by a circle radius 2 (centered at the origin). Area = R da = = = 2 = [2 ] = 4 This equivalence also tells us how to change integrals between rectangular and polar coordinates. R f dx dy = R f r dr dθ On the left side the function is expressed in x & y, f=f(x,y), and on the right side the function is expressed in r & θ, f=f(r,θ). Ex) calculate R x 2 dx dy where R is the top half of the circle radius 1 centered at the origin With rectangular coordinates: R x 2 dx dy = = [ ] With polar coordinates: = 1 and this is possible but hard R x 2 dx dy = ( ) = = = = 1 + cos2 = + sin2 = Ex) find the area of a three-quarter-annulus, part of the region between two circles, radii 2 & 3 Notice that, if you try to set up the integral using dx dy, there is no good way to describe R or slice R. But if you use polar coordinates, then R is 2 3, 0 and: / Area = = = / = / = This region is made from elementary shapes, so you could find its area using basic area formulas: Area = (π(3) 2 π(2) 2 ) = π For alternate proofs of the formula to convert integrals between rectangular and polar coordinates, see the appendix. Triple Integrals Previously, we integrated over a line segment in R using dx, and over a region in R 2 using da (which could be dxdy or rdrdθ). Now we will integrate over a solid region in R 3. Break the solid region R into little boxes. Each one has width Δx, length Δy, height Δz, and volume ΔV. / =

83 83 First, note that adding up the volume of the little boxes yields the total volume. Ex) find the volume of the region R, a cylinder centered around the z-axis, radius=3, from z=0 to z=2 V = R dv, or = [ ] = dy dz (note that this dy integral is the area of a circle with radius 3) = 0 2 9π dz = [9πz] 0 2 = 18π Of course, we could also find the volume with the formula, V = πr 2 h = π(3) 2 (2) = 18π Next we consider the case of integrating a function f(x,y,z) on that 3-dimensional region. We form the sum i j k f(x i, y j, z k ) ΔV, or i j k f(x i, y j, z k ) Δx Δy Δz. This is an approximation of the integral, R f(x,y,z) dv, or R f(x,y,z) dx dy dz. Suppose the region R is a solid rectangle, a x b, c y d, p z q. Then the integral of f(x,y,z) over R is p q c d a b f(x,y,z) dx dy dz Note: the integral of a 2-variable function can represent a volume in 3 dimensions. So the integral of a 3- variable function can represent a volume in 4 dimensions. This is impossible to draw! So we must be satisfied with understanding the idea as an extension from lower dimensions. Other interpretations are easier to draw, and we will look at some of those. In fact, all of the applications from 2-variable functions generalize to 3-variable functions. The best way to keep track of what is being calculated is to note the units of each component. For example, for R dx dy dz, if x & y & z are lengths, then the result is (length) 3, or 3D volume. Also, for R f dx dy, if f & x & y are lengths, then the result is (length) 3, or 3D volume. Math Application: Density, mass, moment, center of mass for 3-variable functions For a solid object, if its density at a point (x,y,z) in a region R is r(x,y,z), measured in kg/m 3, then the mass is: m = R r(x,y,z) dv, or R r(x,y,z) dx dy dz The moment is: For the x-direction, M x = R x r(x,y,z) dx dy dz For the y-direction, M y = R y r(x,y,z) dx dy dz For the z-direction, M z = R z r(x,y,z) dx dy dz The center of mass is (,, )=,,

84 84 Vector Fields How would you mathematically represent wind patterns? How would you represent water currents? How about the force on an electron in the presence of a magnet? For wind and water, we would measure the velocity of the particles at any point. This is a velocity field. For the magnet, we would measure the force exerted on a charged particle at any point. This is a force field. In either case, at every point in space, there is a magnitude and direction, so we have a vector field. A vector field is a function which takes a point in R n (or a region of it) and assigns a vector in R n. (We will usually work in R 2 or R 3.) In each of these pictures, vectors at various points have been drawn. Notation: We can write the formula for an arbitrary vector field. In R 3, F = P(x,y,z)i + Q(x,y,z)j + R(x,y,z)k or Ex) for the vector field F = (y/x)i + (x 2 y+sin(πx))j, find the vector at the point (4,-2)

85 85 We have already calculated one type of vector field before, the gradient vector field. For a function f(x,y), at every point (x,y), you get the vector f=(f x, f y ). At any point, we know the vector direction is the direction of the greatest rate of change of the function and the vector magnitude is that maximum rate of change. Ex) find the gradient vector field for f(x,y) = 2e xy +y 2 Vector fields are one of the most important objects of study in physics. Here are some examples. Fluid velocity field The velocity vector of a fluid at any point is described by the vector field F. Gravitational force field Newton s law of gravitational attraction says that the force between two masses is F =, where G is a known constant, m & M are the two masses (in kg) and r is the distance between them (in m). The force also has a direction. If mass M is fixed at the origin (0,0) and mass m is at the point = (x,y), then the force on mass m is from to the origin. That direction vector is (0,0)-(x,y) = (-x,-y), or -. Using the notation r=, the unit direction vector is Now we have the magnitude and the direction. Therefore the gravitational force is described by a vector field called the gravitational force field, = Electric force field Coulomb s law of attraction for charged objects says that the force between two charges is F =, where k is a known constant, q & Q are the two charges (in Coulombs), and r is the distance between them (in m). The force has a direction. Assume charge Q is fixed at the origin (0,0) and charge q is at the point = (x,y). If the charges have the same sign (so qq>0), the force on charge q is away from the origin. If the charges have opposite signs (so qq<0), the force on q is toward the origin. The unit direction vector is ± (positive in the first case, negative in the second Now case). we have the magnitude and the direction. Therefore the electric force is described by a vector field called the electric force field, = The gravitational field is actually a gradient vector field. For the function f(x,y) = calculation shows that f = (where r = = + ). = ( ) / = ( + ) / = ( 0.5)( + ) / (2 ) ( 0.5)( + ) / (2 ) =, the following In general, where f =, f is called the potential function of the vector field, and the vector field is called conservative.

