The Calculus Concept Companion - in progress! Uwe Kaiser

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1 The Calculus Concept Companion - in progress! Uwe Kaiser October 19, 2013

2 Week 1/2 CCC 1 Intro an Syllabus What is a function: basics an key terms. Concepts: Quantities, variables, functions an graphs Key concepts:; Numbers an functions; quantities, measurements an units In applications an science we care about quantities an their measurement. By choosing a unit a number can be assigne to a measurement. So if length is measure in meters m then the length of let s say a stick is compare with a stanar length of 1 m. If a stick is 1 2 m long then we nee two sticks to get the stanar meter length. Time is another example of a quantity an the unit of secon has to be fixe. So on t forget that the numbers associate to a quantity epen on the chosen unit. Other examples of quantities we stuy are temperature, current, work etc. The laws of nature are relations between quantities. They are given by formulas involving the quantities as variables. In practice the result of a measurement is a rational number (like 0.5 in the example above). But we will assume that our quantities can take real number values because this is practical. For instance the length of the iagonal of a unit square is 2. This is not a fraction but we can measure it. In general we use symbols as placeholers for quantities, like t for time, l for length, T for temperature an so on. Sometimes we talk about a constant quantity, which is just a single number, also enote by a symbol. This often is comfortable because whether a quantity is constant or variable can epen on the situation consiere. For example we may set up an experiment where the temperature is fixe. In another experiment we change the temperature between 10 an 150 Fahrenheit an stuy the resulting electric resistance of a cable. So whether temperature T is variable or constant epens on the situation. A typical application situation involves many quantities. We want to know how they relate to each other. Let us consier gas containe in a volume V at a temperature T (measure in Kelvin), which will lea to a pressure p accoring the equation pv = kt, 1

3 (where k is a constant). In a situation like this we call a variable, which we change in an experiment the inepenent variable. For example this can be the temperature changing within some range. Then, if we keep the volume fixe, the resulting pressure is an example of the corresponing epenent variable. For each fixe volume the temperature etermines the pressure, or we say that the pressure p is a function of temperature T. Here the temperature is changing in some interval [0, T max ]. So accoring to the above law p = k V T, where now c = k V is a constant. The graph of p = ct is a line through the origin in a coorinate system where we ientify lengths along the axes with the numbers associate to T an p with respect to units. Problem: Suppose we keep the temperature fixe an change the volume instea. What are the inepenent an the epenent variable in this case? What is the function escribing the relation? In calculus we often call the inepenent variable x an the epenent variable y an enote their relation by y = f(x). This means that for each possible value of x (usually in some interval) there is a uniquely etermine value of y. The function can be given by a formula like y = x 3 or a graph (Give a sketch of the graph of x 3!). Problem: Write up all the functions an classes of functions you recall from your previous courses! Another typical example is a point moving along a line, on which we choose a fixe point 0 an a irection. In this case the inepenent variable is time t an the epenent variable is position s. This is the real number given by the istance to the origin, but with a negative sign for points to the left of 0. Here before/behin are of course etermine by the irection of the line. The position graph s(t) here is not to be confuse with the orbit of the motion. Problem: Describe the motion moele by the functions s = 2t an s = 3t + 2 an sketch the corresponing position graphs. 2

4 CCC 2 Slopes an tangents an Point-slope formula Key concepts: Change an erivative, graient/slope, lim-notation, velocity Differentiation is the technique which enables the velocity or more generally the slope or graient of a graph to be obtaine from an equation without rawing the graph. Before learning those techniques let s try to see what we are calculating. If a tangent is rawn at a point P on a curve then the slope (or graient) of this tangent is sai to be the slope of the curve at P : The slope of the line can be foun as rise over run an is 8 units for the graph above: If we go 0.5 units to the right along the x-axis the graph of the line goes up by 4 units. The equation of the graph above is y = f(x) = 4x 2 1. Next let us calculate the slope of the chor, which joins P (1, f(1)) with Q(1.5, f(1.5)). If we exten the chor to a line we call this a secant line an its slope a secant slope. In our example we get f(1.5) f(1) = = = 10. This slope is also calle the average rate of change of f(x) for the interval [1, 1.5]. If we choose values of x closer an closer to 1 an calculate the corresponing secant slopes we will get secant slopes closer an closer to 8, which is the slope of the tangent line. For example the slope of the secant line 3

5 through (1, f(1)) an (1.1, f(1.1)) is 8.4 (o the calculation!). If we call variable istance on the x-axis h then the secant slope is etermine by the formula f(1 + h) f(1) h an when h approaches zero these number will approximate the slope of y = f(x). This number is enote the erivative of f(x) or y with respect to x at 1, an we write f (1) = f x f(1 + h) f(1) = lim = 8. 1 h 0 h The lim is the mathematical symbol for the limit, the number that the values f(1+h) f(1) h are approaching when h gets smaller an smaller. In general we enote the slope of the graph of f(x) at some point P (a, f(a)) on the graph of f(x) by f (a) = f x f(a + h) f(a) y = lim = lim a h 0 h x 0 x where y is the change of the variable y resulting from a change of the variable x. These symbols of course epen on the point (a, f(a)), where we consier the changes. If s(t) is the position graph of an object moving along a vertical line then s (t) = s t = lim s t 0 t is the velocity because s is the change of position within a time perio t. We call s the isplacement in the time interval [t, t + t]. If t is small enough an the changes take place at some time t we can assume that the velocity of the object is not changing, so that v(t 0 ) = s t s(t 0 + h) s(t 0 ) s = lim = lim t 0 h 0 h t 0 t s t. We say that v(t 0 ) is the instantaneous velocity at time t 0. It is what we get from the average velocity for the interval 4

6 CCC 3 Key concepts: Integral an area, Σ-notation Integration is the technique that allows to calculate easily the area between the x-axis an the graph of a positive function over some interval. Consier the function f(x) = 3x 2 + 8x 1 an the interval [1, 3]. We want to know the area between the x-axis an the graph of the function over the interval [1, 3] (the blue shae area on the left). Of course we coul buil a lake with that shape, fill in water an measure the volume when the water is 1 foot high. Then we know the area in square feet. Fortunately there is are easier (an more practical) methos getting the answer straight from the equation of the function. We have a symbol for the content of the blue area, namely 3 1 3x2 + 8x 1x. This is calle the efinite integral of f(x) over the interval [1, 3]. This number is equal to 56 for the blue area in the picture. We will see next week that numbers like this are remarkably easy to calculate. But first let us try a geometric approach just like we i for the erivative. Let us approximate the area by areas of rectangles. For this ivie the interval [1, 3] into ten equal intervals [1, 1.2], [1.2, 1.4],... [2.8, 3] an use these as base sies. For each of these intervals consier the value of the function at the left en-point an use this for the heights. If we a up the areas of 5

