Solucao Calculo 5ed CAP.3

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(1)Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions. h. e 1 =1 . 1. (a) e is the number such that lim h 0 h (b) x. x. x. (2.7 1)/x. x. (2.8 1)/x. 0.001. 0.9928. 0.001. 1.0291. 0.0001. 0.9932. 0.0001. 1.0296. 0.001. 0.9937. 0.001. 1.0301. 0.0001. 0.9933. 0.0001. 1.0297 h. h. 2.7 1 2.8 1 =0.99 and lim =1.03 . Since From the tables (to two decimal places), lim h h h 0 h 0 0.99<1<1.03 , 2.7<e<2.8 .. 2. (a) The function value at x=0 is 1 and the slope at x=0 is 1 . x. e. (b) f (x)=e is an exponential function and g(x)=x is a power function.. d x x d e e1 (e )=e and (x )=ex dx dx. . x. e. (c) f (x)=e grows more rapidly than g(x)=x when x is large. /. 3. f (x)=186.5 is a constant function, so its derivative is 0 , that is, f (x)=0 . /. 4. f (x)= 30 is a constant function, so its derivative is 0 , that is, f (x)=0 . /. 5. f (x)=5x1 f (x)=50=5 10. /. 10 1. 6. F(x)=4x F (x)=4(10x 2. /. )=40x. 9. 2 1. 7. f (x)=x +3x4 f (x)=2x. +30=2x+3 1.

(2) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions. 8. 5. /. 8 1. 5 1. 7. 4. 8. 5. /. 8 1. 5 1. 7. 4. 8. g(x)=5x 2x +6 g (x)=5(8x 9. g(x)=5x 2x +6 g (x)=5(8x 10. f (t)=. 11. y=x. )2(5x )2(5x. )+0=40x 10x )+0=40x 10x. 1 6 4 1 5 / 3 5 3 t 3t +t f (t)= (6t )3(4t )+1=3t 12t +1 2 2. 2/5. /. y =. 2 (2/5)1 2 7/5 x = x = 5 5. /. x. x. 12. y=5e +3 y =5(e )+0=5e. 2 5x. 7/5. x. 4 3 4 / 2 2 13. V (r)= r V (r)= (3r )=4 r 3 3 14. R(t)=5t. 3/5. /. R (t)=5. 9. /. 15. Y (t)=6t Y (t)=6(9)t 7. 10. 16. R(x)=. x. 3 (3/5) 1 8/5 t =3t 5. . 10. =54t. 8. /. = 10 x R (x)=7 10 x =. 7. 1/2. x. 1/3. /. 18. y= x =x y =. 1 x 2. 19. F(x)=. 20. f (t)= t 2. 21. g(x)=x +. 7 10 x. /. x. 17. G(x)= x 2e =x 2e G (x)= 3. 10. 5. =. 1 2/3 x = 3 1 2. 5 5. 8. 1 1/2 x 1 x x 2e = 2e 2 2 x. 1 2/3. 3x. x=. 1 5 1 5 4 / 4 x F (x)= (5x )= x 32 32 32. 1 1/2 1/2 1 1/2 1 3/2 1 1 / =t t f (t)= t t = + 2 2 t 2 t 2t t 1 2. x. 2. 2. /. 3. =x +x g (x)=2x+(2)x =2x. 2 x. 3. 2.

(3) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions. 22. y= x (x1)=x. 3/2. 1/2. /. x y =. 3 1/2 1 1/2 1 1/2 1 1/2 x x = x (3x1) [ factor out x ] 2 2 2 2. 3x1 . 2 x. /. or y =. 2. x +4x+3 3/2 1/2 1/2 =x +4x +3x 23. y= x 1 1 2 3 / 3 1/2 1/2 3/2 3 x +3 x = x+ y = x +4 2 2 2 2 x 2x x 3/2. 2/2. 1/2. note that x =x x =x x 2. x 2 x 1/2 / =x2x y =12 24. y= x 2. /. . 1 2. x. 3/2. =1+1/(x x ). 2. 25. y=4 y =0 since 4 is a constant. 1 1/2 u = 2 + 3 /(2 u ) 2. /. 26. g(u)= 2 u+ 3u = 2 u+ 3 u g (u)= 2 (1)+ 3 2. /. 27. y=ax +bx+c y =2ax+b v. 28. y=ae +. b c / v 2 3 v b v 1 2 2c + =ae +bv +cv y =ae bv 2cv =ae 2 2 3 v v v v. 4. t 3. 2 3/4. 1. 2. 29. v=t . 3. =t t. 2. 30. u=. 3. 2/3. /. v =2t . t +2 t =t +2t A. 31. z=. y. 10. /. u =. y. t. 7/4. 2 1/3 t +2 3 11. /. 3 2 y. +Be =. y 32. y=e. 3. =2t+. 4t. +Be z =10Ay. +Be =Ay. 10. 3/2. 3 4. t. 7/4. 3. =2t+. 4. 4t. 1/2. =. 2. t. 3. +3 t. 3. 3 t 10A 11. +Be. y. y x 1. x+1. x. /. x. +1=e e +1=e e +1 y =e e =e x. /. x+1. x. 33. f (x)=e 5x f (x)=e 5 . 3.