86 86 Vector Field Operators Now we will introduce some important operators on vector fields. We will learn some amazing general properties. Further, in physics applications, the mathematical results tell us about the physical properties of systems such as fluid flow and electromagnetic fields. An operator is a type of function. For this topic, we don t need the formal definition; just think of it as a function which takes an input and gives an output. Swirl The swirl of a vector field F=(P,Q) in two variables is: swirl(f) = Ex) for F = (y,-x), find the swirl Q=-x, so = -1 & P = y so = 1 swirl(f) = -1 1 = -2 Ex) for F = (3x,2y), find the swirl. swirl(f) = 0 0 = 0 The reason for the name swirl can be understood in the context of fluid flow. Fluid particles near (x,y) tend to go straight or spin in circles (or some combination), and the speed of their spinning is determined by the magnitude of swirl. The vector field in the first example has some circular movement at every point in the plane. If swirl(f) = 0, the fluid flow is called non-swirling. The vector field in the second example has no circular movement. Note that, for swirl, the input is a 2-variable vector field and the output is a scalar which involves derivatives of the components. Divergence The divergence of a vector field F is defined as: div(f) = + + (Note that div can be calculated for a vector field with any number of variables.) There is a shorthand notation for the divergence formula related to the gradient. Recall that f =

87 87 Rewrite this as, and think of as a vector of derivative operators, = or + + Now div(f) can be written as F because = + + Therefore we have: div(f) = F = + + Note that the divergence of a vector field can be defined using the dot product, and the result is a scalar. Note that, for div, the input is a vector field and the output is a scalar whose formula involves derivatives of the components. Ex) for F = (3x 2 y, xy), find div(f) Ex) for F = xyzi + xcos(y)j + tan(x 2 +xy 3 )k, find div(f) The reason for the name divergence can be understood in the context of fluid flow. If F(x,y,z) represents the flow, or velocity, of a fluid, then div(f) is the net flow of the mass of fluid at the point (x,y,z). If div(f) = 0, then the same amount of fluid always comes in as is going out, and the fluid (and its flow F) is called incompressible. If the divergence at a point is positive, more fluid is leaving than entering, so in a sense fluid is gained and the point is called a source. If the divergence at a point is negative, more fluid is entering than leaving, so in a sense fluid is lost and the point is called a sink. Curl The curl of a vector field in 3 variables is defined as: curl(f) = + + OR Again, there is a shorthand notation for the formula related to the gradient. curl(f) = X F because = curl(f) = X F = + + Note that the curl of a vector field can be defined using the cross product, and the result is a vector field. Note that, for curl, the input is a 3-variable vector field and the output is a 3-variable vector field (whose formula involves derivatives of the components). Ex) for F= 3xze 2y i + xyz 2 j + (2x+3yz)k, calculate curl(f)

88 88 Ex) F = yi xj +1k, calculate curl F curl F = (0-0)i + (0-0)j + (-1-1)k = -2k ex) Ex) F = xi + yj + zk, calculate curl F curl F = (0-0)i + (0-0)j + (0-0)k = 0 The reason for the name curl can again be understood in the context of fluid flow. Fluid particles near (x,y,z) tend to rotate around the axis determined by the curl vector, and the speed of their rotations is determined by the magnitude of the curl vector. If curl(f) = 0, the fluid is rotation-free. The vector field from the first example is plotted on the left, and it is easy to see the rotation; the axis of rotation is straight up and down, which is the z-axis. The vector field from the second example is plotted on the right, and it is easy to see there is no rotation. Curl seems like a 3-dimensional version of swirl, and that s not a coincidence. For F = (P,Q,R), we can make a 2-dimensional version by setting R=0. Then curl(p,q,0) = (0-0)i + (0-0)j + k = swirl(p,q)k. The magnitude of the z-component of the curl formula is the swirl formula. This can be written as curl(p,q,r) k = swirl(p,q) Theorem: If f(x,y) has continuous second partial derivatives, then swirl( f) = 0 Proof: by a direct calculation, swirl( f) = swirl = f yx - f xy = 0 Theorem: If f(x,y,z) has continuous second partial derivatives, then curl( f) = 0 Proof: by a direct calculation, curl( f) = X ( f) = (f zy f yz )i + (f xz - f zx )j + (f yx - f xy )k = 0i + 0j + 0k Recall that a vector field F is called conservative when f=f. Therefore, if F is conservative then curl(f) = 0 For certain vector fields, the converse is also true. For a vector field F, if F is defined everywhere in R 3 and the component functions P,Q,R have continuous partial derivatives and curl(f) =0 then F is a conservative vector field. (The proof of this requires Stokes Theorem, which is covered later in the chapter.)

89 89 Ex) determine if F = (y-e z sinx, x, e z cosx) is a conservative vector field F is defined everywhere on R 3 and its component derivatives are continuous curl(f) = + + = (0-0)i + (-e z sinx - -e z sinx)j + (1 1)k = 0i + 0j + 0k OR 0 therefore, F is conservative (in other words, for some function f, f=f) Further, we can find the potential function f: f x = y-e z sinx f = xy + e z cosx +? where? looks like a constant when you take so it is a function of y&z f y = x f = xy +?? where?? is a function of x&z f z = e z cosx f = e z cosx +??? where??? is a function of x&y therefore, f = xy + e z cosx + C Theorem: for a 3-dimensional vector field F, if the components P,Q,R have continuous second partial derivatives, then div(curl(f)) = 0 Proof: by a direct calculation, div (curl(f)) = div,, = + + = + + = 0 where terms cancel because you can switch the order of the derivatives ex) show that F = (2xy, yz, xe z ) is not the curl of another vector field. F certainly has continuous derivatives. So if F = curl(g), then div(f) = div(curl(g)) = 0. But div(f) = 2y+z+xe z 0 Line Integral Previously, we had a function y=f(x) and we could integrate it over an interval of x-values [a,b]. We will now integrate a function f over a parametric curve C, with the interval determined by t-values. After that, we will look at the general formula for a scalar line integral. Then we will integrate a vector field F over a curve C (a vector field line integral). Then we will see the connection between scalar line integrals and vector field line integrals. Line integral of a scalar function with respect to arc length First, lets look at what it means to integrate a function over an interval. We previously saw integrals of y=f(x) over an interval of x-values. Now lets look at integrals of f(x,y) where x=x(t), y=y(t) so the interval is for t-values. Before: Now:

