Math 153 Calculus III Notes

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1 Math 153 Calculus III Notes 10.1 Parametric Functions A parametric function is a where x and y are described by a function in terms of the parameter t: Example 1 (x, y) = {x(t), y(t)}, or x = f(t); y = g(t) Describe the graph given by x = 2 & 2 t, y = 1 2 t2 & 3, for 0 t 4. (Create a table of points.) Also, set your calculator to Parametric mode; enter the functions, and set the appropriate Window. Mathematica Command : ParametricPlot 2 # 2 t, 1 2 t2 # 3, {t, 0, 4}, PlotRange ' {{#6, 2}, {#3, 5}} 4 2 &6 &4 &2 2 &2 Example 2 Find a parametrization of the line y = 2 x + 4 without using fractions. 3 Example 3 t = 0. Find a parametrization for the line in example (2) so that the line passes through the point (3, 6) when Copyright 2015 CStevens

2 2 Calculus III Notes Example 4 Find a parametrization for the path of a particle moving along the parabola f(x) = 9 & x 2 so that it moves from the right x-intercept to the left x-intercept for 0 t 1. Example 5 Find a parametrization for the path of a particle moving from (0, 0) to (4, 0), then along a circular arc from (4, 0) back to (0, 0) for 0 t 3. Example 6 Eliminate the parameter to write the function in rectangular coordinates: 2 cos(t) + 1, 4 cos 2 (t) + 2 and describe the function for all t 0. Example 7 Graph f(t) = {cos(3.5 t), sin(2.8 t)} on the interval [0, 12]

3 Calculus III Notes Calculus With Parametric Functions Tangent Lines If x = f(t) and y = g(t), then 0y 0x = Example 1 Find the equation of the line tangent to the curve x = t 2 & 5 t, y = ln(t) + 1 when t = 1 = 0y 0t 0x 0t g' (t) f' (t) Example 2 Note: 02 y = 0 0y 0x 2 0x 0x = 0 0t Find the concavity, or 02 y, of the above function at t = 2 (and possibly the inflection points). 2 0x 0t 0x 0 0y = 0y 0t 0x 0x 0x 0t Areas To calculate the area under the function y = F(x) where F is parameterized F = {x = f(t), y = g(t)} as t transverses from 3 to 4, we can adapt the definite integral as such, b b 4 F(x) 0x = y 0x = g(t) f ' (t) 0t Example 3 Find the are under the curve F = 6 t & t, 4 6 t72 & 4, for 0 t a a 3 (Can you set up the integral for the area using an explicit function?)

4 4 Calculus III Notes Example 4 A cow is tied to a circular silo with just enough rope to reach the other side. Find the area of grass the cow has to graze. (The figure only shows half of the total area.) Arc Length b Recall the arc-length formula L = a 1 + 0y 0x 2 L = a b 0x. We can rewrite this as y 0x 2 0x = y 0t 0x 0t 2 0x 0t 0t If 0x > 0 we have 0t 4 L = y 0t 0x 0t 4 2 0x 0t = 0t 3 0x 0t 2 + 0y 0t 2 0t Example 5 Find the length of the spiral t cos(t), t sin(t) for 0 t 8 9. You may need an integration table to evaluate the integral. Also, estimate the length using four circles each with an average radius &1 & &1.0 &1.5 &2.0 Surface Area Surface area can be derived in a similar way and results in Example 6 0 t 9 73 about the x-axis. 4 S = 2 9 y 0x 0t 2 + 0y 0t t = 2 9 y 0s 3 Set up the integral to calculate the surface area obtained by revolving x = sin 2 (t), y = sin(3 t) for

5 Calculus III Notes Polar Coordinates Polar Function r = f(:) Converting Between Polar and Rectangular Coordinates Example 1 x = r cos(:) x 2 + y 2 = r 2 y = r sin(:) tan(:) = y x (a) Convert the rectangular point R(5, &2) to polar coordinates (r, :). (b) Convert the polar point P(2, 310 ) to rectangular. Example 2 Show that r = tan(:) sec(:) is the equation of a parabola. Types of Polar Curves Circles: r = a; r = a sin(:) ; r = a cos(:); r = a cos(:) + b sin(:) Cardioid: r = a ± b sin(:), a = b limaçon: r = a ± b sin(:), a > b limaçon with a loop: a < b rose: r = a sin(b :) or r = a cos(b :); If b is even> 2 b petals; If b is odd> b petals Example 3 Plot the polar function r = sin(:) Show

6 6 Calculus III Notes Example 4 Plot the rose r = 5 cos(2 :), and find the slope of the tangent line when : = (Find the equation of the tangent line in polar form.) Show Derivatives of Polar Functions The slope, 0y 70x, of polar function r = f(:) can be found by using the parametric form {x = r cos(:), y = r sin(:)} 0y = D :[r sin(:)] 0x D : [r cos(:)] = D :[f(:) sin(:)] = = D : [f(:) cos(:)] f' (:) sin(:)+f(:) cos(:) f' (:) cos(:)&f(:) sin(:) r' sin(:)+r cos(:) r' cos(:)&r sin(:) Example 5 Find the slope of the line tangent to r = 3 & 2 cos(:) when : = 9 3.