(4) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions. /. Notice that f (x)=0 when f has a horizontal tangent, f negative when f is decreasing. 5. 3. /. 4. /. is positive when f is increasing, and f. /. is. 2. 34. f (x)=3x 20x +50x f (x)=15x 60x +50 .. /. Notice that f (x)=0 when f has a horizontal tangent and that f odd function. 15. 3. /. 14. /. is an even function while f is an. 2. 35. f (x)=3x 5x +3 f (x)=45x 15x .. /. Notice that f (x)=0 when f has a horizontal tangent, f negative when f is decreasing. 1. /. 2. /. is positive when f is increasing, and f. /. is. 2. 36. f (x)=x+1/x=x+x f (x)=1x =11/x .. /. Notice that f (x)=0 when f has a horizontal tangent, f. /. is positive when f is increasing, and f. /. is 4.

(5) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions. negative when f is decreasing. /. 2. 3. 37. To graphically estimate the value of f (1) for f (x)=3x x , we’ll graph f in the viewing rectangle [10.1,1+0.1] by [ f (0.9),f(1.1)] , as shown in the figure. If we have sufficiently zoomed in on the graph of f , we should obtain a graph that looks like a diagonal line; if not, graph again with 10.01 and 1+0.01 , etc. 2.2991.701 0.589 / Estimated value: f (1) = =2.99 . 1.10.9 0.2 2. 3. /. 2. /. Exact value: f (x)=3x x f (x)=6x3x , so f (1)=63=3 .. 38. See the previous exercise. Since f is a decreasing function, assign Y (3.9) to Y 1. Y. min. and Y (4.1) to 1. .. 0.493860.50637 0.01251 = =0.06255 . 4.13.9 0.2 1 3/2 1 3/2 1 1/2 / / Exact value: f (x)=x f (x)= x , so f (4)= (4 )= 2 2 2 /. Estimated value: f (4). 4. x. /. 3. 1 8. =. 1 =0.0625 . 16. /. x. 39. y=x +2e y =4x +2e . At (0, 2) , y =2 and an equation of the tangent line is y2=2(x0) or y=2x+2. 2. 2. /. /. 40. y=(1+2x) =1+4x+4x y =4+8x. At (1, 9), y =12 and an equation of the tangent line is y9=12(x1) or y=12x3. 2. 3. /. 2. /. 41. y=3x x y =6x3x . At (1, 2) , y =63=3 , so an equation of the tangent line is y2=3(x1) , or y=3x1 .. 5.

(6) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions. 3 1/2 / 3 x . At (4, 8) , y = (2)=3, so an equation of the tangent line is 2 2 y8=3(x4), or y=3x4 .. 42. y=x x =x. 3/2. /. y =. 43. (a). (b) From the graph in part (a), it appears that f negative (so f. /. /. is zero at x 1.25 , x 0.5 , and x 3 . The slopes are 1. 2. 3. is negative) on ( , x ) and (x , x ). The slopes are positive (so f 1. 2. 3. /. is positive) on. (x , x ) and (x , ). 1. 2. 3. 4. 3. 2. /. 3. 2. (c) f (x)=x 3x 6x +7x+30 f (x)=4x 9x 12x+7. 44. (a). 6.

(7) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions. (b) From the graph in part (a), it appears that f (so f. /. /. is zero at x 0.2 and x 2.8 . The slopes are positive 1. 2. is positive) on ( , x ) and (x , ) . The slopes are negative (so f 1. x. 2. /. 2. /. is negative) on (x , x ) . 1. 2. x. (c) g(x)=e 3x g (x)=e 6x. 3. 2. /. 2. 45. The curve y=2x +3x 12x+1 has a horizontal tangent when y =6x +6x12=0 2. 6(x +x2)=0 6(x+2)(x1)=0 x=2 or x=1 . The points on the curve are (2, 21) and (1, 6) . 3. 2. /. 2. 46. f (x)=x +3x +x+3 has a horizontal tangent when f (x)=3x +6x+1=0 6 3612 1 x= =1. 6 . 6 3 3. /. 2. 2. 47. y=6x +5x3 m=y =18x +5 , but x 0 for all x , so m 5 for all x . x. /. x. 48. The slope of y=1+2e 3x is given by m=y =2e 3 . The slope of 3xy=5 y=3x5 is 3 . x. x. m=3 2e 3=3 e =3 x=ln 3 . This occurs at the point (ln 3, 73ln 3) (1.1, 3.7) . 7.