7 these rectangles (see picture above) we get for each rectangle 0.2 f(left enpoint), an thus the sum of the areas (see picture on the right) is 0.2 (f(1) + f(1.2) + f(1.4) + f(1.6) f(2.8)), which can be calculate to (o it!), so slightly smaller than the actual value (because the rectangles miss some of the area uner the curve). In orer to simplify expressions as above we introuce the so calle Σ-notation, where Σ stans for sum. In the above example we woul write i=0 f( i). The numbers f(1), f(1.2),... f(2.8) for a sequence of numbers a 0, a 1,... a 9, where a i = f(1+0.2 i). In general, given a sequence of numbers a m, a m+1,..., a n we enote n a i = a m + a m a n. Let s try the efinition: 7 i=3 i=m i 2 = = = 131. The Σ-symbol satisfies some nice obvious properties like (a i +b i ) = a i + b i or ca i = c a i (see book p. 289). We can now escribe b a f(x)x geometrically for each positive function f(x) an a < b. It is the area A between the x-axis an the graph over [a, b]. We can compute it approximately by iviing the interval [a, b] into small intervals of length x, so have points a, a + x, a + 2 x,..., a + (N 1) x, b = a + N x iviing [a, b] into N small intervals. Then we approximate the area A by x f(a) + x f(a + x) + x f(a + 2 x) x f(a + (N 1) x). Using Σ-notation we have N 1 f(a + i x) x, i=0 an when x approaches 0 the corresponing numbers above approach A. We write this as A = b a f(x)x = lim x 0 i f(a + i x) x. 6

8 CCC 4 Key concepts: Anti-erivatives, Funamental theorem of Calculus Inefinite Integrals as anti-erivatives Let s consier the situation of CCC 3 in the case where the function f(x) is replace by the velocity function v(t) of an object moving along a line. We know that if v(t) is constant it is easy to etermine the istance riven. Then we fin the position if we know where we are at let s say t = 0. When we rive with 30 miles per hour for 30 minutes we have riven 15 miles. Sure! Well this is the basic equation for constant velocity motion: change in position = velocity time or in formulas: s(t) = s 0 + v t where v is a constant an s 0 = s(0) is initial position. Next suppose our velocity v(t) is changing, lets say it looks like on the left below. The position graph s(t) is on the right han sie. In fact these are the graphs for a boy falling in vacuum (actually in vacuum all boies fall the same) with t in secons an s in feet. Now s(t) = 16t 2 has erivative v(t) = 32t. How oes this relate to the constant velocity motion? Well assume that the motion has constant velocity v(t i ) on small intervals of length t with left enpoints t i. For example ivie the interval [0, 3] into ten intervals of length 3/10 = 0.3. So we will get t 0 = 0, t 1 = 0.3, t 2 = 0.6,... t 9 = 2.7, or t i = 0.3 i for i = 0,..., 9. Then using the basic formula above for each of the intervals we get for the total istance 7

9 covere: v(0) t + v(0.3) t + v(0.6) t v(2.7) t, an so 32( ) 0.3 = ( ) = = 129.6, which is not 144 = s(3). This is smaller than s(3) = 144 because the velocity is increasing on each of the intervals. If we make t smaller an smaller then the numbers above will approximate s(3). In fact they o. Of course these numbers also approximate the area uner the graph of v(t), In fact from the area for a triangle we get for the area over [0, 3]: v(3) = = = 9 16 = 144 = i=0 v(t i ) t, where we use Σ-notation from CCC 3. Thus for t very small we get the precise area, or using the integral notation from CCC 3 In general we have 0 3 v(t)t = lim t 0 i s(t 2 ) s(t 1 ) = t 1 v( t i) t. t 2 v(t)t is the isplacement for the interval [t 1, t 2 ] from the velocity. Now v(t) = s t by the very efinition of velocity (see CCC 2). Just as v(t) is the erivative of s(t) we say that s(t) is an anti-erivative of v(t). Of course we can t etermine position from velocity because position always epens on an initial position. The formula s(t 2 ) = s(t 1 ) + t 1 t 2 v(t)t escribes precisely this. Thus the integral of a function over some interval is the ifference of values of an anti-erivative at the en points. But what happens if we move backwars. Let s look at Definite Integrals. The formula for isplacement s(t 2 ) s(t 1 ) = t 1 t 2 v(t)t also hols when we are allowe to go backwars. Geometrically then the integral is the net or signe area. If v(t) is negative the position gets smaller an we have to subtract. Now the integral symbol picks up area below the x- or t-axis with a minus sign. 8

10 CCC 5 Key concepts: Derivatives an anti-erivatives, polynomials Take a look at: Derivatives of polynomials, Linearity of the erivative an Anti-erivatives of polynomials If f(x) is a function then we enote the graient function or erivative f (x). It measures the steepness of the graph at all points. Let s calculate the erivative of f(x) = x 3 using the approach of CCC 2. We want to use the formula from CCC 2 Now ( ) f f(x + h) f(x) (x) = lim. h 0 h f(x + h) f(x) = (x + h) 3 x 3 = x 3 + 3x 2 h + 3xh 2 + h 3 x 3 = 3x 2 h + 3xh 2 + h 3 an thus f(x + h) f(x) = 3x 2 + 3xh + h 2. h Now whatever x is, when h gets small, 3x 2 h+h 2 gets small (for example take x = 10. Then we have 300h + h 2. If h = =.001 is number is so small). So we see f (x) = x 3. This formula generalizes to the famous Power Rule: x xn = nx n 1 for all numbers n (mostly we will have n a positive integer but in fact the formula hols for all n). So x x1 = 1 x 0 = 1 1 = 1 x x2 = 2 x 2 x x3 = 3x 2. x x0 = x1 = 0, horizontal graph has steepness (Maybe you shoul recall zero!) Using the power rule we get in fact erivatives of all polynomials. If f(x) = x 3 + 4x 2 x + 6 9

11 then f (x) = 3x 2 + 8x 1. Do you see how this works? You take the erivative of each term in the sum separately an a them together. The erivative of x 3 is 3x 2. The erivative of x 2 is 2x, we take this an just multiply by the factor 4 to get 8x. The erivative of x is 1. Finally the erivative of the constant 6 is zero, 6 oes not change! Note that if we multiply a function by a positive constant it s graient will change by that constant: 2x 2 is twice as steep as x 2 at every point. If you a two functions their graients will a up too. We call these properties linearity. f cf(x) = c x x, f (f(x) + g(x)) = x x + g x, If f(x) is the position function s(t) of a moving object the linearity is more obvious: Suppose I cover 10 miles in an hour. Then if I cover 20 miles in an hour I am twice as fast. Also if s 1 (t) is the motion of the back of a train, an a man moves within the train an the position relative to the back is s 2 (t) then the combine motion is s 1 (t) + s 2 (t). The velocity of the absolute motion is s 1 t + s 2 t. Recall if f(x) is a function then a function F (x) such that F (x) = f(x) is calle anti-erivative of f(x). We use the notation f(x)x for anti-erivatives of f(x). The symbol isn t really a function but stans for all possible functions. For example 2xx = x 2 + C, where C stans for a constant. The power rule above tells us formulas for anti-erivatives x n 1 x = n + 1 xn+1. Using the anti-erivatives an the Funamental Theorem of Calculus we can now for example fin change in position from a velocity function. 10