7 Calculus III Notes Areas and Lengths in Polar Coordinates Areas of Polar Functions The area of a circle A = 9 r 2 The area of a sector of angle! A = : r2 A = : 2 r2 and If r = f(#) and # = Δ#, then A i = Δ: 2 f(: i) 2 A = 1 2 : 1 : 2f(:) 2 0: Example 1 Find the area of the rose r = 4 cos (3:) Example 2 Find the area outside the cardioid r = 1 + cos(:) and inside the circle r = 3 sin(:).

8 8 Calculus III Notes Example 3 Find the area outside the cardioid r = cos(:) and inside the limaçon r = sin(:). Arc Length of Polar Curves b L = a r 2 + 0r 0: 2 0: Example 4 Find the length of the limaçon r = sin(:).

9 Calculus III Notes Sequences and Series A sequence is a list of numbers, i.e, {1, 3, 5, 7, 9, }, or 1, &1 2, 1 4, &1 8, 1 16,... Example 1 Find the first four terms of the infinite sequence: n+1 n! n=1 Example 2 Find the nth term of the sequence: {&7, &3, 1, 5, } Example 3 Find the nth term in the sequence: 2, &2, 4, &2, 4, &4, 8, Recurrence relation: a n+1 = f(a n ) (Not an explicit formula for each a n ) Example 4 Find the first 5 terms of the sequence defined by: a n+1 = 2 a n & 3 if a 1 = 2, and find an explicit form for a n.

10 10 Calculus III Notes Convergence of a Sequence A sequence {a n } has the limit L, written lim a n = L or a n C L as n C nc if we can make the terms a n as close to L by taking n sufficiently large. If the limit exists we say the sequence converges, otherwise the sequence diverges. Example 5 For the sequence 1 2 n find N such that a n < for all n > N. Example 6 Find the limit of the sequence a n = 2 n 3 n 2 +1 Example 7 Determine the limit of the sequence (&1)n n n+1. Monotonic Sequences Sequences are monotonic if a n < a n+1 for all n 1, or b n > b n+1 for all n 1. Example 8 Show that the sequence a n = n+1 is decreasing 2 n Bounded Sequences A sequence is bounded above if there exists an M such that a n M for all n 1, or bounded below if there exists an m such that m a n for all n 1. Every bounded monotonic sequence is convergent. Why?

11 Calculus III Notes Series The sum of an infinite sequence {a n } is called an infinite series, or just a series. That is a series is ai = a 1 + a 2 + a 3 i=1 Example 1 Find i = i=1 Example 2 1 Find = i=1 2 i To determine whether a series has a value, or limit, we can use the idea of limits of partial sums. Given a sequence {a n } or the series i=1 a n, a partial sum, s n, is denoted as n s n = ai = a 1 + a 2 + a a n i=1 If the sequence of partial sums {s n } is convergent, and lim nc s n = s exists as a real number, then the series a n is called convergent, and the sum s is called the sum of the series. Otherwise, the series is called divergent. The Geometric Series The geometric series is a series that can be written as Example 3 a r i&1 = a + a r + a r 2 + a r a r n&1 + ; a 0 i=1 Show that the geometric series converges to a when r < 1, and divergent if r 1. 1&r Example 4 Consider the Sequence a n = 4 (0.95) n&1. a) Find s 8 b) Find s = s

12 12 Calculus III Notes Example 5 Find the value of the series & & & Example 6 4 k&1 Evaluate k=1 5 k+1 Example 7 1 Show that the Harmonic Series: s = i=1 i is divergent. Example 8 1 Find the value of the telescoping series: k=1 k 2 +k Test for Divergence If lim a n does not exist, or if lim a n 0, then the series an is divergent. nc nc Example 9 2 i+3 Show the series diverges. i=1 5 i n=1

13 Calculus III Notes The Integral Test and Estimates of Sums Properties of Convergent Series 1) Let A = a k, then c a k = c a k = c A. 2) Let A = a k and B = b k, then (a k ± b k ) = a k ± b k = A + B 3) Let k=m a k be a convergent series, then k=n a k converges. That is, the addition or subtraction of a finite number of terms does not affect the convergence of a series. Divergence Test If a series a k is convergent then lim kc a k = 0. Therefore, if lim kc 0 the series diverges. Example 1 Determine the convergence of k=1 2 k k 3 k The Harmonic Series 1 Recall = k=1 k > > > 1 2 is a divergent series even though lim kc 1 k = 0 Another way to show it diverges is to compare the series to the area under the function f(x) = 1 x from x = 1 to x = k 1 x &x= k= The Integral Test Suppose f is a continuous, positive, decreasing function for x 1 and let a n = f(n) for n = 1, 2, 3. Then, ak and f(x) 0x either both converge or both diverge. In the case of convergence, the value of the integral is not usually equal 1 to the value of the series. k=1 Example 2 Determine the convergence of 5 k=1 k 3