(8) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions. 49.. 2. Let (a, a ) be a point on the parabola at which the tangent line passes through the point (0, 4). The 2. tangent line has slope 2a and equation y(4)=2a(x0) y=2ax4 . Since (a, a ) also lies on the line, 2. 2. a =2a(a)4, or a =4 . So a= 2 and the points are (2, 4) and (2, 4). 2. /. 2. 50. If y=x +x , then y =2x+1 . If the point at which a tangent meets the parabola is (a, a +a), then the slope of the tangent is 2a+1 . But since it passes through (2, 3), the slope must also be 2. y a +a+3 = . x a2 2. a +a+3 2 2 . Solving this equation for a we get a +a+3=2a 3a2 Therefore, 2a+1= a2 2. a 4a5=(a5)(a+1)=0 a=5 or 1 . If a=1 , the point is (1, 0) and the slope is 1, so the equation is y0=(1)(x+1) or y=x1. If a=5, the point is (5, 30) and the slope is 11, so the equation is y30=11(x5) or y=11x25. 2. /. 2. /. /. 51. y= f (x)=1x f (x)=2x, so the tangent line at (2, 3) has slope f (2)=4. The normal line has 1 1 1 1 7 = and equation y+3= (x2) or y= x . slope 4 4 4 4 2. /. 52. y= f (x)=xx f (x)=12x. So f (1)=1, and the slope of the normal line is the negative 8.

(9) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions. reciprocal of that of the tangent line, that is, 1/(1)=1. So the equation of the normal line at (1, 0) is 2. y0=1(x1) y=x1. Substituting this into the equation of the parabola, we obtain x1=xx x= 1. The solution x=1 is the one we require. Substituting x=1 into the equation of the parabola to find the y coordinate, we have y=2. So the point of intersection is (1, 2), as shown in the sketch.. 53. 1 1 x(x+h) x+h x =lim h h 0 hx(x+h). /. f (x) =lim f (x+h) f (x) =lim h h 0 h 0 =lim h 0. h 1 1 =lim = 2 hx(x+h) h 0 x(x+h) x 2. 54. Substituting x=1 and y=1 into y=ax +bx gives us a+b=1 (1) . The slope of the tangent line /. /. y=3x2 is 3 and the slope of the tangent to the parabola at (x, y) is y =2ax+b . At x=1 , y =3 3=2a+b (2) . Subtracting (1) from (2) gives us 2=a and it follows that b=1 . The parabola has 2. equation y=2x x . 2. 55. f (x)=2x if x

(10) 1 and f (x)=x 2x+2 if x>1 . Now we compute the right and lefthand derivatives defined in Exercise : f (1+h) f (1) 2(1+h)1 h / f (1)=lim =lim =lim =lim 1=1 and h h h h 0. h 0. h 0. h 0. 2. 2. f (1+h) f (1) (1+h) 2(1+h)+21 h f (1)=lim =lim =lim =lim h=0 . + h h + + + h + /. h 0. h 0. h 0. /. /. /. h 0. /. Thus, f (1) does not exist since f (1) f (1) , so f is not differentiable at 1 . But f (x)=1 for x<1 . +. /. and f (x)=2x2 if x>1 .. 9.

(11) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions. 56. g(x)= lim. . h 0. {. 12x if x<1 2 if 1

(12) x

(13) 1 x if x>1 x. g(1+h)g(1) 12(1+h) 1 2h =lim =lim =lim ( 2 ) =2 and h h h h 0. h 0. h 0. 2. lim +. h 0. 2. g(1+h)g(1) (1+h) 1 2h+h =lim =lim =lim (2+h)=2, h h h + + + h 0. h 0. h 0. /. so g is differentiable at 1 and g (1)=2. 2. 2. g(1)g(1) (1) 1 2h+h lim =lim =lim =lim (2+h)=2 and h h h . h 0. h 0. h 0. h 0. g(1)g(1) (1)1 h / lim =lim =lim =lim 1=1 , so g (1) does not exist. h h + + + h +. h 0. h 0. h 0. h 0. /. Thus, g is differentiable except when x=1 , and g (x)=. 2. {. 2 if x<1 2x if 1

(14) x<1 1 if x>1. 2. 57. (a) Note that x 9<0 for x <9 | x|<3 3<x<3 . So f (x) =. {. 2. x 9. if x

(15) 3. x +9. if 3<x<3. x 9. ifx 3. 2. 2. /. f (x) =. {. 2x if x<3 2x if 3<x<3 2x if x>3. =. {. 2x if | x|>3 2x if | x|<3. To show that 10.