90 In the previous case, as x changes our position on the horizontal axis changes. In the new case, as t changes our position on the curve changes. (C is defined parametrically by r(t)=(x(t), y(t) ) Expression Classic: y=f(x) Parametrized: f=f(t) = f(x(t), y(t) ) = f(r(t)) variable x t Rectangle height f f Rectangle width Δx Δs = ( ) + ( ) Area, approximate f Δx f Δs Area, exact ( ) (, ) = ( ( ), ( ) ) + 90 Ex) find the area under f=x 3 y -1 over the curve C defined by y=x 2 for 2 x 3. The first step is to convert all expressions so that the variable is the parameter t. For the curve: x=t, y=t 2 and 2 t 3. Also, dx/dt=1 & dy/dt=2t. For the function, f=(t 3 )(t 2 ) -1 = t. Now: = ( ) u=1+4t 2 du=8t dt = ( ) / = / = (1 + 4 ) / = 37 / 17 / 24.6 It seems like we didn t really need to use t, we could just have replaced y with x 2 and then done the calculation. That s because y is a function of x. But parametric curves include cases more general than that. Ex) find the area under f=3x-4y over C defined by the left half of the circle centered at the origin with radius 1. Rewrite in terms of the parameter t. For the curve, x=cos t, y=sin t for t. For the function, f=3cost 4sint. Now, / / = (3 cos 4 sin) ( sin) + (cos ) / = 3 cos 4 sin / / = [3 sin + 4 cos ] / = (3+0) - (-3+0) = 6 Note that we chose to start at the top of the curve and go to the bottom. We could also have started at the bottom. That would switch the boundaries of integration, which would switch the sign of the result. A selected direction for a curve is called an orientation. So changing the orientation changes the sign but not the magnitude of the result. Note that the classic form is a special case of the parametrized form. For the classic integral, the parametrized curve is the straight line from (a,0) to (b,0), the parametrization is x=t, y=0. Algebraically, + = (, 0) 1 = ( ). Note: there are many equivalent notations for f, which can be confusing. The variable t is plugged into functions x(t) & y(t) to get values for x & y, which together form the vector r. Those values are plugged into f and generate a scalar. If you write f(t), it is simpler but you omit the role of the vector r or coordinates x & y. If you write f(r(t)) or f( x(t), y(t) ), it is more complete but also more complicated. Line integral of a scalar function in 3 dimensions The line integral formula extends from 2 dimensions to 3 dimensions in the natural way. To integrate a function f over a curve C (with respect to arc length):

91 91 (,, ) = ( ( ), ( ), ( )) + + Ex) find the area under the function f=y sin(z) along the curve C defined by r = (cos t, sin t, t) for 0 t 2π C y sinz ds = 0 2π sin(t) sin(t) [ ( )] + [ ( )] + [ ( )] = 0 2π sin 2 (t) = 2 (1 cos(2 )) = sin(2 ) = 2 Note: this could represent the area under the function along the curve. However, since for functions of 2 variables you need 3 dimensions to see the area, then for functions of 3 variables you need 4 dimensions to see the area, and we cannot draw that. Line integral of scalar functions in coordinate form So far we only looked at a specific type of line integral, integrals which can be written with respect to ds. Since these parametric curves live in the xy-plane, a line integral could involve integrating with respect to x (with dx) or with respect to y (with dy). The coordinate form for a line integral is C P dx + Q dy To calculate it, we can make the conversion to using t as the variable: x=x(t), dx = or y=y(t), dy =. Ex) calculate C xy 2 dx + y dy where C is the curve x-2=y 2 from (3,-1) to (6,2) First, write the parametrizations. On the curve C, y=t, x=t 2 +2, and -1 t 2. Also, dx=2t dt, dy=dt. Now, C xy 2 dx + y dy = (t + 2)t 2t dt + t dt = = + + = = 34.5 To calculate a coordinate scalar line integral in 3 variables, C P dx + Q dy + R dz, for the curve C where x=x(t), y=y(t), z=z(t), rewrite everything in terms of the parameter t. Ex) calculate C dx + xy dy + z dz where C is defined by =(t 3,2t,sin t) for 0<t<2 Application in Physics: Density and Mass Recall that if a linear object covers the line segment a<x<b and has a linear density p(x) (in units kg/m), then the mass is = ( ) Also, its moment is M = ( ) Now, let s extend this to parametric equations. If a linear object s shape is the curve C defined by r=(x(t), y(t) ) for a<t<b, with linear density p(x,y), then the mass is m = Further, its moments are M x = and M y = Finally, its center of mass is (, )=, Ex) Find the center of mass of the wire with density p=x+2y which forms a quarter-circle center (0,0) radius 1 in the first quadrant. First make the parametrizations: For C, x=cos t, y=sin t, 0 t π / 2 For p, p=cos(t) + 2sin(t) For the differentials, x =-sin(t) & y =cos(t) & ds = ( sin( )) + (cos( )) = dt