14 14 Calculus III Notes The p-series 1 The p-series converges when p > 1 and diverges when p 1. p k=1 k Example 3 5 k 3 Determine the convergence of k=4 k 9+1 Estimating a Series with Positive Terms Let f be a continuous, positive, decreasing function for x 1 and let a n = f(n) for n = 1, 2, 3, 4. Also, let s = convergent and let s n = n k=1 a k. Define the remainder R n = s & s n, then R n is bounded as follows: f(x) 0x Rn f(x) 0x n+1 n an be n=1 Example 4 4 How many terms of the series are needed to approximate the series to within ? 3 k=1 k N k=1 k, The actual value... 4 N k=1 k,

15 Calculus III Notes Comparison Test and the Limit Comparison Test The Comparison Test Let a k and b k be series with positive terms. 1. If 0 < a k b k and b k converges, then a k converges. 2. If 0 < b k a k and b k diverges, then a k diverges. Example 1 Determine the convergence of k=2 k k 3 &1 The Limit Comparison Test a Let a k and b k be series with positive terms and let lim k = L. kc b k 1. If 0 < L < (that is, L is a finite positive number), then a k and b k either both converge or both diverge. 2. If L = 0 and b k converges, then a k converges. 3. If L = and b k diverges, then a k diverges. Example 2 3 k Determine the convergence of 3 &5 k 2 +1 k=1 2 k 5 +2 k 3 &15

16 16 Calculus III Notes Estimating Sums When using the comparison test, if a k b k,and if b k is a convergent p-series or geometric series, we can estimate remainder R n using the n th partial sum of a k with the remainder T n from the n th partial sum of b k. If a k b k then s n = n k=1 a k n k=1 b n = t n, the remainders are R n T n where R n = s & s n and T n = t & t n. Example 3 2 Estimate the value of the series k+1 &k using the first 20 terms. k=1 3 k Remember, for a geometric series t = a and t 1&r n = a 1&rn. 1&r

17 Calculus III Notes Alternating Series Recall the harmonic series: = 1 diverges k=1 k Now consider the alternating harmonic series: 1 & & 1 + = (&1) k& k=1 k (#1) k#1 k=1 k Log[2] Alternating Series Test The alternating series (&1) k a k converges provided 1. the terms of the series are non-increasing in magnitude, or a k a k+1 for all k 2. lim kc a k = 0 Example 1 Determine the convergence of k=2 (&1) k ln(k). k 2 Example 2 Does the series in example 1 converge if it is not alternating? Log[k] k=2 k 2 $ 1 6 π2 EulerGamma + Log[2] $ 12 Log[Glaisher] + Log[π] % // N

18 18 Calculus III Notes Absolute Convergence Consider an convergent alternating sequence (&1) k a k. If the corresponding non-alternating sequence a k also converges we say the series is absolutely convergent. When the corresponding non-alternating sequence diverges we say the original sequence converges conditionally. Example 3 Determine if the series converges absolutely: k=1 (&1) k&1 k+1 k+2 Error or Remainder for a Partial Sum of an Alternating Series Let S = k=1 (&1) k&1 a k and S n = n k=1 (&1) k&1 a k. The remainder (or error) R n = S & S n in estimating S with S n is such that R n a n+1. That is, the error is less than the magnitude of the next term in the sequence. Example 4 Find the error in estimating S = k=1 estimate S to within 0.001? (&1) k&1 1 k 3 with the first 5 terms. How many terms are needed to Recall for a non-alternating convergent series the error or remainder R n in estimating S with S n is R n n a x 0x. How many terms would be needed to estimate k=1 1 k 3 to within

19 Calculus III Notes The Ratio and Root Test The Ratio Test a Let a k be an infinite series with positive terms and let r = lim k+1 kc a k. 1. If r < 1, the series converges. 2. If r > 1, the series diverges. 3. If r = 1, the test is inconclusive. Example 1 Determine the convergence k 2 k k! k=1 Example 2 Determine whether each series converges or diverges. n a. 2 2 n+1 n=0 3 n b. n=1 n n n! Example 3 Determine the convergence of (&1) n n=1 n n+1.

20 20 Calculus III Notes The Root Test Let a k be an infinite series and let I = lim kc 1. If I < 1, the series converges absolutely. 2. If I > 1, the series diverges. 3. If I = 1, the test is inconclusive. e Example 4 Determine the convergence or divergence of 2 k k=1 k. k k a k Example 5 k Determine the convergence of 3 +2 k 3 k k=2 3 &1 k

