2.5 SOME APPLICATIONS OF THE CHAIN RULE

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1 2.5 SOME APPLICATIONS OF THE CHAIN RULE The Chain Rule will help us etermine the erivatives of logarithms an exponential functions a x. We will also use it to answer some applie questions an to fin slopes of graphs given by parametric equations. DERIVATIVES OF LOGARITHMS D( ln(x) ) = x an D( ln( g(x) ) ) = g '(x) g(x). Proof: We know that the natural logarithm ln(x) is the logarithm with base e, an e ln(x) = x for x > 0. We also know that D( e x ) = e x, so using the Chain Rule we have D( e f(x) ) = e f(x) f '(x). Differentiating each sie of the equation e ln(x) = x, we get that D( e ln(x) ) = D( x ) use D( e f(x) ) = e f(x). f '(x) with f(x) = ln(x) e ln(x). D( ln(x) ) = replace e ln(x) with x x. D( ln(x) ) = an solve for D( ln(x) ) to get D( ln(x) ) = x. The function ln( g(x) ) is the composition of f(x) = ln(x) with g(x), so by the Chain Rule, D( ln( g(x) ) = D( f( g(x) ) ) = f '( g(x) ). g '(x) = g(x). g '(x) = g '(x) g(x). Example : Fin D( ln( sin(x) ) ) an D( ln( x ) ). Solution: (a) Using the pattern D( ln( g(x) ) = g '(x) g(x) with g(x) = sin(x), then D( ln( sin(x) ) ) = g '(x) g(x) = D( sin(x) ) sin(x) = cos(x) sin(x) = cot(x). (b) Using the pattern with g(x) = x 2 + 3, we have D( ln( x ) ) = g '(x) g(x) = 2x x We can use the Change of Base Formula from algebra to rewrite any logarithm as a natural logarithm, an then we can ifferentiate the resulting natural logarithm. Change of Base Formula for logarithms: log a x = log b x log b a for all positive a, b an x. Example 2: Use the Change of Base formula an your calculator to fin log π 7 an log 2 8. Solution: log π 7 = ln 7 ln π (Check that π.7 7) log 2 8 = ln 8 ln 2 = 3. Practice : Fin the values of log 9 20, log 3 20 an log π e.

2 2 Putting b = e in the Change of Base Formula, log a x = log e x log e a = ln x ln a, so any logarithm can be written as a natural logarithm ivie by a constant. Then any logarithm is easy to ifferentiate. D( log a ( x ) ) = x ln(a) an D( log a ( f(x) ) ) = f '(x). f(x) ln(a) Proof: D( log a ( x ) ) = D( ln x ln a ) =. ln(a) D( ln x ) = ln(a). x = x ln(a). The secon ifferentiation formula follows from the Chain Rule. Practice 2: Calculate D( log 0 ( sin(x) ) ) an D( log π ( e x ) ). The number e might seem like an "unnatural" base for a natural logarithm, but of all the logarithms to ifferent bases, the logarithm with base e has the nicest an easiest erivative. The natural logarithm is even relate to the istribution of prime numbers. In 896, the mathematicians Haamar an Valle Poussin prove the following conjecture of Gauss: (The Prime Number Theorem) For large values of x, x {number of primes less than x} ln(x). DERIVATIVE OF a x Once we know the erivative of e x an the Chain Rule, it is relatively easy to etermine the erivative of a x for any a > 0. D( a x ) = a x. ln a for a > 0. Proof: If a > 0, then a x > 0 an a x = e ln( ax ) = e x. ln a. D( a x ) = D( e ln( ax ) ) = D( e x. ln a ) = e x. ln a. D( x. ln a ) = a x. ln a. Example 3: Calculate D( 7 x ) an ( 2 sin(t) ) Solution: (a) D( 7 x ) = 7 x ln 7 (.95) 7 x. (b) We can write y = 2 sin(t) as y = 2 u with u = sin(t). Using the Chain Rule, = u. u = 2u. ln(2). sin(t) cos(t) = 2. ln(2). cos(t). Practice 3: Calculate D( sin( 2 x ) ) an ( 3( t2 ) ).