92 92 Mass m = C p ds = 0 π/2 cos(t) + 2sin(t) dt = [sin(t)-2cos(t)] 0 π/2 = = 3kg x-moment M x = C x p ds = 0 π/2 cos(t) (cos(t) + 2sin(t)) dt = 0 π/2 cos 2 t + 2sin(t)cos(t) dt = 0 π/ cos(2t) + sin(2t) dt = [0.5t sin(2t) 0.5cos(2t)] 0 π/2 = (0.25π ) (0.5) = 0.25π 0.79 y-moment = M y = C y p ds = 0 π/2 sin(t) (cos(t) + 2sin(t)) dt = 0 π/2 2sin 2 t + sin(t)cos(t) dt = 0 π/2 1 - cos(2t) + 0.5sin(2t) dt = [t - 0.5sin(2t) 0.25cos(2t)] 0 π/2 = (0.5π ) (0.25) = 0.5π 1.57 Center of mass = (π/12, π/6) = (0.26, 0.52) Line integral in a vector field The idea behind a line integral in a vector field can be seen from the physics application of work. Recall the basic case, in which quantities are scalar and constant, W = F x, where W is work (in Joules), F is force (in Newtons) and x is distance (in meters). Where the quantities are scalar but varying, W = b a F dx. Where the quantities are vectors but constant, =. Now we look at the final case where the quantities are vector and varying. The answer, in one form, is =. For the position function, we can write instead of, so the integral is This formula also makes sense if you construct an approximation. A short piece of the path is Δ, the value of the vector field on that segment is, and the work done over that short piece of the path is approximately. The total work is approximately. The total work is exactly =. To actually calculate this, we will rewrite using the parameter t. Recall that ( ) describes a path in space C, so = (,, )= ( ( ), ( ), ( )). For the boundaries of integration, let ( )= and ( )=. For the force vector function, = (,, )= ( ), ( ), ( ) = ( ( )). The following formulas for are all equivalent: = = ( ), ( ), ( ) also, = = = There are so many equivalent notations because you can write it in terms of or x/y/z or t. Thus, = = ( ) = ( ), ( ), ( ) ( ), ( ), ( ). This is the formula to find Work. Further, this formula is the general definition of the line integral of a vector field F over a curve C defined by ( ). Note that there are several forms of the equation; use whichever works best for you. Ex) Find the work done by the force vector field = (x+y, x-2z, 2x-y) over the straight line C defined by r = (1+3t,3-4t,-2+4t), for 0 t 1 First, rewite using the parameter. C is already parametrized. d =(3,-4,4)dt. For the vector field, = (4-t,5-5t,-1+10t). So, W = ( ) = (4, 5 5, ) (3, 4,4) = = = 12 + = 16.5 joules Ex) find the line integral of F=(xy,2-yz, xz) over the curve C which is the straight line from (2,1,2) to (5,2,-3) First rewrite parametrically. C is a straight line, the direction vector is (5,2,-3) (2,1,2) = (3,1,-5), so its parametrization is:

93 93 r = (2,1,2) + t(3,1,-5) = (2+3t, 1+t, 2-5t) with 0<t<1. dr = (3,1,-5)dt F = ( (2+3t)(1+t), 2-(1+t)(2-5t), (2+3t)(2-5t) ) = (2+5t+t 2, 3t+5t 2, 4-4t-15t 2 ) The line integral is = 0 1 (2+5t+t 2, 3t+5t 2, 4-4t-15t 2 ) (3,1,-5) dt = t+3t 2 + 3t+5t t + 75t 2 dt = = t + 83t 2 dt = [-14t + 19t t 3 ] 0 1 = Note that, for a vector field = (P,Q,R), where P=P(x,y,z)=P(x(t),y(t),z(t)), and similarly for Q & R C = C (P,Q,R) (dx, dy, dz) = C P dx + Q dy + R dz Thus, C = C P dx + Q dy + R dz a vector field line integral has an equivalent scalar line integral. This makes sense because the vector field line integral is an integral (or sum) of magnitudes (or scalars), as you can see from the dot product. Also, this form of a scalar line integral is called the coordinate form; it involves dx,dy,dz instead of ds. As you can see in the example, we can still rewrite it with a parametrization to calculate it. Also, = = = which provides an alternate way to write a line integral of a vector field, with respect to ds. Fundamental theorem of line integrals Previously, we saw the fundamental theorem of calculus for one variable, ( ) = f(b) f(a). Roughly speaking, the integral of the derivative is the original function. Now, we will see a verision called the fundamental theorem for line integrals. The key step is to identify the appropriate derivative it is the gradient. = ( ) ( ( )) where the curve C is defined by ( ) for a t b Proof: = ( ) = + + = ( ) = ( ) ( ( )) Ex) calculate C where =(y-e z sinx, x, e z cosx) and C is defined by = (t,5t-t 2,2t) for 0 t 2π In a previous example, we found that this vector field has potential function f= xy + e z cosx. So the integral is: = = f( (2π)) f( (0)) = f(2π,10π-4π 2,4π)- f(0,0,0) = (2π)(10π-4π 2 )+e 4π (0 0+e 0 cos0) = 20π 2-8π 3 + e 4π 1 Note that typically, a line integral calculation begins with parametrizing the vector field and the curve C, then an expression is integrated. By the Fundamental Theorem of Line Integrals, we do not have to do all that work; once we have = we can just use the potential function and the endpoints. Further, this implies something remarkable: in calculating this line integral, the path does not matter, only the endpoints. This property is called path independence. This property could be incredibly useful in calculating a line integral. For example, if the path is very twisted or has a very complicated expression, it doesn t matter, you only need the two endpoints to complete the calculation. Now we need to know which line integrals have this property. We have just shown that if the vector field is conservative, then the line integral is path independent. It turns out that if the line integral is path independent, then the vector field is conservative. (The proof is in the appendix.)