21 Calculus III Notes Strategy for Testing Series In general, there is no set method or set list of tests to apply until one works. It s much more efficient to know what the form of the series is in order to test for convergence. Here are some generalities to keep in mind when testing a series: 1. If lim kc a k 0 the series diverges. 2. If the terms have a common ratio, or of the form a r k&1 it s a geometric series and converges if r < 1 and diverges if r If the series is of the form 1 k p and converges if p > 1 and diverges if p If the series looks similar to a geometric or p-series, a comparison test should be considered. The comparison tests are only good for series with positive terms, so a test for absolute convergence may be necessary. 5. If a series can be written in the form (a k & a k+1 ) it may be a telescoping series. Find lim nc S n. 6. If a series alternates, use the alternating series test; (a k C 0). Also can test for Absolute Convergence. 7. If a series contains factorials the Ratio Test is a good options. Note the ratio test does not work for geometric or p-series. 8. When a series is of the form (a k ) k the Root test may be helpful. 9. If a k = f(x) for each x = k, and f(x) is integrable, the Integral Test may be useful. Summary of Series Tests Test Series Converges when Diverges when Note nth ' term a k lim k( a k 0 Cannot show convergence Geometric Series a r k'1 r < 1 r > 1 sum : S = a 1'r p'series 1 k p p > 1 p 1 Integral Test a k ; and a k = f(k) 0 f(x) & x converges 1 Direct Comparison (a n, b n > 0) Limit Comparison (a n, b n > 0) a k a k 0 < a k b k, and b k converges a lim k k( = L > 0, and b k b k converges Alternating Series ('1) k'1 a k 0 < a k+1 a k and lim k( a k = 0 f(x) &x diverges 1 0 < b k a k, and b k diverges a lim k k( = L > 0, and b k b k diverges Ratio Test a k lim n( a k+1 a k < 1 lim n( a k+1 a k > 1 Root Test (a k ) k k lim n( a k k < 1 lim n( a k > 1 R n = f(x) &x n R n a n+1 Inconclusive if lim n( a k+1 a k = 1 Inconclusive if lim n( ( a k )^(11k) = 1

22 22 Calculus III Notes Section 11.7 Exercises (from your text) Answers to odd problems are in the back of your text.

23 Calculus III Notes Power Series A Power Series is a series of the form cn x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + n=0 Example 1 For what values of x does the series x n converge? n=0 A more general power series centered at a is given by Example 2 cn (x & a) n = c 0 + c 1 (x & a) + c 2 (x & a) 2 + c 3 (x & a) 3 + n=0 3 Find the values for which the power series (x & n=0 n 2)n converges. Example 3 For what values does the series n! (x & 2) n converge? n=0 Example 4 x n Find the values for which the series converge. n=0 n!

24 24 Calculus III Notes Theorem For a given power series cn (x & a) n there are only three possibilities: n=0 1. The series converges only at x = a. 2. The series converges for all x. 3. The series converges on a set of values such that x & a < R and diverges when x & a R. The value R is called the radius of convergence with the center at a. Example 5 x Find the interval of convergence for 3 & n=o 2 4 n. Example 6 Find the interval of convergence for (&1) n x n n=0 n!

25 Calculus III Notes Representing Functions as Power Series Example 1 Use the sum of a geometric series to find the function represented by x n = 1 + x + x 2 + x 3 +, and give n=0 its interval of convergence. Example 2 Find a power series for f(x) = x3 1&x N Example 3 Find the power series for f(x) = 2 x 1+x 2 Integration and Differentiation of Power Series Theorem If the power series c n (x & a) n has a radius of convergence R > 0 then the function f defined by f(x) = c 0 + c 1 (x & a) + c 2 (x & a) 2 + = cn (x & a) n n=0 is differentiable (and therefore continuous) on the interval (a & R, a + R) and (i) (ii) 0 0 x f(x) = f ' (x) = n cn (x & a) n&1 f(x) 0x = n=0 n=1 cn (x&1) n+1 n+1 + C The radii of convergence for (i) and (ii) are both R.

26 26 Calculus III Notes Example 4 Find the derivative of the function in Example 2. Example 5 Find a power series for ln(x + 1). Example 6 Find a power series for tan &1 (x). Example Use a power series to evaluate 0x to x 0 Mathematica Command Mathematica will calculate a partial sum of a power series of degree n centered at x = 0 as follows: Series[Cos[x], {x, 0, 10}] 1 $ x2 2 + x4 24 $ x x $ x O[x]11 To remover the other terms, O[x] 11, use the Normal command Plot[Evaluate@Normal[Series[Cos[x], {x, 0, 14}]], {x, #8, 8}, PlotRange ' {#2, 2}] 2 1 &5 5 &1

27 Calculus III Notes Taylor and Maclaurin Series Theorem If f has a power series expansion centered at a with radius of convergence R, that is then the coefficients are given by the formula The Taylor Series of the function f at a is f(x) = cn (x & a) n x & a < R n=0 c n = f(n) (a). n! f f(x) = (n) (a) (x & a) n f'' (a) = f(a) + f ' (a) (x & a) + (x & a) 2 f''' (a) + (x & a) 3 + f(4) (a) (x & a) 4 + n=0 ni 2! 3! 4! f Note: if a = 0 the series is called a Maclaurin Series: f(x) = (n) (0) x n f'' (0) = f(0) + f ' (0) x + x 2 f''' (0) + x 3 + f(4) (0) x 4 + n=0 ni 2! 3! 4! However, does this mean that every function can be represented by a power series? Theorem: Taylor s Formula with Remainder Let f be a function whose (n + 1)st derivative f (n+1) (x) exists for all x in an open interval I containing a. Then, for each x in I, f(x) = f(a) + f ' (a) (x & a) + the formula f'' (a) 2! (x & a) 2 + and c is some number between a and x. f''' (a) 3! (x & a) f(n) (a) n! R n (x) = f(n+1) (c) (n+1)! (x & a) 4 + R n (x) where the remainder (or error) is given by (x & a) n+1 Theorem: Convergence of a Taylor Series If a function f has derivatives of all orders in an open I = (a & R, a + R), and if lim nc R n (x) = 0 for all x for all x in I, then for all x in I. N Example 1 Find the Maclaurin series for f(x) = 6 x. f f(x) = (n) (a) (x & a) n n=0 n!