3 3 SOME APPLIED PROBLEMS Now we can examine applications which involve more complicate functions. Example 4: A ball at the en of a rubber ban (Fig. ) is oscillating up an own, an its height (in feet) above the floor at time t secons is h(t) = sin( t/2 ). (t is in raians) (a) How fast is the ball travelling after 2 secons? after 4 secons? after 60 secons? (b) Is the ball moving up or own after 2 secons? after 4 secons? after 60 secons? (c) Is the vertical velocity of the ball ever 0? Solution: (a) v(t) = D( h(t) ) = D( sin( t/2 ) ) = 2 cos( t/2 ) D( t/2 ) = cos( t/2 ) feet/secon so v( 2 ) = cos( 2/2 ) ft/s, v( 4 ) = cos( 4/2 ) 0.46 ft/s, an v( 60 ) = cos( 60/2 ) 0.54 ft/s. (b) The ball is moving upwar when t = 2 an 60 secons, ownwar when t = 4. (c) v( t ) = cos( t/2 ) an cos( t/2 ) = 0 when t = π ± n. 2π (n =, 2,... ). Example 5: If 2400 people now have a isease, an the number of people with the isease appears to ouble every 3 years, then the number of people expecte to have the isease in t years is y = t/3. (a) How many people are expecte to have the isease in 2 years? (b) When are 50,000 people expecte to have the isease? (c) How fast is the number of people with the isease expecte to grow now an 2 years from now? Solution: (a) In 2 years, y = /3 3,80 people. (b) We know y = 50,000, an we nee to solve 50,000 = t/3 for t. Taking logarithms of each sie of the equation, ln(50,000) = ln( t/3 ) = ln(2400) + (t/3). ln(2) so 0.89 = t an t 3.4 years. We expect 50,000 people to have the isease about 3.4 years from now. (c) This is asking for / when t = 0 an 2 years. = ( t/3 ) = t/3. ln(2). (/3) t/3. Now, at t = 0, the rate of growth of the isease is approximately people/year. In 2 years the rate of growth will be approximately /3 880 people/year. Example 6: You are riing in a balloon, an at time t (in minutes) you are h(t) = t + sin(t) feet high. If the temperature at an elevation h is 72 T(h) = + h egrees Fahrenheit, then how fast is your temperature changing when t = 5 minutes? (Fig. 2)

4 4 Solution: As t changes, your elevation will change, an, as your elevation changes, so will your temperature. It is not ifficult to write the temperature as a function of time, an then we coul calculate T(t) = T '(t) an evaluate T '(5), or we coul use the Chain Rule: T(t) = T( h(t) ) h(t). h(t) = T( h ) h. h(t) = When t = 5, then h( t ) = 5 + sin( 5 ) 4.04 so T '(5) 72 ( + h) 2. ( + cos( t ) ). 72 ( ) 2. ( +.284) 3.64 o /minute. Practice 4: Write the temperature T in the previous example as a function of the variable t alone an then ifferentiate T to etermine the value of T/ when t = 5 minutes. Example 7: A scientist has etermine that, uner optimum conitions, an initial population of 40 bacteria will grow "exponentially" to f(t) = 40. e t/5 bacteria after t hours. (a) Graph y = f(t) for 0 t 5. Calculate f(0), f(5), f(0). (b) How fast is the population increasing at time t? (Fin f '(t).) (c) Show that the rate of population increase, f '(t), is proportional to the population, f(t), at any time t. (Show f '(t) = K. f(t) for some constant K.) Solution: (a) The graph of y = f(t) is given in Fig. 3. f(0) = 40. e 0/5 = 40 bacteria. f(5) = 40. e 5/5 09 bacteria, an f(0) = 40. e 0/5 296 bacteria. (b) f '(t) = ( f(t) ) = ( 40. e t/5 ) = 40. e t/5 ( t/5 ) = 40. e t/5 ( /5 ) = 8. e t/5 bacteria/hour. (c) f '(t) = 8. e t/5 = 5. ( 40. e t/5 ) = 5 f(t) so f '(t) = K. f(t) with K = /5. PARAMETRIC EQUATIONS Suppose a robot has been programme to move in the xy plane so at time t its x coorinate will be sin(t) an its y coorinate will be t 2. Both x an y are functions of the inepenent parameter t, x(t) = sin(t) an y(t) = t 2, an the path of the robot (Fig. 4) can be foun by plotting (x,y) = ( x(t), y(t) ) for lots of values of t. t x(t) = sin(t) y(t) = t 2 plot point at (0, 0) (.48,.25).0.84 (.84, ) (, 2.25) (.9, 4)

5 5 Typically we know x(t) an y(t) an nee to fin /, the slope of the tangent line to the graph of ( x(t), y(t) ). The Chain Rule says that =., so, algebraically solving for, we get = / /. If we can calculate / an /, the erivatives of y an x with respect to the parameter t, then we can etermine /, the rate of change of y with respect to x. If then x = x(t) an y = y(t) are ifferentiable with respect to t, an 0, = / /. Example 8: Fin the slope of the tangent line to the graph of (x,y) = ( sin(t), t 2 ) when t = 2? Solution: / = cos(t) an / = 2t. When t = 2, the object is at the point ( sin(2), 2 2 ) (.9, 4 ) an the slope of the tangent line to the graph is = / / = 2t cos(t) = 2. 2 cos(2) Practice 5: Graph (x,y) = ( 3cos(t), 2sin(t) ) an fin the slope of the tangent line when t = π/2. When we calculate an, the slope of the tangent line to the graph of ( x(t), y(t) ), we use the erivatives, an each of these erivatives also has a geometric meaning: measures the rate of change of x(t) with respect to t -- it tells us whether the x-coorinate is increasing or ecreasing as the t-variable increases. measures the rate of change of y(t) with respect to t. Example 9: For the parametric graph in Fig. 5, tell whether, is positive or negative when t=2. an Solution: As we move through the point B (where t=2) in the irection of increasing values of t, we are moving to the left so x(t) is ecreasing an Similarly, the values of y(t) are increasing so line,, is negative. (As check on the sign of we can also use the result is negative. is positive. Finally, the slope of the tangent / = / = positive negative = negative.)