94 94 Ex) calculate C where C is =(t,t 2,t 3 ) for 1<t<2 and = (4xyz 3, 2x 2 z 3, 6x 2 yz 2 ) This will be a much simpler calculation if is conservative, with potential function f. f x = 4xyz 3 f = 2x 2 yz 3 +? f y = 2x 2 z 3 f = 2x 2 yz 3 +? f z = 6x 2 yz 2 f = 2x 2 yz 3 +? therefore f = 2x 2 yz 3 (note that since this is a definite integral, we don t need the constant) so C = = ( ) ( ( )) = f(2,4,8) - f(1,1,1) = = Notation: For a 2-variable function, df = f x dx + f y dy = f d. (Also, for a 3-variable function, df = f x dx + f y dy f z dz = f d.) So written this way, the Fundamental Theorem of Line Integrals becomes C df = f( ) f( ), where and are the endpoints of the curve. Let s find another property of path independent line integrals. Let be a path-independent vector field. Let C be a path which begins and ends at a point A (C is called a closed path). Let B be another point on the path, and C 1 and C 2 are the two paths from A to B. = by the definition of path independence = 0 by getting one side to be 0 = 0 by rewriting with the curve endpoints + = 0 because switching endpoints switches the sign = 0 because the combined paths of the two integrals form the closed path The line integral C is path independent in a domain D precisely when C = 0 for every closed path. To summarize: is conservative C is path independent C = 0 for all closed paths This will be useful for any conservative force, and there are many important conservative forces in physics. We have already seen the two most important, the gravitational force and the electromagnetic force. Ex) How much work is done by the gravitational field = (2,3) to (4,-3) to (-6,1) back to (2,3)? is conservative, and the path is closed, so W = C = 0. to move a 5 kg mass along the triangle from Ex) How much work does the earth s gravitational field do to bring a 70000kg satellite from an elevation of 100,000m to 40,000m? (earth mass = 5.97 X kg, G = 6.67 x Nm 2 /kg 2 ) For the gravitational field, its potential function f =, so W = C = = f(40000) f(100000) = 6.97 x x = 4.19 x joules Application in Physics: Conservation of Energy Let s apply these formulas to motion. Suppose is a force field which moves an object. (This could be through gravitation, electromagnetism, etc.) The object will move from point A to point B along a path C with the formula ( ) where (a)=a and (b)=b. Newton s Second Law of Motion says that (force) = (mass)(acceleration), F=ma (in scalar form) and = (in vector form). Acceleration is the second derivative of position, so = ( ). Lets calculate how much work the force field does to move the object from A to B. W = C = ( ) ( ) = ( ) ( ) = [ ( ) ( )] = ( )

95 95 = ( ) = ( ) ( ) Note that = = which is the speed. The expression is called the kinetic energy K. So the work done by the field on the object represents the object s change in kinetic energy: W = K(B) K(A). If is a conservative field (such as gravity or electromagnetic), then =. The potential energy of an object is defined as P=-f, so = P. The total energy is the kinetic energy and potential energy combined, E = K + P. W = C = C - P d = -P( (b)) + P( (a)) = P(A) P(B) Using the two equations for W: K(B) K(A) = P(A) P(B) K(A) + P(A) = K(B) + P(B) E(A) = E(B) Therefore, the total energy E is constant, or conserved. This is the Law of Conservation of Energy, and it is an example of why gradient fields are called conservative. Green s Theorem Previously, the Fundamental Theorem of Calculus and its vector version, the Fundamental Theorem of Line Integrals gave an equation between the integral of a derivative expression over a line (which is 1-dimensional) and the evaluation of the original expression at the line s boundary endpoints (which are 0-dimensional). FTC: ( ) = f(b) f(a) FTLI: = ( ) Next, we find a relationship for two-variable functions between an integral over a 2-dimensional graph (a surface) and an integral over a 1-dimensional graph (a line). Amazingly, it s a similar relationship, as long as we properly use the idea of boundary, and carefully define the proper derivative expression the one we want is swirl(f). When the curve C is the boundary of the surface S: S swirl(f) da = C F dr Written out in coordinates, using F=(P,Q) and dr=(dx,dy), this is the same as: = + Note that, in order for C to be the boundary of a surface S, it must be a closed path. Typically the line integral is hard to calculate but the surface integral is easy; sometimes it s the other way. ex) calculate C (x-y)dx+(x+y)dy where C is the circle centered at the origin with radius 2 Method 1: calculate directly as a line integral Parametrize: x=2cos(t) & y=2sin(t) dx = -2sin(t)dt dy = 2cos(t)dt 0<t<2π C (x-y)dx+(x+y)dy = 0 2π (2cos(t)-2sin(t))(-2sin(t))dt + (2cos(t)+2sin(t))2cos(t)dt = 0 2π -4cos(t)sin(t) + 4sin 2 t + 4cos 2 t + 4cos(t)sin(t) dt = 0 2π 4 dt = [4t] 0 2π = 8π Method 2: since C is a closed path we can use Green s Theorem F = (x-y, x+y) C (x-y)dx+(x+y)dy = S swirl(f) da = S 2 da = 2 Area(S) = 8π where we used the fact that swirl(f) = Q x P y = = 2 In that example, both methods will solve the problem. That is not always the case. Ex) calculate C x 2 cosx dx + e y cos y dy where C is the circle radius 1 centered at the origin. If we calculated this as a line integral, we would parametrize: x=cos(t), y=sin(t) and dx=-sin(t)dt,dy=cos(t)dt C x 2 cosx dx + e y cos y dy = 0 2π cos 2 t cos(cos(t))dt + e sin(t)cos(sin(t)) cos(t)dt It is not clear how to do that integral. Fortunately, with Green s Theorem, we don t have to. C x 2 cosx dx + e y cos y dy = S swirl(f)da for F=(x 2 cosx, e y cos y ) swirl(f) = Q x - P y = 0 0 = 0 so the integral = 0 Notice that Green s Theorem tells us that for a conservative field F = f, the line integral over closed loop is 0. Recall that swirl(f) = swirl( f) = 0. Therefore, C F dr = S swirl(f) da = 0