28 28 Calculus III Notes N Example 2 Find the power series for f(x) = sin(x) centered at x = 0. N Example 3 Differentiate the Power series for sin(x) to find the series for cos(x). Example 4 Find the Taylor polynomial for f(x) = 2 x 2 & 4 x + 6 centered at x = 3. Example 5 Find the power series for sinh(x) centered at x = 0. Example 6 Find the Taylor series for f(x) = x + 1 center at x = 3 Example 7 Find the first four terms in the Maclaurin series for e &x ln(x + 1) by multiplying the two series together. Example 8 1 Evaluate x 2 6 &x2 0x to 5 decimal places. 0

29 Calculus III Notes The Binomial Theorem Recall Pascal s Triangle and how it can be used to find the coefficients of a binomial raised to a positive integer: The rows are numbered k = 0, k = 1, k = 2, and the elements in each row are numbered n = 0, n = 1, n = 2. The first 56 in the bottom row corresponds to k = 8 and n = 3. Also, the polynomial (a + b) 8 expanded is a a 7 b + 28 a 6 b a 5 b a 4 b a 3 b a 2 b a b 7 + b 8 The coefficients can be represented in binomial form k n, i.e., 8 = 56. The binomial coefficient can be written and 3 calculated using factorials: k n = k! (k&n)! n! ; therefore 8 3 = 8! 5! 3! = = 56 The binomial coefficient can also be written in an expanded simplified form as: We can now write the binomial theorem in the form: If we let a = 1 and b = x, then we have: k k(k&1) (k&2) (k&3) P(k&n+1) = n n! k (a + b) k = n=0 (1 + x) k = k n ak&n b n k n=0 k n xn Newton expanded the binomial theorem so that k was no longer a positive integer, which turns the sum into an infinite series. Example 1 Find the Maclaurin series for f(x) = (1 + x) k. Copyright 2015 CStevens

30 30 Calculus III Notes The Binomial Series If k is any real number and x < 1, then (1 + x) k = 1 + k x + k(k&1) x 2 + k = n=0 n xn 2! k(k&1) (k&2) 3! x 3 + P where k k(k&1) (k&2) (k&3) P(k&n+1) = ; converges when n n! x < 1; diverges when x > 1; converges at x = 1 if &1 < k 0, and converges at both ±1 when k 1. Example 2 Expand the power series for f(x) = 1 (1+x) 3 Example 3 Find a Maclaurin series for x 3 25&x using a binomial series.

31 Calculus III Notes Vectors A vector is a quantity that has two characteristics: (1) magnitude, (2) direction, and is often represented as an arrow. Two vectors are equal if they have the same direction and magnitude. Example 1 form, A vector with initial point P 1 (3, 2) and terminal point P 2 (&4, 7) is written in component form, or standard Magnitude of a Vector: Norm of v P 2 & P 1 = (&4, 7) & (3, 2) = (&7, 5) v = &7, 5 The length, or magnitude, of a vector v in S 2 or S 3 is denoted as TvU or v, TvU = v v 2 2 TvU = v v v 3 2 Direction Angle of a vector tan(:) = v 2 v 1 Example 2 Find the magnitude and direction of the vector: &20, 50. Example 3 Find a vector with direction angle : = 65 and magnitude 200. Scalar Multiplication The vector v = v 1, v 2 can be multiplied by the scalar c c v = c v 1, v 2 = c v 1, c v 2 Unit Vector: a vector, u, of length 1 in the direction v is given by u = v v Example 4 Find a unit vector in the direction 3, &8. Example 5 Find a unit vector with direction : = 225.

32 32 Calculus III Notes The vectors i and j, and k Two special unit vectors in S 2 : i = 1, 0 and j = 0, 1, or in S 3 : i = 1, 0, 0, j = 0, 1, 0, and k = 0, 0, 1. Example 6 If u = 2 i & 3 j and v = &5 i + 7 j, find w = 4 u & 6 v. Example 7 Find a force vector with magnitude 50 Newtons in the direction 6, 1. Addition of Vectors (the Resultant) Let u = u 1, u 2 and v = v 1, v 2, then N u + v = u 1 + v 1, u 2 + v 2 Example 8 Two ropes are tied to weight. The first rope is parallel to the x-axis and pulled with a force of 60-pounds, and the second is at an angle of 35 to the first with a force of 40-pounds. Find the direction of motion of the box and the total force acting on the box. Example 9 An airplane needs to maintain a course with bearing N63 E and a speed relative to the ground of 200 mph. It s flying in a wind of 60 mph with a direction of S20 E. Find the speed and direction of the plane relative to the wind in order for the desired path. Example 10 Two ropes attached to a ceiling are suspending a 250 pound box. The rope on the left has an angle of depression 70, and the rope on the right has an angle of depression of 25. Find the tension on each rope.