12 Week 3 CCC 6 Key concepts: Computations with tangent Lines; linear motion Take a look at: Slope an Rate of Change, Velocity an Rate of Change an an Position by integration Consier the graph of y = x 3 3x below. We woul like to answer the following question: At what points on the graph is the slope 9? What are the equations of the tangent lines at those points. It seems to be ifficult to get precise answers from the graph. procee from the equation. From CCC 5 we get y = 3x 2 6x for the slope at (x, y) on the graph. So the slope is 9 gives the equation or 3x 2 6x = 9, 3x 2 6x 9 = 3(x 2 2x 3) = 3(x 3)(x + 1) So we So the first tangent line is through the point ( 1, 3) on the graph an has the equation: y ( 3) = 9(x ( 1)) or y = 9x

13 The secon tangent line is through the point (3, 1) on the graph an has the equation: y 1 = 9(x 3) or y = 9x 26. Now let s state the general case. If y = f(x) is a function then the tangent line at the point (x 0, y 0 ) on the graph has the equation: y y 0 = f (x 0 ) (x x 0 ). So for example the tangent line to f(x) = x 4 + 6x for x = 1 can be calculate as follows: f (x) = 4x 3 +18x 2 so f (1) = 14, an f(1) = = 6. The equation of the tangent line is: y 6 = 14(x 1) or y = 14x 8. Now let s recall our formula for computing position from velocity: s(t) = s(t 0 ) + t 0 t v(u)u to calculate the position from some initial position s(t 0 ) an the velocity function using integration. Using the Funamental Theorem of Calculus we know we fin the integral from the anti-erivative of v. Thus there is a way to o the calculation avoiing the integral symbol. Let s just o an example: Suppose that a motion is escribe by v(t) = 10t 6t 2. We have the graph of v(t) on the left an the graphs of two corresponing position functions (anti-erivatives) on the right, with s(0) = 0 respectively s(0) = 3. 12

14 Using our knowlege about anti-erivatives of polynomials it is easy to fin formulas for the two position functions. We know that s(t) = 5t 2 2t 3 + C, where C is a constant. If s 1 (0) = 0 then we calculate 0 = C an get s 1 (t) = 5t 2 2t 3, the top graph. If s 2 (0) = 3 we calculate 3 = C an so s 2 (t) = 5t 2 2t 3 3. Note that s 1 (1) = 5 2 = 3 but s 2 (1) = = 0. We can set the initial position for any point in time. Then the velocity function etermines the position. Note that if the velocity function is constant v then s(t) = vt + C is a special case of the power formula for anti-erivatives. Since s(t 0 ) = vt 0 + C we can calculate C = s(t 0 ) vt 0 an get the general formula for constant velocity motion: s(t) = s(t 0 ) + v(t t 0 ) 13

15 CCC 7 Key concepts: The erivative of exponential functions; More on linear motion Take a look at: Position, velocity an acceleration Last time we euce the formula for constant velocity motion. A motion where the velocity is changing in such a way that it changes by equal amounts in equal time intervals is calle a constant acceleration motion. In general, the instantaneous change of the velocity is calle acceleration an can change with time. It is again a function of time an we have a(t) = v t. By applying integration as before we get the general formula v(t) = v(t 0 ) + t 0 t a(u)u. Let s call v(t 0 ) = v 0 an s(t 0 ) = s 0 because then we see better that it is a constant. We can avoi the integral by just using that v(t) is an antierivative of a(t). If we can fin the anti-erivative of a(t) then we can etermine the constant by some initial conition v(t 0 ). Then knowing v(t) we can fin s(t) as before. For the constant acceleration motion the antierivative of the constant a is v(t) = at + C an so from v(t 0 ) = at 0 + C we get v(t) = v 0 + a(t t 0 ) The anti-erivative of v(t) is To see this just ifferentiate s(t) = v 0 t + a 2 (t t 0) 2 + C. t (v 0t + a 2 (t2 2t 0 t + t 2 0) + C ) = v 0 + a 2 (2t 2t 0) = v 0 + a(t t 0 ) Finally from s(t 0 ) = s 0 = v 0 t 0 + C we get C = s 0 v 0 t 0 an the law for the general constant acceleration motion: s(t) = s 0 + v 0 (t t 0 ) + a 2 (t t 0) 2. 14

16 For the motion in constant gravity a = g an t 0 = 0 we get the simpler formula: s(t) = s 0 + v 0 t g 2 t2. Suppose that a = 5, v 0 = 1 an s 0 = 2 (units coul be for example here feet/sec 2, feet/sec). Then for t 0 = 0 we get s(t) = 2 + t t2. The formulas above are sometimes comfortable. But you o not have to remember them. Just recall that s(t) v(t) a(t) is given by taking erivatives an thus going backwars we take anti-erivatives. Here is an example: Suppose that a(t) = t, an at time t = 0 we have initial velocity v 0 = 1 an initial position s 0 = 0. Then v(t) = 1 2 t2 + C is the velocity, an 1 = v(0) = C shows that v(t) = t2 is the formula for velocity (Check from this formula again that v(0) = 1 an v t = t). By taking the anti-erivative again we get s(t) = t t3 + C an from s(0) = 0 we get C = 0. Thus the solution is s(t) = t t3. Take a look at: Derivatives of exponential equations In many applications we nee to know how the exponential function is changing. This turns out to be more ifficult that our iscussion of polynomials. We stuy first the slope of e t at t = 0 when e 0 = 1. So imagine a motion with s(t) = e t an we want to know the velocity at t = 0. So let s calculate the numbers e t 1 an e t 1 t for t = 0, 0.1, 0.01, an Guess the value of e t 1 t for infinitesimal t, so answer the question: What value o we approach when t gets smaller an smaller? t e t 1 (e t 1)/ t not efine We see: When t gets smaller then the value of e t 1 t approaches 1. 15

17 Now we fin for the function s(t) = e t the isplacement s in terms of t, an thus s t. The isplacement is (using the law of exponents) s = s(t + t) s(t) = e t+ t e t = e t (e t 1), which will approach 0 when t approaches 0. The average velocities s t = s(t + t) s(t) t = e t e t 1, t which will approach e t 1 = e t when t approaches 0. Thus we get for the instantaneous velocity v(t) = s t = et 1 = e t. Thus we know that t et = e t or x ex = e x. A small algebraic trick actually allows to get the general formula for any constant k. In fact x ekx = ke kx e x ekx = lim k(x+ x) e kx x 0 x = lim x 0 ekx ek x 1 x = lim x 0 (kekx ) ek x 1 k x Now when we calculate the lim x 0 we only consier the change of x, not any change in x, here x is fixe but arbitrary. But x 0 certainly precisely when k x 0 (except maybe when k = 0 but that will give the constant function e 0 = 1). So we can call k x = h an get So for example x e5x = 5e 5x. x ekx = ke kx e lim h 1 = ke kx h 0 h 16