(16) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions. /. f (3) does not exist we investigate lim h 0. f (3+h) f (3) by computing the left and righthand h. derivatives defined in Exercise . 2. f (3+h) f (3) [(3+h) +9]0 =lim =lim (6h)=6 and f (3)=lim h h /. h 0. h 0. h 0. 2. 2. f (3+h) f (3) (3+h) 9 0 6h+h =lim =lim =lim (6+h)=6 . f (3)=lim + h h h + + + + /. h 0. h 0. h 0. Since the left and right limits are different, lim h 0. h 0. f (3+h) f (3) / does not exist, that is, f (3) does not h. /. exist. Similarly, f (3) does not exist. Therefore, f is not differentiable at 3 or at 3 .. (b) 58. If x 1 , then h(x)=| x1|+| x+2|=x1+x+2=2x+1. If 2<x<1 , then h(x)=(x1)+x+2=3. If x

(17) 2, then h(x)=(x1)(x+2)=2x1. Therefore,. h(x)=. /. To see that h (1)=lim x 1. {. 2x1 if x

(18) 2 3 if 2<x<1 2x+1 if x 1. /. h (x)=. {. 2 if x<2 0 if 2<x<1 2 if x>1. h(x)h(1) h(x)h(1) 33 does not exist, observe that lim =lim =0 but x1 x1 31 x 1. x 1. h(x)h(1) 2x2 / lim =lim =2 . Similarly, h (2) does not exist. x1 + + x1 x 1. x 1. 11.

(19) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions. 2. /. 59. y= f (x)=ax f (x)=2ax. So the slope of the tangent to the parabola at x=2 is m=2a(2)=4a. The 1 slope of the given line, 2x+y=b y=2x+b, is seen to be 2, so we must have 4a=2 a= . So 2 1 2 when x=2, the point in question has y coordinate 2 =2. Now we simply require that the given 2 line, whose equation is 2x+y=b, pass through the point (2, 2): 2(2)+(2)=b b=2. So we must have 1 a= and b=2. 2 /. /. 60. f is clearly differentiable for x<2 and for x>2. For x<2, f (x)=2x, so f (2)=4. For x>2, . /. /. /. /. f (x)=m, so f (2)=m. For f to be differentiable at x=2, we need 4= f (2)= f (2)=m. So . +. +. f (x)=4x+b. We must also have continuity at x=2, so 4= f (2)=lim f (x)=lim (4x+b)=8+b. Hence, +. x 2. +. x 2. b=4. 3. 2. /. 2. 61. y= f (x)=ax +bx +cx+d f (x)=3ax +2bx+c. The point (2, 6) is on f , so f (2)=6 8a+4b2c+d=6 (1) . The point (2, 0) is on f , so f (2)=0 8a+4b+2c+d=0 (2) . Since there are /. /. horizontal tangents at (2, 6) and (2, 0), f ( 2)=0. f (2)=0 12a4b+c=0 (3) and /. f (2)=0 12a+4b+c=0 (4) . Subtracting equation (3) from (4) gives 8b=0 b=0. Adding (1) and (2) gives 8b+2d=6, so d=3 since b=0. From (3) we have c=12a, so (2) becomes 8a+4(0)+2(12a)+3=0 3 3 9 3=16a a= . Now c=12a=12 = and the desired cubic function is 16 16 4 3 3 9 y= x x+3. 16 4 c c 62. (a) xy=c y= . Let P= a, a x. /. . The slope of the tangent line at x=a is y (a)=. c 2. . Its. a c 2c 2c c c equation is y = (xa) or y= x+ . so its y intercept is . Setting y=0 gives x=2a, so 2 2 a a a a a 2c the x intercept is 2a. The midpoint of the line segment joining 0, and (2a, 0) is a c a, =P. a (b) We know the x and y intercepts of the tangent line from part (a), so the area of the triangle 1 1 1 bounded by the axes and the tangent is (base)(height)= xy= (2a)(2c/a)=2c, a constant. 2 2 2 12.

(20) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions. 1000. 63. Solution 1: Let f (x)=x. . Then, by the definition of a derivative, 1000. /. f (1)=lim x 1. f (x) f (1) x 1 =lim . But this is just the limit we want to find, and we know (from x1 x1 x 1 /. the Power Rule) that f (x)=1000x. 999. /. 1000. 999. , so f (1)=1000(1). =1000 . So lim x 1. 1000. 1)=(x1)(x. Solution 2: Note that (x 1000. lim x 1. x. 1 (x1)(x = lim x1 x 1 =. lim (x. 999. +x. 999. +x. 998. +x. 998. 999. +x. 998. +x. 997. 997. x. 1 =1000 . x1. 2. + +x +x+1). So. 2. +x + +x +x+1) x1. 997. 2. + +x +x+1) = 1+1+1+...+1+1+1. x 1. 1000 ones. = 1000, as above.. 64. In order for the two tangents to intersect on the y axis, the points of tangency must be at equal 2. distances from the y axis, since the parabola y=x is symmetric about the y axis. Say the points of 2. 2. 2. tangency are (a, a ) and (a, a ), for some a>0. Then since the derivative of y=x is dy/dx=2x, the 2. 2. lefthand tangent has slope 2a and equation ya =2a(x+a), or y=2axa , and similarly the right hand tangent line has equation 2. 2. 2. ya =2a(xa), or y=2axa . So the two lines intersect at (0, a ). Now if the lines are perpendicular, 1 2 1 then the product of their slopes is 1, so (2a)(2a)=1 a = a= . So the lines intersect at 4 2 1 0, . 4. 13.