33 Calculus III Notes The Dot Product The Dot Product The dot product (or scalar product) of two vectors a = a 1, a 2, a 3 and b = b 1, b 2, b 3 is given by a b = a 1 b 1 + a 2 b 2 + a 3 b 3 Example 1 Find the dot product of the vectors 2, &3, &4 and 8, 3, &2. Properties of the Dot Product If a, b, and c are vectors and c is a scalar, then 1. a a = TaU 2 2. a b = b a 3. a (b + c) = a b + a c 4. (c a) b = c(a b) 5. 0 a = 0 Geometry of the Dot Product If : is the angle between vectors a and b, then: a b = TaU TbU cos(:) Therefore, the angle between two vectors a and b can be found: cos(:) = Example 2 a b TaU TbU Draw the vectors a = 4 i + 6 j +2 k and b = &3 i +4 j + 6 k and find the angle separating the vectors. Example 3 vector. Find c such that the vector 4, &2, c is orthogonal to the vector 3, 1, 2, and write each vector as a unit

34 34 Calculus III Notes Direction Angles Let 3, 4, and V be the angles between a vector a = a 1, a 2, a 3 and the unit vectors i, j, and k. Find these angles. Vector Projections Scalar projection: Vector projection: The scalar projection of u onto v is: The vector projection u onto v is: comp v u = u v TvU proj v u = u v TvU 2 v Example 4 Find the scalar and vector projection of u = 3, 6 onto v = 8, 2, and draw a picture of the projections. Work: w = F D Example 6 Calculate the work done in moving a point mass from (2, 1) to (10, 3), in meters, with a 20-Newton force in the direction 1, 3.

35 Calculus III Notes The Cross Product The cross product of two vectors u and v in S 3 is given by u v = u 2 v 3 & u 3 v 2, u 3 v 1 & u 1 v 3, u 1 v 2 & v 1 u 2 To see where this comes from, we need to look at the determinant of a matrix. The Determinant of a Matrix Example 1 Find the determinant of the matrix: Example 2 Find the determinant of the matrix: & &4 Remember: + & + & + & + & + The Cross Product Using the Determinant u v = i j k u 1 u 2 u 3 v 1 v 2 v 3 Example 3 Find the cross product of the vectors u = 2 i &3 j +k and v = 3 i + 4 j & 2 k. Example 4 Show that the vector u v in Ex. 3 is orthogonal to both u and v.

36 36 Calculus III Notes Theorem If : is the angle between u and v then: Tu v T = TuU TvU sin(:) (0 : 9) Example 5 Find the area of the parallelogram defined by the vectors u = 8 i + 3 j +k and v = 2 i + 6 j & 2 k. Properties of Vector Multiplication 1. u v = &v u 2. (c u) v = c(u v) = u (c v) 3. u (v + w) = u v + u w 4. (u + v) w = u w + v w 5. u (v w) = (u v) w 6. u (v w) = (u w) v & (u v) w Property 5 is called the scalar triple product and can be calculated as a determinant: u (v w) = u 1 u 2 u 3 v 1 v 2 v 3 w 1 w 2 w 3 Show that absolute value of the scalar triple product is the volume of a parallelepiped with edges defined by the vectors u, v, and w. Example 6 Find the volume of the parallelepiped with adjacent edges defined by the points (2, &1, 3), (4, 0, 2), (&2, 3, 1), and (3, 2, &2). Torque Torque is a measure of the tendency of a body to rotate about the origin, e.g., turning a bolt with a wrench. The torque vector is defined as $ = r F and the magnitude of the torque is T$U =U r FU = TrU TFU sin(:). Example 7 A torque of at least 10 joules is required to tighten a bolt. If a 25 centimeter length wrench has a force of 50 newton applied to it that makes an angle of 70 to the wrench, will this provide enough torque to tighten the bolt?

37 Calculus III Notes Equations of Lines and Planes Vector Equation of a Line The equation of a line through the point (x 0, y 0, z 0 ) in the direction of the vector v = a, b, c is given by r = r 0 + t v where r 0 is the initial position vector r 0 = x 0, y 0, z 0, and r is the position vector for any point on the line. Example 1 A line contains the point (4, 3, 5) and direction vector v = 1, 2, &2. Write the equation of the line; find three other points on the line; plot the points and line. Parametric Form of a Line in S 3 Since, (x, y, z) = x 0, y 0, z 0 + t a, b, c, the parametric form for the equation of a line is: x = x 0 + a t y = y 0 + b t z = z 0 + c t Symmetric Form of a line in S 3 Solving the above equations for t we get the symmetric form of the equation of a line: x&x o a = y&y o b = z&z o c Example 2 Find the parametric and symmetric form for the line passing through the points (4, &2, 1) and (2, 8, &3), and find the point where the line intersects the x-z-plane. Example 3 Find the distance between the line r = 3, 2, 8 + t &2, 1, 3 and the point (4, &2, &4). Example 4 Find the distance between the lines. (First show that they are skew and non-intersecting.) L 1 : x = 2 t & 3 y = &3 t + 4 z = t + 2 L 2 : x = 3 s & 1 y = 4 s + 1 z = 2 s & 5