18 CCC 8 Key concepts: Prouct Rule an Quotient Rule Take a look at: Differentiating factore polynomials (This actually assumes that you know that x sin(x) = cos(x). We will see this soon!) an When to use the quotient rule (well the quotient rule is actually not our last rule, so here you have to know that x sin(4x) = 4 cos(4x). We will see this also soon!). Derivatives can be calculate from few basic examples like t (t) = 1 an t et = e t using a collection of Rules. Here we will justify the Prouct rule, Reciprocal rule, Quotient rule an Power rule from the very efinition that s t is calculate from s t by taking the limit. So s t = lim s t 0 t. Recall that the limit symbol means that s t is the value that is approache by consiering the average velocities s t when t gets smaller an smaller. Throughout for all functions f(t) (or g(t),s(t),...) of t let us use the symbol f = f(t + t) f(t), which stans for a change of the value of the function originating from a change from t to t + t. Note that this also implies: f(t + t) = f(t) + f. We will assume throughout that f approaches 0 when t approaches 0 an that the limits always exist. Prouct Rule Let s(t) = f(t)g(t) be a prouct of the functions f(t) an g(t). Then f(t + t)g(t + t) = (f(t) + f)(g(t) + g) = Thus = f(t)g(t) + ( f) g(t) + f(t) ( g) + ( f) ( g) s t = s(t + h) s(t) t = f g g(t) + f(t) g + ( f) t t t 17

19 an it follows for infinitesimal t (an thus g t number while f is approaching 0): just approaching a fixe t f (f(t)g(t)) = g(t) + f(t)g t t Let s(t) = 1 g(t). Then Reciprocal Rule an so s = s(t + t) s(t) = = s t = 1 g(t + t) 1 g(t) = 1 g(t) + g 1 g(t) = g(t) (g(t) + g) g(t)(g(t) + g) = g g(t)(g(t) + g)) g t g(t)(g(t) + g) = g t ( 1 g(t)(g(t) + g) ). If t becomes infinitesimal also g becomes infinitesimal (we assume this!) an thus t ( 1 g(t) ) = g t ( 1 g(t) ) 2 Quotient Rule Exercise: (a) Show that the quotient rule: f g t (f(t) g(t) ) = tg(t) f(t) t g(t) 2 hols as follows: Let s(t) = f(t) g(t) = f(t) 1 g(t). Then apply the prouct rule an the reciprocal rule, an simplify algebraically. (b) Apply the quotient rule to the special case f(t) = 1 an euce the reciprocal rule from the quotient rule. Remarks: Note that in the formulas above we usually write f t for the erivative function when we consier the function f(t) of t. Actually this 18

20 shoul be f t (t), but looks clumsy. You may want to rewrite the rules using the alternative notation f t = f (t). Note that also all rules can be expresse using the inepenent variable x: x (f(x)g(x)) = f (x)g(x) + f(x)g (x) an x (f(x) g(x) ) = f (x)g(x) f(x)g (x) g(x) 2. Write the rules for a function P of a variable T. How o the rules look like? or or The rest is to o examples, as many as you can. Here are a couple. x ( x 1 x ) = 1 (1 x) x ( 1) (1 x) 2 = 1 (1 x) 2 x (xex ) = ( x x ) ex + x ex x = ex + xe x = (x + 1)e x u ( 1 u 2 + 4u + 2 ) = u (u2 + 4u + 2) 2u + 4 = (u 2 + 4u + 2) 2 (u 2 + 4u + 2) 2 What are the rules I use in the three cases above? 19

21 Week 4 CCC 9 Key concepts: Derivatives an anti-erivatives of trigonometric functions Take a look at: Calculating erivatives of trigonometric functions (Their graphs of the trigonometric functions are not correct. The graphs are not built from half-circles.) A simple harmonic motion is the projection of the motion of a point on a circle rotating with constant angle velocity onto the x- or y-axis. Recall that the coorinates of a point moving in a plane are given by (x(t), y(t)). If the point moves on a circle let s say of raius 1 then x(t) 2 + y(t) 2 = 1. Thus we can write for each time t, x(t) = sin α(t) an y(t) = cos α(t), where α(t) is the angle at time t. The angle velocity is the rate of change of the angle (which we usually measure in raians) with respect to time. If we assume that tα(t) = ω is constant then the y-coorinate of a simple harmonic motion is escribe by y(t) = A sin ωt, where A is the amplitue (just the raius of the circle). Note that the motion is perioic. After a time T = 2π ω the motion repeats because ω(t + 2π ω ) = ωt + 2π an the sinus function is 2π-perioic. If we look at how the velocity is changing it looks like it follows the same pattern. The graph to the left shows the position y(t) = sin(t) in blue an the graph of cos(t) in the same coorinate system (so we just put ω = 1 for simplicity). Note that where sin(t) has horizontal tangents (velocity 0) the function cos(t) is zero. The velocity is π 2 behin the position corresponing to the trigonometric ientity: sin(α) = cos(α π 2 ). 20

22 In fact in the interval [0, π 2 ] the sinus function has positive slope (positive velocity) an the cosine function is positive there. Also when the harmonic motions goes through its center the velocity is maximum. Let us use our efinition of the erivative to fin the velocity of the simple harmonic motion from the limit of average velocities efinition. So let s put y(t) = sin(t) an calculate y(t + t) y(t) t = sin(t + t) sin(t) t = sin(t) cos( t) + sin( t) cos(t) sin(t). t Here we use the angle aition formula How to get the angle aition formula sin(α + β) = sin(α) cos(β) + sin(β) cos(α) There s also a corresponing formula for the cosine function: cos(α + β) = cos(α) cos(β) sin(α) sin(β). We will also nee this formula a lot, that s why it s mentione here. Let s look back at our calculation an break things apart into factors containing only t an others containing only t: y(t + t) y(t) t = cos(t) sin( t) t + sin(t) cos( t) 1 t Let s calculate how the expression sin( t) behaves when t gets small: Note that actually the values for negative t on t have to be compute sin( t (Why?). Looks like lim t 0 t = 1. We will give a more precise argument for this later on when we unerstan better how to fin limits. Thus we see that cos(t) sin( t) t will approach cos(t). Let s stuy the secon term: sin( t t t cos( t) 1 t = (cos( t) 1)(cos( t) + 1) t(cos( t) + 1) = sin( t) t sin( t) cos( t) + 1. But in the last expression the factor sin( t t approaches 1 as before, cos( t)+1 approaches 2 an sin( t) approaches 0. Thus the whole will approach 0 an 21