(21) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates. dV dV dx 2 dx = =3x dt dx dt dt. 3. 1. V =x . 2. 2. (a) A= r . dA dA dr dr = =2 r dt dr dt dt. dA dr 2 =2 r =2 (30m)(1m/s)=60 m / s dt dt. (b). 3. 3. y=x +2x. dy dy dx dy 2 2 = =(3x +2)(5)=5(3x +2) . When x=2 , =5(14)=70 . dt dx dt dt. dx dy dx dy dx y dy +2y =0 x =y = . dt dt dt dt dt x dt dy dx 4 2 2 =6 , = (6)= 8 . When y=4 , x +4 =25 x= 3 . For dt dt 3 2. 2. 4. x +y =25 2x. dz dx dy dz 1 dx dy 2 2 2 =2x +2y = x +y . When x=5 and y=12 , z =5 +12 dt dt dt dt z dt dt dx dy dz 1 46 2 z =169 z= 13 . For =2 and =3, = (5 2+12 3)= . dt dt dt 13 13 2. 2. 2. 5. z =x +y 2z. dy dy dx 1 3 1/2 2 dx = = (1+x ) (3x ) = 6. y= 1+x dt dx dt 2 dt 3. we have 4=. 2. 3x. 2 1+x. 3. dx dy =4 when x=2 and y=3 , . With dt dt. dx 3(4) dx =2 cm / s. dt 2(3) dt. 7. (a) Given: a plane flying horizontally at an altitude of 1 mi and a speed of 500 mi / h passes directly over a radar station. If we let t be time (in hours) and x be the horizontal distance traveled by the plane (in mi), then we are given that dx/dt=500 mi / h. (b) Unknown: the rate at which the distance from the plane to the station is increasing when it is 2 mi from the station. If we let y be the distance from the plane to the station, then we want to find dy/dt when y=2 mi. (c) 2. 2. (d) By the Pythagorean Theorem, y =x +1 2y(dy/dt)=2x(dx/dt) . 3 dy x dx x dy 2 2 = = (500) . Since y =x +1 , when y=2 , x= 3 , so (e) = (500)=250 3 433 mi / h. dt y dt y 2 dt 8. (a) Given: the rate of decrease of the surface area is 1 cm 1.

(22) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates. 2. 2. / min. If we let t be time (in minutes) and S be the surface area (in cm ), then we are given that 2. dS/dt=1 cm / s. (b) Unknown: the rate of decrease of the diameter when the diameter is 10 cm. If we let x be the diameter, then we want to find dx/dt when x=10 cm.. (c) (d) If the radius is r and the diameter x=2r , then r=. 1 2 x and S=4 r =4 2. 1 x 2. 2. 2. = x . dS dS dx dx . = =2 x dt dx dt dt 1 dS dx dx 1 dx 1 (e) 1= . When x=10 , . So the rate of decrease is cm / =2 x = = 20 dt dt dt 2 x dt 20 min. 9. (a) Given: a man 6 ft tall walks away from a street light mounted on a 15 fttall pole at a rate of 5 ft / s. If we let t be time (in s) and x be the distance from the pole to the man (in ft), then we are given that dx/dt=5 ft / s. (b) Unknown: the rate at which the tip of his shadow is moving when he is 40 ft from the pole. If we let y be the distance from the man to the tip of d his shadow (in ft), then we want to find (x+y) when x=40 ft. dt. (c) 15 x+y 2 = 15y=6x+6y 9y=6x y= x . 6 y 3 d d 2 5 dx 5 25 (e) The tip of the shadow moves at a rate of ft / s. (x+y)= x+ x = = (5)= dt dt 3 3 dt 3 3 (d) By similar triangles,. 10. (a) Given: at noon, ship A is 150 km west of ship B; ship A is sailing east at 35 km / h, and ship B is sailing north at 25 km / h. If we let t be time (in hours), x be the distance traveled by ship A (in km), and y be the distance traveled by ship B (in km), then we are given that dx/dt=35 km / h and dy/dt=25 km / h. (b) Unknown: the rate at which the distance between the ships is changing at 4:00 P.M. If we let z be the distance between the ships, then we want to find dz/dt when t=4 h. 2.