96 96 Application in Physics: Work Ex) find the work done by the force F = (cosy-y,- xsiny) to move a particle once counterclockwise around the rectangle with corners at (2,1) (5,1) (5,7) (2,7) W = C F dr Method 1: calculate directly as a line integral W = C cos(y)-y dx + -xsin(y)dy To parametrize, break up C into 4 line segments. On the first, x=t, y=1, t goes from 2 to 5, dx=dt, dy=0. On the second, x=5, y=t, t goes from 1 to 7, dx=0, dy=dt. On the third, x=t, y=7, t goes from 5 to 2, dx=dt, dy=0. On the fourth, x=2, y=t, t goes from 7 to 1, dx=0, dy=dt. = 2 5 cos(1)-1 dt sin(t)dt cos(7)-7dt sin(t)dt [note that when dx=0 or dy=0 on a given piece, then that piece disappears] = 3cos(1)-3 + 5cos(7)-5cos(1) + 2cos(7)-5cos(7) cos(1)-2cos(7) = 18 Method 2: since C is a closed path, we can use Green s Theorem W = C F dr = S swirl(f) da [where curve C is the boundary of region S] Now calculate: swirl(f) = (-xsiny) x (cosy-y) y = -siny siny = 1 = S 1 da = 1 Area(S) = 18 Note that if swirl(f) is a function (not a constant), it still may be easier to calculate the integral that way. Application in Math: using Green s Theorem to calculate the area ex) find the area enclosed by the ellipse + = 1 Recall that the area of the region S is S 1 da Here, S is the interior of the ellipse. We can convert that to a line integral using Green s Theorem, as long as we can find F=(P,Q) so that swirl(f) = = 1. There are many possibilities, let s use F=(-0.5y,0.5x) [in this problem that choice makes the calculations the easiest]. Area = S 1 da = C -0.5y dx + 0.5x dy where C is the ellipse, the boundary of the enclosed area. The ellipse can be parametrized with x = acos(t), y=bsin(t) for 0 t 2π, and then dx=-a sin(t)dt, dy=b cos(t)dt Area = 0 2π -0.5(bsin t)(-a sin t)dt (a cos t)(b cos t)dt = 0 2π 0.5ab sin 2 t + 0.5ab cos 2 t dt = 0 2π 0.5ab dt = 0.5abt 0 2π = πab What does Green s Theorem mean? Why is it true? Previously, we used the fundamental theorem of calculus, and we saw that the way to prove it was to divide the line from a to b into segments. Then, the sum of the little changes is the same as the total change between the endpoints. So = = ( ) ( ) and in the limit ( ) = ( ) ( ). We want to translate this result to a function whose domain is a 2-dimensional region so the graph is a 2- dimensional area. For simplicity, consider the case where the region is a rectangle. Start with an oriented closed path C which is the boundary of the region. Then divide the region up, for example into four pieces, with each one enclosed by an oriented closed path. Notice that, because of the orientation, the interior path segments cancel, so C equals the sum of the four (oriented) loops. So the line integrals are equal as well, = + + +

97 97 When you divide the region futher and further, you get a sum of many integrals over tiny loops. It turns out that, in the limit, this expression becomes and the integral is over the entire region since the paths now cover the entire region. To understand why the expression is, and for a more complete proof, see the appendix.

98 98 Parametric Surfaces Previously, we described a curve using parametric equations. The formula for the curve C was f = f(t) = f(x(t),y(t),z(t)) = f(r(t)). Note that for a curve in 2 dimensions or 3 dimensions (or any number of dimensions), no matter the number of coordinates, the function could be expressed as a function of one variable, the parameter t. If you zoom in on the graph, you see a straight line. This resembles the real number line, which can be described with one variable, and we have typically used x for that. Now, we will describe surfaces using parametric equations. When we zoom in on a surface, it looks like a plane. This resembles the R 2 plane, which requires two variables to describe it; we have typically used x & y. To describe our surface, we will use the parameters u & v. So the formula for a surface S is (u,v)=(x(u,v),y(u,v),z(u,v)). For the function, the input is a 2-dimensional point and the output is a 3- dimensional point. As a result, the surface is 2-dimensional and it exists in 3 dimensions. Ex) graph the surface r = (2cos(u),2sin(u),v) Previously, for the surface z=f(x,y), traces were lines contained in the surface which represented z as a function of one variable (the other variable was constant) and created a sort of coordinate grid on the surface itself. Now, we make a coordinate grid for parametrized surfaces. We must use our variables u & v. Note that slicing the surface with a plane for a constant x or y does not work, since z is not a function of x & y. For fixed v the height is fixed, and as u varies it traces out a circle of radius 2. For fixed u, a place on the circle is fixed, and as v varies it traces out a vertical line. The graph of the surface was redrawn to include these lines, called grid lines. As before, parametric equations are more general and versatile than a function, and allow for varied graphs.