23 thus also if we multiply it by sin(t). Now by checking the graphs it shoul be obvious that t cos(t) = sin(t) (notice that cos(t) is escening in [0, π 2 ] but sin(t) is non-negative there). So we can a to our collection of erivatives (for fun using the x variable): sin(x) = cos(x), (cos(x) = sin(x) x x That s the point to start using our other rules: x tan(x) = x sin(x) cos(x) = cos(x) cos(x) sin(x)( sin(x) cos 2 (x) = 1 cos 2 (x) = sec2 (x) an all the other formulas for erivatives of trigonometric functions. Also, as usual we get the formulas for anti-erivatives at the same time: sin(x)x = cos(x) + C, cos(x)x = sin(x) + C Now we can use our other rules to calculate more erivatives, for example: α (α2 sin(α)) = 2α sin(α) + α 2 cos(α) or u (tan(u) ) = tan(u) u sec2 (u) u u 2 an also anti-erivatives: (2 sin(x) + cos(x) + x 3 )x = 2 cos(x) + sin(x) + x4 4 + C an we also can calculate areas with this: 0 π sin(x)x = cos(x) π 0 = cos(π) ( cos(0)) = 2 22

24 Finally let s just at this point that (sin(kx)) = k cos(kx), x (cos(kx)) = k sin(kx) x for constants k. This is not surprising. For large k sin(kx) will oscillate faster so the slopes will have bigger magnitue. 23

25 CCC 10 Key concepts: Integration by parts; higher erivatives Take a look at: Integration by parts If we want to calculate anti-erivatives of proucts we have a problem because there is no rule that applies. In fact the prouct rule tells that that in general x (f(x)g(x)) f (x)g (x). But we can still use the prouct rule to calculate anti-erivatives. Let s look at: (f (x)g(x) + f(x)g (x)) x = f(x)g(x) + C this is true because if we take the erivative of the right han sie by the prouct rule we get the integran, or we are just saying that f(x)g(x) is anti-erivative of f (x)g(x) + f(x)g (x). Usually integrals o not come in the form above an the rule is use as follows: f (x)g(x)x = f(x)g(x) f(x)g (x)x We omit the constant because each inefinite integral sign contains constants. The iea is f(x)g (x) might be easier than f (x)g(x)x. Here is an example with f (x) = e x an g(x) = x: e x xx = e x x e x 1x = xe x e x + C = e x (x 1) + C Here we use that e x x = e x + C. Similarly we can take f (x) = cos(x) an thus f(x) = sin(x) is an anti-erivative, an g(x) = x an get: cos(x) xx = sin(x) x sin(x) 1x = x sin(x) + cos(x) + C Don t forget that there are many ways to write own the formula for integration by parts, epening on what names we give to functions an variables. Also the orer in which the functions appear can change: u v t t = u(t)v(t) u t vt Here u, v are functions of t. But we i not write this explicitly uner the integral sum to avoi clumsy notation like u(t) v t (t). Of course the 24

26 Funamental Theorem of Calculus tells us that efinite integrals are just given by evaluating anti-erivatives an the formula becomes a b u(x)v (x)x = u(x)v(x) b a a b u (x)v(x)x Here is an example with u(x) = x an v (x) = e 4x so that 1 4 e4x is an antierivative: 3 xe 4x x = x e4x e4x x. This can now be evaluate further: 1 4 (3e12 0 e 0 ) e 4x x = 3 4 e ( e4x ) = e (e12 1) = e = 1 16 (11e12 +1) This looks like a huge number, is that possible. Well look at the area: 25

27 Take a look at: Higher Derivatives In applications higher erivatives often are important. The meaning of these higher erivatives, both geometrically an in physics, is of its own interest. At this point we just want to introuce the corresponing notation. Of course we can take the erivative of a erivative. If we write y = f(x) then the notation is y for the first (usual) an y = (y ) for the secon erivative. In Leibniz notation it is f x for the first an 2 f x for the secon erivative. Here 2 is an example: If f(x) = sin(x) then f (x) = cos(x) an f (x) = sin(x). A typical example is when we start with a position function, let s say x(t). Then x (t) is the velocity an x (t) is the acceleration. So for a simple harmonic motion the acceleration is just the negative of the position at each point in time. This explains why we en up with a perioic motion as we see. If the position gets large positive the acceleration will be negative an get the velocity to become negative an the motion turn aroun. 26

28 Key concepts: The Chain Rule CCC 11 Take a look at: Using the chain rule to ifferentiate (But on t believe that hanging ropes hang in the form of parabolas!) an at Chain rule y x x Suppose that you are climbing a hill with constant slope (see left picture above), let s say 0.3 meters rising vertically per one meter horizontally, an the x-coorinate of your position is escribe by x(t) = 0.5t + 1 with x(t) is horizontal meters an t in secons. It is now easy to calculate the rate of change of your altitue (your y-coorinate) with respect to time by plugging in: y(t) = 0.3(0.5t + 1) = t an we see that y (t) = 0.15 vertical meters per secon. Note how the units fit: We multiply 0.3 vertical meters/horizontal meters with 0.5 horizontal meters/secon to get 0.15 vertical meters/secon. This seems to be obvious. A little less obvious example is given when we slie own a hill escribe by the graph of a function f(x) = x 2 with both x, y measure in feet an x(t) = 1+1.5t 2 escribes our horizontal motion (see right picture). (Note that the unit of 0.1 here actually has to be in vertical feet/square of horizontal feet to get the right units for y.) Then of course we can again calculate our 27

29 vertical motion: y(t) = ( t 2 ) 2 = (1 + 3t t 4 ) = t t 4. Thus our vertical velocity at time time t is v(t) = 0.6t 0.9t 3. Now our horizontal velocity is x (t) = 3t an the change of our vertical position relative to our horizontal position is 0.2x vertical feet/horizontal feet. Note that we have the right units. The number 0.2 has the above units of vertical feet/square of horizontal feet an is multiplie by x with units horizontal feet. How can both these numbers together give irectly the answer above. Well the answer is that: 0.2 ( t 2 ) 3t The general case is that of a function y(t) = f(x(t)) with erivative y (t) = y t = f (x(t))x (t). The prime notation has to be taken carefully here because f (x(t)) = f x x(t) while x (t) = x t, so we are actually taking erivatives with respect to ifferent variables. The easiest way to memorize (an euce by cheating) the chain rule is: f t = f x x t. In the mathematics literature the above is seen as a composition (f g)(t) an written as: t (f g) = (f g) (t) = f (g(t)) g (t) Note that on the right han sie the is a prouct an not composition, so is often omitte. Instea of the variable t also often the variable x is use an then x (f g) = f (g(x))g (x) Here is a first example: Take f(x) = sin(x) an g(x) = x 2. Then f(g(x)) = sin(x 2 ). We have f (x) = cos(x) an g (x) = 2x. Thus x (sin(x2 )) = cos(x 2 ) 2x = 2x cos(x 2 ). 28