38 38 Calculus III Notes Equations of Planes The equation of a plane in S 3 can be written in linear from as: a x + b y + c z = d, (e.g., 2 x + 3 y + 4 z = 12). This equation can also be written as a(x & x 0 ) + b(y & y 0 ) + c(z & z 0 ) = 0 for any point (x 0, y 0, z 0 ) on the plane. This is called the scalar equation of a plane. Example 5 Verify the point (&2, 4, 1) is on the plane above and rewrite the equation of the plane using this point. If we let n = a, b, c, r 0 = x 0, y 0, z 0, and r = x, y, z, the equation of a plane can be written as n (r & r 0 ) = 0 or n r = n r 0 N These are called the vector equations of the plane. Also, the vector n = a, b, c is the normal vector of the plane which is a vector orthogonal to the plane. Example 6 Find the equation of the plane containing the line x&3 = y+2 = z&4 and the point (&2, 1, 3). 4 &3 2 Example 7 Find the distance from the 3 x + 5 y & 2 z = 8 and the point (2, 2, 5). Example 8 Find the angle between the planes 2 x + 5 y + 3 z = 8 and 4 x & y + z = 2

39 Calculus III Notes Cylinders and Quadric Surfaces A cylinder is a surface that consists of all lines (called rulings) that are parallel to a given line and pass through a given plane curve. An example is the graph if z = x 3 & 2 x. Notice that the variable y is missing from the equation of the surface. This is typical of a cylinder; any one if the variables, x, y, or z can be missing. N Example 1 Graph and describe the cylinder x = y 2 & 3 Quadric Surfaces A quadric surface is a surface that is at most quadratic in x, y, and z. That is, in general form: A x 2 + B y 2 + C z 2 + D x y + E x z + F y z + G x + H y + I z + J = 0 N These surfaces are analogous to the conic sections in two dimensions. Using translations and rotations these can be rewritten into one of two standard forms: A x 2 + B y 2 + C z 2 + J = 0 or A x 2 + B y 2 + I z = 0 The best way to visualize or graph the surfaces is to look at the traces, i.e., x = k, y = k, and/or z = k, for various values of k. Example 2 Describe the traces and sketch the quadric surface x z 2 = y; an elliptic paraboloid.

40 40 Calculus III Notes N Example 3 Use traces to sketch the quadric surface 2 x y 2 & z 2 = 8; a hyperboloid of one sheet. N Example 4 Sketch the quadric surface & x2 4 & y2 9 + z2 = 1; a hyperboloid of two sheets. N Example 5 Sketch the quadric surface x 2 2 & y2 = z; a saddle. 4

41 Calculus III Notes Cylindrical and Spherical Coordinates Cylindrical Coordinates Cylindrical coordinates are written (r, :, z), and the conversion to rectangular coordinates (x, y, z) is x = r cos(:) y = r sin(:) z = z Converting from rectangular to cylindrical use x 2 + y 2 = r 2 tan(:) = y x z = z z (r, θ, z) z x : r y Example 1 Plot the point with cylindrical coordinates 4, 5 9 3, 6 and write the point in rectangular coordinates. Example 2 Describe the cylindrical constant equations: r = k, : = k, z = k. Example 3 Describe the graph of z = r, and rewrite it in rectangular coordinates.

42 42 Calculus III Notes Spherical Coordinates Spherical coordinates are written (I, :, X), and the conversion to rectangular coordinates (x, y, z) is x = I sin(x) cos(:) y = I sin(x) sin(:) z = I cos(x) Converting from rectangular to cylindrical use I 2 = x 2 + y 2 + z 2 z X (ρ, θ, ϕ) ρ x : r y Example 4 Find the rectangular coordinates for the spherical point 4, 9 3, Example 5 Identify the surface given by I sin(x) = 2

43 Calculus III Notes Vector-Valued Functions A vector-valued function is a function of t resulting in a vector r where each component in r is a function of t. That is, r(t) = f(t), g(t), h(t) = f(t) i + g(t) j + h(t) k The Limit of a Vector-Valued Function If r(t) = f(t), g(t), h(t), then provided the limits of the component functions exist. lim r(t) = lim f(t), lim g(t), lim h(t) tca tca tca tca Example 1 Find the domain of the function r(t), and lim r(t) for the function r(t) = 3 sin(t), ln(t + 1),. tc0 t&3 t The graph of the vector-valued function r(t) = f(t) i + g(t) j + h(t) k is a space curve with parametric equations {x = f(t), y = g(t), z = h(t)}. N Example 2 Sketch the function (space curve) r(t) = 2 cos(t) i + 2 sin(t) j + t k.

44 44 Calculus III Notes N Example 3 Find the equation of the curve of intersection of the surface x = y 2 and the plane 2 x + 4 y + 3 z = 3. N Example 4 Describe the curve given by r(t) = 4 cos(t) i + 4 sin(t) j + 3 cos(7 t) k. N Example 5 Sketch the space curve r(t) = t i + t 2 j + t k.

45 Calculus III Notes Derivatives and Integrals of Vector Functions Differentiation The derivative of a vector function is: r ' (t) = f ' (t) i + g' (t) j + h' (t) k. Example 1 Find r ' (t) for r(t) = sin &1 (t) i + 6 &2 t j + 1 & t k. Example 2 (a) Graph the function r(t) = (3 & t) i + t j. (b) Find r ' (t) and plot r(1) and r ' (1). Example 3 Find the unit tangent vector T(t) = r' (t) Tr' (t)u for r(t) = 62 t i + sin(t) j + t k for t = 0.