99 99 Ex) This is a picture of fusilli bucati. It s the shape of a corkscrew, but hollow on the inside. With parametric equations, we can write the formula for this tasty pasta. (u,v) = (u+cosv,(2+sinv)cosu, (2+sinv)sinu+3) In the above fusili graph, notice that as u increases, for constant v, the x-value increases linearly while y and z values oscillate. As v increases, for constant u, y & z will oscillate like a sin curve, while x will oscillate like a cos curve, making loops. The graph of the surface can be redrawn to include these lines. Tangent Plane for Parametric Surfaces Previously, we saw that for the surface z=f(x,y), the tangent plane at the point (x 0,y 0,z 0 ) was z z 0 = f x (x 0,y 0,z 0 ) (x-x 0 ) + f y (x 0,y 0,z 0 ) (y-y 0 ). Now we will write the equation of the tangent plane for a parametric surface r=r(u,v)=(x(u,v),y(u,v),z(u,v)) at the point r(u 0,v 0 )=(x 0,y 0,z 0 ). First, we need two direction vectors in the plane. Once again, we form them by taking the derivative with respect to the two variables; in this case those are u & v. The two tangent vectors are r u = (x u,y u,z u ) & r v = (x v,y v,z v ). We can either take the cross-product to get a normal vector and write the standard equation, or write the parametric equation of the plane. Ex) find the equation of the tangent plane to r = (,,) at the point r(,)=(,,) r u = r v = Method 1 (normal vector): Normal vector n = r u X r v = Tangent plane: (x- x 0,y- y 0,z- z 0 ) (,,) = 0 Method 2 (parametric): (x,y,z) = (x 0,y 0,z 0 ) + m(,,) + n(,,)

100 100 (Note that we already used u & v as parameters for the surface, so we need different parameters for the plane.) Surface Integrals Previously, we saw how to take the integral over a parametrized curve. First we learned how to rewrite in terms of the parameter in order to convert from ds to dt. The basic line integral is first, and the result is the arc length L. Further, this conversion allowed us to calculate any path integral, as in the second equation. Now we use the same idea to calculate the integral over a parametrized surface. First we must learn how to rewrite in terms of the parameters in order to convert from ds to da (or du dv). Doing this will allow us to calculate surface area. Then that conversion will allow us to calculate any surface integral. We will develop the formula as we investigate the first application. Application in Math: Surface Area for Parametric Surfaces Previously we saw that the surface area for a surface z=f(x,y) was Area = ( ) + ( ) + 1 The expression to be integrated came from a X b, the norm of the cross-product of the two tangent vectors. Now, we develop the formula for surface area for parametric surfaces using the same idea. Over each rectangle ΔA with sides Δx and Δy in the domain is a piece of the surface ΔS, and a parallelogram in the tangent plane. For each parallelogram, the tangent vectors are r u Δu =,, and r v Δv =,,, and the area is r u Δu X r v Δv = r u X r v Δu Δv. So the total area of the surface is: Approximately: Area = ΔS r u X r v Δu Δv Exactly: Area = ds = r u X r v du dv OR r u X r v da Ex) find the surface area of r(u,v)=(,,) for u v Now we know the formula for conversion for integrals over parametric surfaces: Application to Physics: Density (per area), mass and center of mass

101 101 Surface Integral in a Vector Field Application in Physics: Electrodynamics Application in Physics: Thermodynamics

102 102 Limits If a function is continuous and well-defined at a point, finding the limit at that point is as easy as plugging the point into the function. That was true for one-variable functions, and it s true for two-variable functions. Ex) find lim (, ) (, ) = = 1.5 In general, however, limits for functions of two variables are much more challenging to calculate than with functions of one variable. Instead of one variable approaching one value, two variables are each approaching a value. To think of this graphically, one variable point is approaching a fixed point. But it is not specified how that approach is made. Perhaps x stays the same and y is changing. Perhaps y stays the same and x is changing. Perhaps they are both changing. It turns out, this choice may affect the limit value. We will explore limits in the same 3 ways as before numerically, graphically, and algebraically. Numerical limits Previously, to find the limit of a function of one variable, we made a list of function values. We need a way to represent function values of two variables. A table does this nicely. Ex) find lim (, ) (, ) ( ) (, )= sin( + ) + Notice that, no matter how the values of x & y approach 0 & 0, the function value approaches 1. Ex) find lim (, ) (, )

103 For this function, it matters very much how the values for x & y approach 0 & 0. If y is fixed at 0 and x approaches 0, the limit is 1. If x is fixed at 0 and y approaches 0, the limit is -1. If x=y and they approach 0 together, the limit is 0. Clearly, the limit does not exist. For one variable functions, we need to take a two-sided limit. For more variables, we need an every-sided limit. In general, formally proving that a limit exists from any direction is hard. Ex) (, )= is undefined at (0,1). Using the table below, find various limits for f(x,y) as (x,y) (0,1) 103 Graphical limits As with one-variable limits, another way to investigate the limit is with the graph of the function. Ex) find lim (, ) (, ) ( ) Based on the graph, it seems that any way to approach (x,y) = (0,0) causes the function to approach 1. Ex) find lim (, ) (, ) Multiple views:

104 104 The graph allows us to see that different approaches by (x,y) to (0,0) produce different function values, ranging anywhere from 1 to -1. It would help if we could identify a path along which all function values are 1 (or another constant). The representation which allows us to do that is the contour plot. Now it is obvious how the path to (0,0) determines the limit. Next we see the contour plot of the earlier function with a well-defined limit at (0,0), f(x,y) = ( ) Note that every path to (0,0) crosses level curves while approaching 1.