30 Week 5 Reviews an test 29

31 Week 6 CCC 12 Key concepts: Taking ifficult erivatives; Derivatives of inverse functions Take a look at: Applying the rules of ifferentiation, Derivatives of inverse functions an Derivatives of inverse trigonometric functions When taking erivatives things often get more ifficult in the sense that we have to apply many rules at the same time an also apply the same rule multiple times. An easy case is application of the prouct rule to a prouct of multiple factors. Of course this can always be one by two applications of the prouct rule in a row. But it is worth to fin the general pattern: x (f(x)g(x)h(x)) = x (f(x)(g(x)h(x))) = f (x)g(x)h(x) + f(x) x (g(x)h(x)) = f (x)g(x)h(x) + f(x)g (x)h(x) + f(x)g(x)h (x) So you just work through all factors ifferentiating one factor at a time an summing up the results. Example 1: For f(x) = x 2 e 8x sin(7x) we get f (x) = x2 x e8x sin(7x) + x 2 e8x x sin(7x) + x2 8x sin(7x) e x = 2xe 8x sin(7x) + x 2 8e 8x sin(7x) + x 2 e 8x 7 cos(7x) = xe 8x (2 sin(7x) + 8x sin(7x) + 7x cos(7x)) The chain rule gives more subtle situations. Let s look at F (x) = f(g(h(x)), which is a composition of three functions. In establishing the general principle we think of this as f((g h)(x)) an apply the chain rule to the inner function g h an the outer function f to get so a further application gives F (x) = f (g(h(x)) (g h) (x), F (x) = f (g(h(x)) g (h(x)) h (x). 30

32 So we work from the outsie to the insie. Another way to think of this is to introuce variables u = g(v) an v = h(x) an write: F x = F u u v v x. The best is not to use the resulting rule here but to o explicitly what we showe for the general case: Example 2: Let f(x) = cos(sin(cos(x))). Then f (x) = sin(sin(cos(x))) x (sin(cos(x)) = sin(sin(cos(x))) cos(cos(x)) ( sin(x)) = sin(sin(cos(x)) cos(cos(x)) sin(x) The chain rule is the main tool in fining erivatives of inverse functions. This comes from the ientity f f 1 )(x) = f(f 1 (x)) = x. Example 3: Recall that the natural logarithm ln(x) is the inverse function of the natural exponential function. So we have e ln x = x Now take erivatives on both sies, using the chain rule on the left sie an x x = 1 (This is not algebraic cancellation but the fact that a line of slope 1 has erivative 1 because erivative=slope). so because e ln(x) = x we get e ln(x) (ln x) = 1, x an thus x (ln x) = 1 x x (ln x) = 1 x 31

33 Of course by using the chain rule this gives us: x ln(kx) = 1 k ln x + C an by taking the anti-erivatives we get 1 x = ln(x) + C. x Of course since the omain of ln(x) is only (0, ) the formula will only work for x > 0. Example 4: The inverse function of the function sin x is often enote sin 1 (x), which has to be taken very carefully. Note that we write sin 2 x to mean (sin(x)) 2. This will be that way for all sin k (x) an positive integers k. But for k = 1, sin 1 x 1 sin x but enotes the inverse function. This is stanar in most text books. Sometimes people use arcsin x instea to enote the inverse function of sin x to avoi this problem. We start with an apply erivatives on both sies: sin(sin 1 (x)) = x x sin(sin 1 (x)) = x x Now procee as in Example 1. or cos(sin 1 (x)) x sin 1 (x) = 1 1 x sin 1 (x) = cos(sin 1 (x)) Now we raw an angle θ in a right triangle with hypotenuse length 1, opposite length x an thus ajacent length 1 x 2 (Do it!). Then sin(θ) = x so theta = sin 1 (x) an cos(θ) = 1 x 2. Thus we get the formula x sin 1 (x) = Again by taking anti-erivatives we get 1 1 x 2. x 1 x 2 = sin 1 (x) + C 32

34 CCC 13 Key concepts: Calculating integrals; Substitution Take a look at: Definite Integrals, Aition property of efinite integrals an How to solve integrals using substitution Recall that for a < b the symbol a b f(x)x just represents the signe area for f(x) over the interval [a, b]. This is the sum of the signe areas over intervals where f(x) > an f(x) < 0. Here signe means that we count the area content positively if the graph of f(x) is above the x-axis, an negative when the graph of f(x) is below the x-axis. Not surprisingly the following rules are compatible with the above interpretation: for all a. a b a a f(x)x = 0 f(x)x = b a f(x)x Note that here if a < b then b < a, so this extens our efinition. a b f(x)x = a c f(x)x + c b f(x)x This is easy to unerstan geometrically when a < c < b but follows using the previous formula for all cases. So if a > b an f(x) > 0 we will have b a f(x)x < 0, so we always shoul remember that the interpretation of the integral as signe area refers to a irection of the x-axis. If we have a > b we are integrating in the irection of the negative x-axis an that reverses the signs of the areas (in fact in our approximation formulas x are negative quantities). Note that all of the above is also compatible with the Funamental Theorem of Calculus. If F (x) is an anti-erivative of f(x) then a a f(x)x = F (x) a a = F (a) F (a) = 0 33

35 Also an a b f(x)x = F (b) F (a) = (F (a) F (b)) = b a f(x)x a b f(x)x = F (b) F (a) = (F (b) F (c)) + (F (c) F (a)) So here is an example: 1 1 = (F (c) F (a)) + (F (b) F (c)) = x 2 x = 0 1 x 2 x x 2 x = a c f(x)x + x 2 x = x3 = c b f(x)x where because of the symmetry of x 2 we know that the signe areas over [ 1, 0] an [0, 1] are the same. The aition property also is important to fin integrals of piecewise efine functions. For example consier f(x) = x 2. Then So f(x) = { 2 x x 2 x 2 x > f(x)x = = f(x)x + 2 (2 x)x + = 2x x f(x)x 3 + x2 2 2x 3 (x 2)x = (4 2) + ( 9 6 (2 4)) 2 2 = =

36 Calculating integrals can be ifficult. Let s consier We know that by chain rule an so 2x cos(x 2 )x x sin(x2 ) = cos(x 2 ) 2x 2x cos(x 2 )x = sin(x 2 ) + C Of course, in this way we can fin many anti-erivatives. But usually we have given a specific integran. So we want to see whether we can euce antierivative in a systematic way by thinking the chain rule backwars. This leas to a new integration technique, which is calle substitution. Just begin with the chain rule in the form: x f(g(x)) = f (g(x)) g (x) an take anti-erivatives on both sies. Note that the anti-erivative of a erivative gives back the function up to a constant. So we get f(g(x)) + C = f (g(x))g (x)x. But how o we spot that an integran has this form? The way to o this systematically is to go an use a substitution u = g(x). Then g (x) = u x. Using the formal notation u = g (x)x we write f (g(x))g (x)x = f (u)u = f(u) + C = f(g(x) + C 35