46 46 Calculus III Notes Example 4 Find the tangent line to the function r(t) = t i + t 2 j + t 3 k when t = 2. Example 5 A function is smooth if r ' (t) 0 for all t. Where is the function r(t) = 2 + t 2 i + t 3 j not smooth? Differentiation Rules - Page 859 Integration b b b b The definite integral of a vector function is: r(t) 0t = f(t) 0t i + g(t) 0t j + h(t) 0t k a a a a Example 6 4 Evaluate t i + t 6 &t j + 1 k 0t 2 t 1

47 Calculus III Notes Arc Length - Curvature - TNB Vectors Arc Length Recall the arc length formula for a parametric function in S 2 b is L = a the length of a vector function in S 3 : b L = a (x ' (t)) 2 + (y ' (t)) 2 + (z' (t)) 2 0t b = Tr ' (t)u 0t a (x ' (t)) 2 + (y ' (t)) 2 0t. This can be extended to find Example 1 Find the arc length of the helix r(t) = 3 cos(t) i + 3 sin(t) j + t k from t = 0 and t = An arc length function to calculate the length on the interval [a, t] is Example 2 Find the arc length function for example (1). t s(t) = Tr ' (u) T0u a Solving for t we get t = 5 s 226, which means we can reparametrize r(t) using arc length: r(s) = 3 cos 5 s i + 3 sin 5 s j + s k Curvature The curvature of a curve compares the rate the unit tangent vector changes with respect to the change in arc length: An easier way to compute it is: _ = 0T 0s _ = "T' (t)u Tr' (t)u

48 48 Calculus III Notes This still requires finding the unit vector T. An even easier method using only r(t) is _ = Tr' (t) r'' (t)u Tr' (t)u 3 Example 3 Find the curvature of the twisted cubic r(t) = t, t 2, t 3 when t = 1 Example 4 For the case when the function is a plane curve y = f(x), show that _ = f'' (x) 1+(f' (x)) and find the curvature of f(x) = x 2 when x = 1. Osculating Circle N An osculating circle (kissing circle) is a tangent circle that best approximates a curve at a given point. The radius of the osculating circle is r = 1 _. Example 5 Find the radius and center of the circle osculating circle for the function in example (3). The Unit Normal and Binormal Vectors N Along with the unit tangent vector T,two other unit vectors are the unit normal, N, and the unit binormal, B. These three vectors form an orthogonal basis at any point along a smooth curve, similar to the vectors i, j, and k at the origin: N(t) = T' (t) TT' (t)u and B(t) = T(t) N(t) where T(t) = r' (t) Tr' (t)u N Example 6 Find the trihedral TNB for the twisted cubic from example (3) at the point (1, 1, 1).

49 Calculus III Notes Motion in Space: Velocity and, Acceleration Velocity and Acceleration If r(t) represents the position function of a particle in space, then r ' (t) is the velocity and Tr' (t)u represents the speed of the particle. Similarly, the acceleration of the particle is r '' (t) = v' (t) = a(t). Example 1 A force F (t) = 3 (2 & t) i + 6 t j + 3 k is acting on a 3 kg object. Find its path if it has an initial position of (20, 5, 2) and an initial velocity of v = 8, 0, 0. Recall F = m a. Projectile Motion Consider a projectile shot at an angle of 3 from the origin with an initial speed of v 0 = Tv(0)U. v 0 3 Since F = m a, and the only force acting on the projectile is gravity, we have F = m a = &m g j, thus a = &g j. Integrating gives Integrating again gives Since r 0 = 0 and v 0 = v 0 cos(3) i + v 0 sin(3) j, we get Example 2 v(t) = &g t j + v 0. r(t) = & 1 2 g t 2 j + v 0 t + r 0 r(t) = & 1 2 g t2 j + [v 0 cos(3) i + v 0 sin(3) j] t + 0 r(t) = v 0 cos(3) ti + v 0 sin(3) t & 1 2 g t2 j A projectile is shot with an initial velocity of 80 m/s and angle of elevation 40, from a position 15 meters above ground level. Where does the projectile hit the ground, and with what speed does it hit the ground (assuming no wind resistance)?

50 50 Calculus III Notes Tangential and Normal Components of Acceleration It s often useful to write the acceleration vector in terms of T and N. Since T = Also, _ = TT'U (1) becomes = TT'U Tr'U v a = v' = v ' T + vt ' (1) r' = v, we have v = v T, and hence "r'u v T', so TT 'U = _ v. The unit normal is defined as N =, which gives T ' = N TT 'U = N _ v, so equation TT'U a = v ' T + _ v 2 N We can also write the tangential component a T = v ' and the normal component a N = _ v 2 in terms of r, a T = r' (t) r'' (t) Tr' (t)u and a N = Tr' (t) r'' (t)u Tr' (t)u Example 3 Find the tangential and normal components of acceleration for the particle with position function r(t) = t 2, t 3, 3 t 2.