105 105 Algebraic limits Ex) show the limit is not well-defined: lim (, ) (, ) if you approach (0,0) along the line x=y, this limit is: if you approach (0,0) along the line x=0, this limit is: In this problem, we knew to choose the paths x=y and x=0 using numerical and graphical information. To find a limit algebraically, there are two options. One, show it is not well-defined by choosing paths; or two, prove that a limit is well-defined for all paths. In general, knowing which to do and also doing it is very hard. The limit definition of the derivative for a function of several variables Previously, for a one-variable y=f(x), we defined the derivaitive using limits = lim ( ) ( ) Now we will form the limit definition for a derivative of f(x,y). Recall how we discovered partial derivatives. We held one variable constant while the other variable changed. Relative to a given point, for the partial derivative we would hold y constant and vary x to approach the given value. We use this idea to modify the limit definition of the derivative from the one-variable case. ( + h, ) (, ) = lim h (, + h) (, ) = lim h These can be rewritten using vector notation. = lim Now let p = (x,y). Recall that i = (1,0), the unit vector in the x-direction. So the definition in vector notation is: ( + h ) ( ) = lim h Geometrically, what does it mean to let h go to 0? From the point p = (x,y), move in the direction indicated by the unit vector i, but head in that direction a smaller and smaller magnitude. Recall that = D i f (since i is the unit vector in the x-direction). This leads to a way to define the directional derivative with limits. If u=(r,s) is a unit vector, then: ( + h ) ( ) = lim h Let s rewrite this without vector notation. = lim So the limit definition of the directional derivative (without vector notation) is: ( + h, + h ) (, ) = lim h Note: if the direction vector is not a unit vector, then divide by its magnitude, as we have seen before. Note: although p & u are combined in the calculation, do not confuse their meaning. The coordinates of our location are represented by p, and the direction in which we find the derivative is represented by u. It may help to think of the location as a point and the direction as a vector. Ex) For f(x,y) = and = (0.6, 0.8), write down and D u f at the point (a,b) as a limit = D u f =

106 106 The Limit Definition of the Integral for a function of several variables The basic form of the limit definition of the integral, if we make some simplifying assumptions, is: R f(x,y) da = lim m,n (x i, y j )ΔA Notation: R is the region over which we integrate. Constants m (and n) are the number of subintervals for the interval of x (and y). x i is the x-value on the i-th subinterval, y j is the y-value on the j-th subinterval, and ΔA is the area of rectangle formed by Δx and Δy. Note that we have made some simplifications from the general case; we chose Δx and Δy to be constant (so ΔA is also constant); we chose x i and y j to be endpoints on the subinterval. In the general case: R f(x,y) da = (x ij *, y ij * )ΔA ij A ij is the rectangle formed by the i-th subinterval of x-values and the j-th subinterval of y-values and ΔA ij is its area; is the maximum value of all ΔA ij ; x ij * (and y ij * ) is the x-value (and y-value) of the point chosen in A ij. In general, calculating the limit directly (without converting it into an integral and using integration techniques) is very hard. Ex) write down R x 2 tan(xy) da as a limit

107 107 Appendix A proof of the Ratio Test

108 108 A proof of the Taylor Remainder Theorem Theorem: Proof:

109 Here is a proof of the curvature formula in 3 dimensions: 109

110 Here is a proof of the second derivative test for the minimum. (The proof for the maximum is similar.) 110

111 111 For a parametric equation, = Proof: Let x=f(t) and y=g(t)

112 112 A proof of the equivalence of the parametric form and the normal vector form of the equation of a plane. (using Mathematica to do some calculations) In particular, a proof that the two ways to derive the equation of the tangent plane at a point on z=f(x,y), using the slopes in the x- and y-directions, are the same:

113 113 Discussion of the chain rule for a function of several variables, each of which is a function of several variables. Let F be a function of two variables, call them x & y, so F = F(x,y). Let x & y each be functions of two variables, call them t & u, so x=x(t,u) and y=y(t,u). the chain rule says: = + = + The following diagram illustrates how to keep track of all the terms. The top row includes F, the function. The middle row includes x & y, called the intermediate variables. The bottom row includes t & u, the independent variables. Note that a change in t causes a change in x change in x causes a change in F, which contributes the term. A change in F comes from changes in both t & u, etc. Each path in the diagram (from independent to intermediate to function) represents a possible change. Note that every connection between independent and intermediate is possible, so there are 2 2=4 terms total (each is a product of two rates). There are 2 paths from F to t, so has two terms. For F(x,y,z), a function with 3 intermediate variables, which has 3 independent variables, the diagram and equations are below., that = + +

114 114 = + + = + + Alternate proofs of the conversion of integrals between rectangular and polar coordinates. Using differentials: Using Riemann sums:

115 Proof that: path independent -> conservative 115

116 116

117 117 Proof of Greens Theorem for a region S which is a rectangle, a x b & c y d For the curve C which is the boundary of S, consider the pieces Bottom, Top, Left, Right so C = B+R+T+L (all with the appropriate orientation) C P dx + Q dy = B P dx + R P dx + T P dx + L P dx + B Q dy + R Q dy + T Q dy + L Q dy = B P dx + T P dx + R Q dy + L Q dy = b a P(x,c)dx + a b P(x,d)dx + d c Q(b,y)dy + c d Q(a,y)dy = b a P(x,c)dx - b a P(x,d)dx + d c Q(b,y)dy - d c Q(a,y)dy = b a P(x,c) - P(x,d) dx + d c Q(b,y) - Q(a,y) dy = b a [ c d P y (x,y)dy ] dx + d c [ b a Q x (x,y) dx] dy = b a [ d c -P y (x,y)dy ] dx + d c [ b a Q x (x,y) dx] dy = b a d c Q x - P y dy dx = S Q x - P y da To understand why the expression is, consider C P dx and C Q dy separately, and assume the region is a rectangle with a x b & c y d. For C Q dy, on the horizontal edges B&T, dy=0 so the integral is 0. On the right edge R where x=b, the line integral goes from y=c to y=d, making a positive contribution. On the left edge L where x=a, the line integral goes from y=d to y=c, and switching the boundary values means the integral makes a negative contribution. The plugged-in endpoints are x=a & x=b, so this comes from integrating from x=a to x=b. For C P dx, the argument is similar, with one extra step. On B where y=c, the path orientation goes from x=a to x=b, so that will make a positive contribution, and on T where y=d, the path orientation goes from x=b to x=a, so when we switch the boundaries of integration it makes a negative contribution. Thus, it is the equivalent of integrating boundary values (for y), so we must change the sign. from y=d to y=c. Then, we need to switch those Proof of FTC using discrete approximation and canceling Proof of Greens Theorem using discrete approximation and canceling

118 Now, 118