37 Of course f(u) is just an anti-erivative of f (u) so if F (u) is an antierivative of f(u) we usually apply the technique as follows: f(g(x))g (x)x = f(u)u = F (u) + C = F (g(x) + C. Let s see how this formalism works in a more ifficult situation: 5x x 3 x. We write u = 7 + x 3. Then we get u = 3x 2 x. We can solve for x = u 3x 2 plug this back in to get an 5x x x = 3 5 u u 3 We cancele the x 2. Make sure that when you take the anti-erivative with respect to u no x-variable anymore appears in the integran. Note that x is not a constant with respect to u. So we get 5 3 u u = 5 3 ln(u) + C = 5 3 ln(7 + x3 ) + C This shows that we foun the anti-erivative correctly. You can finally check that the erivative of the result is the original integran. 36

38 Key concepts: More Substitution Take a look at: CCC 14 More solving Integrals using substitution Let s first look at a case where the substitution has to be set-up slightly ifferently to be successful: x 1 + x x 4 Here the first guess u = 1 + x 4 will not be successful. We substitute instea u = x 2 so that 2xx = u or xx = u 2. Then x 1 + x x = u 1 + u = 2 tan 1 u + C = tan 1 (1 + x 4 ) + C There are cases where the integran oes not obviously have the form f(g(x))g (x) but the substitution metho is still working. Let s consier: x 2 1 x x We use the substitution u = 1 x. Then u x = 1 If we plug back into the integran we get x 2 ( 2u) u = 2x x 1 ( 1) = 2u or x = 2uu. How o we get ri of the x 2? We solve u = 1 x for x: u 2 = 1 x an thus x = 1 u 2. So we finally en with the integral: 2(1 u 2 ) 2 u = 2 1 2u 2 +u 4 u = 2 2u 2 u 4 1u = 2( 2 3 u3 1 5 u5 u)+c Of course this is not the answer. We always have to substitute back to get the answer: x 2 1 x x = 4 3 (1 x)3/2 2 5 (1 x)5/2 2(1 x) 1/2 + C It woul have been quite ifficult to guess the answer in this case. Of course we can now calculate efinite integrals by first calculating anti-erivatives 37

39 an apply the Funamental Theorem of Calculus. In many situations it is more comfortable to avoi the back-substitution an instea transform the limits. The iea is to note that a b f (g(x))g (x)x = g(b) g(a) f(u)u To see this just use the anti-erivative F (u) of f(u) to see that the right han sie gives F (g(b)) F (g(a)). But this is also the answer we get by first fining the anti-erivative F (u), plugging back u = g(x) an then calculate F (g(x)) b a. Example: We want to calculate xx We use the substitution u = g(x) = 4 + 3x an so u = 3x or x = u 3. Then g(0) = 4 an g(7) = 25. So xx = 4 25 u u 3 = u3/ = 2 9 ( ) = 26 In applications the integral appears as the mathematical tool to accumulate instantaneous information to calculate the total change of a quantity. We have seen this for change of position over a time interval as a result of the function of instantaneous velocities. Suppose we stuy any quantity f(t) of time an consier a b f (t)t. The inepenent variable coul be time or any other quantity, for example the change in the length of a stick cause by a change in temperature. The Funamental Theorem of Calculus tells us that f(b) f(a) = a b f (t)t Here is how we may unerstan in a ifferent way why this is the case referring back to the subivie-approximate-sum up iea. a b f (t)t f (t i ) t f t 38 t = f,

40 where each f is the change of the variable f over the interval corresponing to a rectangle of the subivie proceure of base length t. Now suppose we are in the situation of the chain rule, so the integral we consier has the form With u = g(t) an g (t) = u t a b a b f (g(t))g (t)t we get similarly to the above: f (g(t))g (t)t f u u t = f, t The change in the variable y = f(u) is happening from u = g(a) to u = g(b). Application: Suppose that a person exhales 1 6 (t 1.5) 2/3 liter/sec for 1 secon. How much air has exite his lungs. The total amount of air will sum up from the amounts exite over small intervals of time t. For those time intervals the value of 1 6 (t 1.5) 2/3 evaluate at some point in the interval multiplie by t is an approximation, which becomes the exact value as t gets smaller an smaller. Thus what we are looking for is (t 1.5) 2/3 t, which we can calculate by the substitution metho using the substitution u = t 1.5 so u = t (t 1.5) 2/3 t = u 2/3 u = 3u 1/ = 3(1.5 1/ /3 ) 1.05 liter 39

41 Week 7 CCC 15 Key concepts: Hyperbolic functions, Logarithmic ifferentiation Take a look at: Logarithmic ifferentiation, Derivatives of hyperbolic functions an Derivatives of inverse hyperbolic functions First note that the erivative x ln x = 1 x only makes sense for x > 0 since the omain of ln x is [0, ). There is an easy way to exten this to a very important result using an easy application of the chain rule. In fact the function ln x is efine for all x 0.r x > 0 it is the function ln x consiere before. But for x < 0 ln x = ln( x). From the chain rule we get x ln( x) = 1 x ( 1) = 1 x. In the following we assume that the values of the inepenent variables are always chosen such that a corresponing argument of ln is not 0. So we can summarize: x ln x = 1 x. In logarithmic ifferentiation we use the property that x ln f(x) = f (x) f(x). If we use y = f(x) then the basic formula reuces to y ln y = 1 y y x ln y = y. Note that. This shows that we always have to pay attention to the variable with respect to which we ifferentiate. Recall the laws of logarithms. Since we will mostly use them for the natural logarithm we only state them here in this case: ln( A B ) = ln A ln B, ln(a B) = ln A + ln B, ln(ar ) = r ln A. The best is to show the metho with an example. In the following we start with an equation, then take absolute values on both sies, then we expan 40

42 the right han sie using the laws of logarithms. Then we take erivatives using necessary rules, finally we get the answer for the erivatives of the function. x (x2 1) 5 y = f(x) = (x + 2)(x 4) 3 x x2 1 5 y = x + 2 x 4 3 ln y = ln ( x1/2 x x + 2 x 4 3 ) = 1 2 ln x + 5 ln x2 1 ln x ln x 4 Now we take the erivatives with respect to x on both sies: So the final answer is: y y = x (1 2 ln x + 5 ln x2 1 ln x ln x 4 ) = 1 2x + 10x x x x 4 y = y ( 1 2x + 10x x x x 4 ) This coul be multiplie out after substituting back y = f(x). This woul be still quite some work an you might woner whether logarithmic ifferentiation is easier than using quotient, prouct an power rule (Note that the laws of logarithm are precisely replacing the use of those rules by corresponing laws of logarithms.) But suppose we want to know f (2). Then f(2) = ( 2) = an we get f (2) = ( ) = It woul have been much harer to get this result without logarithmic ifferentiation. 41