(23) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates. (c) 2. 2. 2. (d) z =(150x) +y 2z. dz =2(150x) dt. . dx dt. +2y. dy dt 2. 2. (e) At 4:00 P.M., x=4(35)=140 and y=4(25)=100 z= (150140) +100 = 10,100 . So dz 1 dx dy 10(35)+100(25) 215 = (x150) +y = = 21.4 km / h. dt z dt dt 10,100 101. 11. dx dy dz dx dy 2 2 2 =60 mi / h and =25 mi / h. z =x +y 2z =2x +2y dt dt dt dt dt dz dx dy dz 1 dx dy . z =x +y = x +y dt dt dt dt z dt dt. We are given that. 2. 2. After 2 hours, x=2(60)=120 and y=2(25)=50 z= 120 +50 =130 , so dz 1 dx dy 120(60)+50(25) = x +y = =65 mi / h. dt z dt dt 130. 12. dx y 2 24 dy 24 dx 24 =1.6 m / s. By similar triangles, = y= = = (1.6) . 2 dt 2 dt 12 x x dt x x dy 24(1.6) When x=8 , = =0.6 m / s, so the shadow is decreasing at a rate of 0.6 m / s. dt 64 We are given that. 13. We are given that. dx dy dz 2 2 2 =4 ft / s and =5 ft / s. z =(x+y) +500 2z =2(x+y) dt dt dt. dx dy + dt dt. . 15 3.

(24) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates. minutes after the woman starts, we have x=(4ft/s)(20min)(60s/min)=4800 ft and y=5 15 60=4500 2. 2. z= (4800+4500) +500 = 86,740,000 , so dz x+y dx dy 4800+4500 837 = + = (4+5)= 8.99 ft / s . dt dt dt z 86,740,000 8674 14. We are given that. dx =24 ft / s. dt. (a) dx . dt dy 90x 2 2 When x=45 , y= 45 +90 =45 5 , so = dt y 2. 2. 2. y =(90x) +90 2y. dy =2(90x) dt. . . dx dt. so the distance from second base is decreasing at a rate of. =. 45 24 , (24)= 45 5 5. 24 10.7 ft / s. 5. (b) Due to the symmetric nature of the problem in part (a), we expect to get the same answer and we dz dz dx 45 24 2 2 2 do. z =x +90 2z =2x . When x=45 , z=45 5 , so = (24)= 10.7 ft / s. dt 45 5 dt dt 5 1 dh dA bh , where b is the base and h is the altitude. We are given that =1 cm / min and =2 2 dt dt dA 1 dh db 2 cm / min. Using the Product Rule, we have . When h=10 and A=100 , we = b +h dt 2 dt dt 1 1 1 db db have 100= b(10) b=10 b=20 , so 2= 20 1+10 4=20+10 2 2 2 dt dt db 420 = =1.6 cm / min. dt 10 15. A=. 16. dy dx dy dx dx y dy y 2 2 when x=8 m. y =x +1 2y =2x =1 m / s, find = = . When x=8 dt dt dt dt dt x dt x , y= 65 , so Given. 4.

(25) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates. 65 dx = . Thus, the boat approaches the dock at 8 dt. 65 1.01 m / s. 8. 17. We are given that. dx dy dz 2 2 2 =35 km / h and =25 km / h. z = ( x+y ) +100 2z =2(x+y) dt dt dt 2. dx dy + dt dt. .. 2. At 4:00 P.M., x=4(35)=140 and y=4(25)=100 z= (140+100) +100 = 67,600 =260 , so dz x+y dx dy 140+100 720 = + = (35+25)= 55.4 km / h. dt dt dt z 260 13 18. Let D denote the distance from the origin (0,0) to the point on the curve y= x . dD 1 2 1/2 dx 2x+1 dx 2 2 2 2 2 . With D= (x0) +(y0) = x +( x ) = x +x = (x +x) (2x+1) = dt dt 2 dt 2 2 x +x dx dD 9 27 =3 when x=4 , = (3)= 3.02 cm / s. dt dt 2 20 4 5. 19. dV 1 2 =C10 , 000 , where V = r h is the volume at dt 3 2 1 1 1 3 dV 2 dh r h time t . By similar triangles, = r= hV = . h h= h = h dt 3 3 3 27 dt 9 2 6 dh 800,000 2 When h=200cm , =20cm/min , so C10 , 000= (200) (20)C=10 , 000+ 289 , dt 9 9 If C= the rate at which water is pumped in, then. 3. 253 cm / min.. 20. By similar triangles, 5.

(26) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates. 3 b 1 dV dh dh 2 2 = , so b=3h . The trough has volume V = bh(10)=5(3h)h=15h 12= =30h = 1 h 2 dt dt dt 5h 1 dh 2 4 = = ft / min. . When h= , 2 dt 1 5 5 2. 21. 1 ( base + base )( height ) , and the 1 2 2 volume V of the 10meterlong trough is 10A . Thus, the volume of the trapezoid with height h is 1 a 0.25 1 V =(10) [0.3+(0.3+2a)]h . By similar triangles, = = , so 2 h 0.5 2 dV dV dh dh dh 0.2 2 . When h=0.3 2a=hV =5(0.6+h)h=3h+5h . Now = 0.2=(3+10h) = dt dh dt dt dt 3+10h dh 0.2 0.2 1 10 , m / min = m / min or cm / min. = = dt 3+10(0.3) 6 30 3 The figure is labeled in meters. The area A of a trapezoid is. 22. 1 (b+12)h(20)=10(b+12)h and, from similar triangles, 2 x 6 y 16 8 8h 11h . Thus, = and = = , so b=x+12+y=h+12+ =12+ h 6 h 6 3 3 3. The figure is drawn without the top 3 feet. V =. 2. dh . When h=5 , dt. 11h 110h dV 220 and so 0.8= V =10 24+ h=240h+ = 240+ h 3 3 dt 3 dh 0.8 3 = = 0.00132 ft / min. dt 240+5(220/3) 2275. 23. dV 1 2 1 3 We are given that =30 ft / min. V = r h= dt 3 3. h 2. 2. 3. h h= 12 6.