6 6 Practice 6: For the parametric graph in the previous example, tell whether is positive or negative when t= an when t=3., an Spee If we know the position of an object at every time, then we can etermine its spee. The formula for spee comes from the istance formula an looks a lot like it, but with erivatives. If x = x(t) an y = y(t) give the location of an object at time t an are ifferentiable functions of t, then the spee of the object is ( )2 + ( )2. Proof: The spee of an object is the limit, as t 0, of change in position change in time. (Fig. 6 ) change in position change in time = ( x) 2 + ( y) 2 t = ( x) 2 + ( y) 2 ( t) 2 = ( x t )2 + ( y t )2 ( )2 + ( )2 as t 0. Exercise 0: Fin the spee of the object whose location at time t is (x,y) = ( sin(t), t 2 ) when t = 0 an t =. Solution: / = cos(t) an / = 2t so spee = (cos(t)) 2 + (2t) 2 = cos 2 (t) + 4t 2. When t = 0, spee = cos 2 (0) + 4(0) 2 = + 0 =. When t =, spee = cos 2 () + 4() 2 = Practice 7: Show that an object whose location at time t is (x,y) = ( 3sin(t), 3 cos(t) ) has a constant spee. (This object is moving on a circular path.) Practice 8: Is the object whose location at time t is (x,y) = ( 3cos(t), 2sin(t) ) travelling faster at the top of the ellipse ( at t = π/2) or at the right ege of the ellipse (at t = 0)?

7 7 PROBLEMS Differentiate the functions in problems 9.. ln( 5x ) 2. ln( x 2 ) 3. ln( x k ) 4. ln( x x ) = x. ln(x) 5. ln( cos(x) ) 6. cos( ln(x) ) 7. log 2 5x 8. log 2 kx 9. ln( sin(x) ) 0. ln( kx ). log 2 ( sin(x) ) 2. ln( e x ) 3. log 5 5 x 4. ln( e f(x) ) 5. x. ln( 3x ) 6. e x. ln(x) 7. ln(x) x 8. x + ln(3x) 9. ln( 5x 3 ) 20. ln( cos(t) ) 2. w cos( ln( w ) ) 22. ln( ax + b ) 23. ln( t + ) 24. D( 3x ) 25. D( 5 sin(x) ) 26. D( x. ln(x) x ) 27. ln( sec(x) + tan(x) ) 28. Fin the slope of the line tangent to f(x) = ln( x ) at the point ( e, ). Fin the slope of the line tangent to g(x) = e x at the point (, e). How are the slopes of f an g at these points relate? 29. Fin a point P on the graph of f(x) = ln(x) so the tangent line to f at P goes through the origin. 30. You are moving from left to right along the graph of y = ln( x ) (Fig. 7). (a) If the x coorinate of your location at time t secons is x(t) = 3t +2, then how fast is your elevation increasing? (b) If the x coorinate of your location at time t secons is x(t) = e t, then how fast is your elevation increasing? 3. Rumor. The percent of a population, p(t), who have hear a rumor by time t is often moele 00 p(t) = + Ae t = 00 ( + Ae t ) for some positive constant A. Calculate how fast the rumor is spreaing, p(t). 32. Raioactive ecay. If we start with A atoms of a raioactive material which has a "half life" (the amount of time for half of the material to ecay) of 500 years, then the number of raioactive atoms left after t years is r(t) = A. e Kt ln(2) where K = 500. Calculate r '(t) an show that r '(t) is proportional to r(t) ( r '(t) = b. r(t) for some constant b).