43 Note that the original function y = f(x) is efine for all x 2, 4. But we shoul expect some problems also for x = 0 because the erivatives of x is 1 2 an thus is not efine for x = 0. When we apply the rules of x logarithms we get terms, which are only efine for x not 0, 1, 2, 1, 4. Just like trigonometric functions parametrize the points on the unit circle x 2 + y 2 = 1 the points on the hyperbola x 2 y 2 = 1 are parametrize by the hyperbolic functions cosh(t) = 1 2 (et + e t ), sinh(t) = 1 2 (et e t ) (see the graph in CCC 16). In fact, an easy calculation gives cosh 2 t sinh 2 t = 1. From the efinitions an t ekt = ke kt for constants k we get immeiately: sinh(t) = cosh(t), cosh(t) = sinh(t) t t an of course the corresponing anti-erivative formulas: cosh tt = sinh t + C, sinh tt = cosh t + C We can use this to practice how to fin erivatives of inverse functions. Starting from (switching to variable x for fun!) cosh(cosh 1 (x)) = x by taking erivatives on both sies an applying the chain rule on the left sie we get sinh(cosh 1 (x)) x cosh 1 (x) = 1 or 1 x cosh 1 (x) = sinh(cosh 1 (x)) From cosh 2 x sinh 2 x = 1 we get sinh 2 x = cosh 2 x 1 an so sinh x = cosh 2 x 1. (Note that cosh 2 x > 1 for all x. So the omain of cosh 1 x is [1, ]. Moreover, in orer to actually have cosh(x) an invertible function we nee to restrict the omain of cosh(x) an so the range of cosh 1 x to x 0. Since we have x 0 the positive root suffices above!). Using cosh 2 (cosh 1 (x)) = (cosh(cosh 1 (x))) 2 = x 2 it follows: x cosh 1 x = 1 x2 1 42

44 CCC 16 Key concepts: Implicit Differentiation Take a look at: How to fin erivatives of implicit functions Often curves are not given as the graph of a function but by an equation in for example x, y. So the equation x 2 y 2 = 1 escribes a hyperbola, which is the set of all points with coorinates (x, y) satisfying the equation. Below we see the graph an the tangent line at the point (2, 3). How o we fin the equation of the tangent line? In this case we can solve for y = ± x 2 1 an analyze the erivatives for each branch separately. Implicit ifferentiation just replaces y = f(x) an uses the chain rule when taking the erivative of the equation: so an we get We can solve for f (x) an get x 2 f(x) 2 = 1 x (x2 f(x) 2 ) = x 1 2x 2f(x) f (x) = 0 f (x) = y = x y 43

45 For y = x 2 1 this gives y x =, an for y = x x 2 1 we get y = x 2 1 These are just the answers we woul have gotten irectly. We can easily fin the equation of a tangent line let s say at (2, 3) by using y = 2 3 an get y = (x 2) or y = 2 3 x 1 3 x 2 1. There are more ifficult curves when it is impossible or impractical to solve for y as a function of x, like x sin x + y sin y = 1 Some iea how ifficult this shoul be you can get from looking at some picture of it For example we might be intereste in fining the points where the tangent line is horizontal. Implicit ifferentiation, which essentially is just application of the chain rule, allows us to o so. We begin with the equation x sin x+y sin(y) = 1 an take erivatives with respect to x on both sies. x x sin x + x y sin y = x (1) 44

46 an get by pretening that y = f(x) so by using the chain rule sin x + x cos x + f(x) sin f(x) = 0, x sin x + x cos x + f (x) sin f(x) + f(x) cos f(x) f (x) = 0 To simplify notation we usually on t even write f(x) but write y for it: We can solve this for y to get sin x + x cos x + y sin(y) + y cos y y = 0. y x cos x + sin x = sin y + y cos y A fractions is zero when the numerator is zero an the enominator is 0. So we have a horizontal tangent whenever x = 0. Here is the graph of y = x cos x + sin x: Note that is a non-trivial task to fin y-values even corresponing to x = 0. We see from the graph that there are many values of y satisfying y sin y = 1 or sin y = 1 y Note that this is also quite clear from the formula: If y gets large then 1 y gets small (in fact also for negative y), an sin y runs through all numbers 45

47 in [ 1, 1] again an again. So no woner that y sin y = 1 occurs again an again. But it is ifficult to fin the y-values exactly. (0, ) is almost on the graph: You can check that sin(1.1111) We will get back to the problem of fining approximate solutions, also using Calculus tools, later in the course. If we try to fin other values of x with x cos x + sin x = 0 we can plot the graph an see that there shoul be an approximate solution for x = 2. By asking a computer we get x = for an approximate solution. 46

48 Key concepts: Relate Rates Take a look at: CCC 17 A point moving on a graph, Hot air balloon problem. Check out more youtube vieos on relate rates! Suppose a particle is moving along the ellipse x 2 +4y 2 = 1 an it s horizontal velocity (the x-component of its velocity vector) is x t = 12 when x = 0.5 an y = Assume that we measure lengths in meters an time in secons. We want to know the vertical velocity at that point. We know that the velocity vector will point in the irection of the tangent line an the x- component of the vector is positive, so the vector will point to the right han sie. So we know some x for the tangent line, an after calculating the slope of the tangent line, we can fin the corresponing y. This is one way to think about it an we will follow this way to check our thinking below. But let s look at the motion of the particle in the plane given by (x(t), y(t)), where x, y are relate by the equation above, but we actually on t have any further information about (x(t), y(t)). But the relation between the position variables will lea to a relation between the corresponing velocities, or rates of change. Problems asking how rates of change are relate for quantities relate by equations are calle Relate rates problems. So the problem above asks to fin y t P if we assume that x t = 12, P where P (0.5, ). By implicit ifferentiation we get 2xx + 8yy = 0 an so y = 1 xx 4 y. Here the prime is ifferentiation with respect to t. So at the point P we get y =

49 Note that if we ha first calculate the slope of the tangent line by implicit ifferentiation 2x + 8y y x = 0 an solve: y x = 1 x 4 y, so we get y x P = We know if there are no forces anymore acting on the moving particle to hol it on the curve it will keep on moving along the line with constant velocity. Then, within t = 1 secon it will move x = 12 meter. so, using that the slope of the tangent line is y x = y P x above we get that the particle moves y = 12 ( 1 2 ) = 6 3 3, as expecte. It is important to note that in the situation above the curve x 2 + 4y 2 = 1 is not the position graph of the function (note that this woul mean the particle have to be at two points at the same time). The horizontal x-axis is not time but a space coorinate. Thus it is important to ifferentiate carefully the meaning of y x from the velocity of the moving particle. Another typical Relate Rates problem is the so calle Laer problem: Here is a typical question: A 12 m laer is leaning against a wall (long laer, maybe someboy cleaning a winow on a 4th floor). The laer 48