(27) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates. 2. dV dV dh h dh dh 120 dh 120 6 = 30= = = 2 = 0.38 ft / min. . When h=10 ft, 2 4 dt dt dh dt dt dt 5 10 h. 24. 2. x dx d sin 2 d x=100cot =100csc = 8 . We are given dx/dt=8 ft / s. cot = 100 100 dt dt dt 2. d (1/2) 1 100 1 = = 8= When y=200 , sin = rad / s. The angle is decreasing at a rate of 100 dt 50 200 2 1 rad / s. 50. 25. 1 h 1 bh , but b=5 m and sin = h=4sin , so A= (5)(4sin )=10sin . We are given 2 4 2 d dA dA d =0.06 rad / s, so = =(10cos )(0.06)=0.6cos . When = , dt dt d dt 3 dA 1 2 =0.6 cos =(0.6) =0.3 m / s. dt 3 2. A=. 26. rad / min. By the Law of Cosines, 90 dx d dx 180sin d 2 2 2 . When x =12 +15 2(12)(15)cos =369360cos 2x =360sin = dt dt dt dt x. We are given d /dt=2 / min =. dx 180sin 60 =60 , x= 369360cos 60 = 189 =3 21 , so = dt 3 21 min.. 3 7 = = 0.396 m / 21 90 3 21. 27. Differentiating both sides of PV =C with respect to t and using the Product Rule gives us dV dP dV V dP dP . When V =600 , P=150 and P +V =0 = =20 , so we have dt dt dt P dt dt 7.

(28) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates. dV 600 3 = (20)=80 . Thus, the volume is decreasing at a rate of 80 cm / min. dt 150 28. PV. 1.4. =C P 1.4V. 0.4. dV dV 1.4 dP +V =0 = dt dt dt. V. 1.4 0.4. dP V dP = . When V =400 , dt 1.4P dt. P 1.4V dV 400 250 dP =10 , so we have = (10)= P=80 and . Thus, the volume is increasing at a dt 1.4(80) 7 dt 250 3 36 cm / min. rate of 7 29. With R =80 and R =100 , 1. 2. 1 1 1 1 180 9 400 1 = + = + = = , so R= . Differentiating R R 9 R 80 100 8000 400 1. 2. dR dR 1 1 1 2 1 1 dR 1 1 = + = with respect to t , we have 2 dt 2 dt 2 dt R R R R R R 1 2 1. dR 2 =R dt. 1 2. R. dR. 1. dt. dR 400 = 2 dt 9. 2. R. 1. 2. +. 1. 2. dR. 2. dt. . When R =80 and R =100 , 1. 2. 2. 1 2. (0.3)+. 80. 1 2. 100. (0.2) =. 107 0.132 / s. 810. dB 2/3 2.53 when L=18 using B=0.007W and W =0.12L . dt dB 2 1/3 2015 dB dW dL 1.53 = = 0.007 W (0.12 2.53 L ) dt 3 10,000,000 dW dL dt 2 2.53 1/3 1.53 8 5 (0.12 2.53 18 ) 1.045 10 g/yr = 0.007 3 (0.12 18 ) 7 10. 30. We want to find. 31. dx x dx d . When = , =2 ft / s. sin = x=10sin =10cos dt 10 dt dt 4 2 d d 2 rad / s. 2=10cos = = 5 4 dt dt 10(1/ 2 ) We are given that. 8.

(29) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates. 32. Using Q for the origin, we are given. dx dy when x=5 . Using the =2 ft / s and need to find dt dt 2. 2. 2. 2. x +12 + y +12 =39 , the total length of the rope. dx dy x y Differentiating with respect to t , we get + =0 , so 2 2 dt 2 2 dt x +12 y +12 Pythagorean Theorem twice, we have. dy x = dt y. 2. 2. y +12 2. 2. x +12 2. dx 2 2 2 2 2 2 . Now when x=5 , 39= (5) +12 + y +12 =13+ y +12

(30) dt. 2. , and y= 26 12 = 532 . So when x=5 ,. 2. 2. y +12 =26. dy (5)(26) 10 = (2)= 0.87 ft / s. So cart B is dt 532 (13) 133. moving towards Q at about 0.87 ft / s.. 33. (a) 2. 2. 2. By the Pythagorean Theorem, 4000 +y = . Differentiating with respect to t , we obtain dy dy d . We know that 2y =2 =600 ft / s, so when y=3000 ft, dt dt dt d y dy 3000 1800 2 2 = 4000 +3000 = 25,000,000 =5000 ft and = = (600)= =360 ft / s. dt dt 5000 5 d d y (b) Here tan = (tan )= dt dt 4000. y 4000. 2. d 1 dy d cos dy . sec = = 4000 dt dt 4000 dt dt 2. 2. d (4/5) dy 4000 4000 4 When y=3000 ft, =600 ft / s, =5000 and cos = = = , so = (600)=0.096 dt 4000 dt. 5000 5 rad / s.. 34. 9.