8 8 In problems 33 4, fin a function with the given erivative. 33. f '(x) = 8 x 34. h '(x) = 3 3x f '(x) = cos(x) 3 + sin(x) 36. g '(x) = x + x g '(x) = 3e 5x 38. h '(x) = e f '(x) = 2x. e (x2 ) 40. g '(x) = cos(x). e sin(x) 4. h '(x) = cos(x) sin(x) 42. Define A(x) to be the area boune between the x axis, the graph of f(x), an a vertical line at x (Fig. 8). The area uner each "hump" of f is 2 square inches. (a) Graph A(x) for 0 x 9. (b) Graph A '(x) for 0 x 9. Problems involve parametric equations. 43. At time t minutes, robot A is at ( t, 2t + ) an robot B is at ( t 2, 2t 2 + ). (a) Where is each robot when t=0 an t =? (b) Sketch the path each robot follows uring the first minute. (c) Fin the slope of the tangent line, /, to the path of each robot at t = minute. () Fin the spee of each robot at t = minute. (e) Discuss the motion of a robot which follows the path ( sin(t), 2sin(t) + ) for 20 minutes. 44. x(t) = t +, y(t) = t 2. (a) Graph ( x(t), y(t) ) for t 4. (b) Fin /, /, the tangent slope /, an spee when t = an t = For the parametric graph in Fig. 9, 46. For the parametric graph in Fig. 0, etermine whether etermine whether, are positive, negative or zero when an t = an t = 3. t = an t = 3., are positive, negative or zero when an

9 9 47. x(t) = R. (t sin(t) ), y(t) = R. ( cos(t) ). (a) Graph ( x(t), y(t) ) for 0 t 4π. (b) Fin /, /, the tangent slope /, an spee when t = π/2 an π. (The graph of ( x(t), y(t) ) is calle a cycloi an is the path of a light attache to the ege of a rolling wheel with raius R.) 48. Describe the motion of two particles whose locations at time t are ( cos(t), sin(t) ) an ( cos(t), sin(t) ). 49. Describe the path of a robot whose location at time t is (a) ( 3. cos(t), 5. sin(t) ) (b) ( A. cos(t), B. sin(t) ) (c) Give the parametric equations so the robot will move along the same path as in part (a) but in the opposite irection. 50. After t secons, a projectile hurle with initial velocity v an angle θ will be at x(t) = v. cos(θ). t feet an y(t) = v. sin(θ). t 6t 2 feet. (Fig. ) (This formula neglects air resistance.) (a) For an initial velocity of 80 feet/secon an an angle of π/4, fin t > 0 so y(t) = 0. What oes this value for t represent physically? Evaluate x(t). (b) For v an θ in part (a), calculate /. Fin t so / = 0 at t, an evaluate x(t). What oes x(t) represent physically? (c) What initial velocity is neee so a ball hit at an angle of π/ will go over a 40 foot high fence 350 feet away? () What initial velocity is neee so a ball hit at an angle of 0.7 will go over a 40 foot high fence 350 feet away?

10 0 Section 2.5 PRACTICE Answers Practice : log 9 20 = log(20) log(9) ln( 20 ) ln( 9 ), log log(20) 3 20 = log(3) ln( 20 ) ln( 3 ) log π e = log(e) log(π) ln(e) ln(π) = ln(π) Practice 2: D( log 0 ( sin(x) ) ) = sin(x). ln(0) D( sin(x) ) = cos(x) sin(x). ln(0) D( log π ( e x ) ) = e x. ln(π) D( e x ) = e x e x. ln(π) = ln(π) Practice 3: D( sin( 2 x ) ) = cos( 2 x ) D( 2 x ) = cos( 2 x ). 2 x. ln(2) 3(t2 ) = 3 (t2 ) ln(3) D( t 2 ) = 3 (t 2 ) ln(3). 2t Practice 4: T = 72 + h = 72 + t + sin(t). T = ( + t + sin(t) ). D( 72 ) 72. D( + t + sin(t) ) ( + t + sin(t) ) 2 = 72( + cos(t)) ( + t + sin(t) ) 2 When t = 5, T = 72( + cos(5)) ( + t + sin(5) ) Practice 5: x(t) = 3cos(t) so / = 3sin(t). y(t) = 2sin(t) so / = 2cos(t). = / / = 2cos(t) 3sin(t). When t = π/2, = 2cos( π/2 ) 3sin( π/2 ) = = 0. (See Fig. 2) Practice 6: When x=: pos., pos., pos. When x=3: pos., neg., neg. Practice 7: x(t) = 3sin(t) an y(t) = 3cos(t) so / = 3cos(t) an / = 3sin(t). Then spee = ( / ) 2 + ( / ) 2 = (3cos(t) ) 2 + ( 3sin(t) ) 2 = 9. cos 2 (t) + 9. sin 2 (t) = 9 = 3, a constant. Practice 8: x(t) = 3cos(t) an y(t) = 2sin(t) so / = 3sin(t) an / = 2cos(t). Then spee = ( / ) 2 + ( / ) 2 = ( 3sin(t) ) 2 + ( 2cos(t) ) 2 = 9sin 2 (t) + 4cos 2 (t). When t = 0, the spee is 9. (0) () 2 = 2. When t = π/2, the spee is 9. () (0) 2 = 3 (faster).