(31) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates. We are given that 2. sec =1+. 1 3. d dx 1 2 d =4(2 )=8 rad / min. x=3tan =3sec . When x=1 , tan = , so dt dt dt 3 2 10 dx 10 80 = =3 (8 )= 83.8 km / min. and dt 9 9 3. 35. dx =300 km / h. By the Law of Cosines, dt 1 dy dx dx dy 2x+1 dx 2 2 2 2 2 y =x +1 2(1)(x)cos 120 =x +12x =x +x+1 , so 2y =2x + = . After 2 dt dt dt dt 2y dt 1 minute, 300 dy 2(5)+1 1650 2 x= =5 km y= 5 +5+1 = 31 km = (300)= 296 km / h. 60 dt 2 31 31 We are given that. 36. dx dy =3 mi / h and =2 mi / h. By the Law of Cosines, dt dt dz dx dy dy dx 2 2 2 2 2 . After 15 minutes z =x +y 2xycos 45 =x +y 2 xy 2z =2x +2y 2x 2y dt dt dt dt dt 1 = h , 4 136 2 3 2 3 2 1 3 2 2 2 2 we have x= and y= = z = and z= + 2 4 4 4 4 4 2 4 4 136 2 3 1 2 3 1 2 dz = 2 3+2 2 2 2 2 3 = = 136 2 4 2 2 4 2 dt 136 2 136 2 2.125mi/h.. We are given that. 37. Let the distance between the runner and the friend be . Then by the Law of Cosines, 10.

(32) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates. 2. 2. 2. =200 +100 2 200 100 cos =50,00040,000cos (*). Differentiating implicitly with respect to t d d , we obtain 2 . Now if D is the distance run when =40,000(sin ) dt dt the angle is radians, then by the formula for the length of an arc on a circle, s=r , we have 1 d d 1 dD 7 . To substitute into the expression for , we must D=100 , so = D = = 100 dt dt 100 dt 100 1 2 know sin at the time when =200 , which we find from (*): 200 =50,00040,000cos

(33) cos = 4 sin =. 1. 1 4. 2. =. 15 15 d . Substituting, we get 2(200) =40,000 4 4 dt. 7 100. . 7 15 6.78 m / s. Whether the distance between them is increasing or decreasing depends on 4 the direction in which the runner is running.. d /dt=. 38. 2 = rad / h. 12 6 The minute hand goes around once an hour, or at the rate of 2 rad / h. So the angle between them (measuring clockwise from the minute hand to the hour hand) is changing at the rate of 11 rad / h. Now, to relate to , we use the Law of Cosines: d /dt= 2 = 6 6 The hour hand of a clock goes around once every 12 hours or, in radians per hour,. 2. 2. 2. =4 +8 2 4 8 cos =8064cos (*). d d . At 1:00, the angle =64(sin ) dt dt 2 between the two hands is onetwelfth of the circle, that is, = radians. We use (*) to find at 12 6 d 11 1:00: = 8064cos = 8032 3 . Substituting, we get 2 =64sin 6 dt 6 6 1 11 64 d 2 88 6 = = 18.6 . So at 1:00, the distance between the tips of the dt 2 8032 3 3 8032 3 hands is decreasing at a rate of 18.6 mm / h 0.005 mm / s. Differentiating implicitly with respect to t , we get 2. 11.

(34) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials. 1. As in Example 1, T (0)=185 , T (10)=172 , T (20)=160 , and T (10)T (20) 172160 / / T (20) = =1.2 F / min. T (30)T (20)+T (20)(3020) 1601.2(10)=148 1020 10 F. . We would expect the temperature of the turkey to get closer to 75 F . . as time increases. Since the temperature decreased 13 F in the first 10 minutes and 12 F in the second 10 minutes, we can assume that the slopes of the tangent line are increasing through negative . values: 1.3,1.2,... . Hence, the tangent lines are under the curve and 148 F. is an underestimate. From the figure, we estimate the slope of the tangent line at t=20 to be 184 147 37 = . 0 30 30 37 2 / Then the linear approximation becomes T (30)T (20)+T (20) 10 160 (10)=147 147.7 . 30 3 /. 2. P (2). P(1)P(2) 87.174.9 = =12.2 kilopascals / km. 12 1 /. P(3) P(2)+P (2)(32) 74.912.2(1)=62.7 kPa. From the figure, we estimate the slope of the tangent line at h=2 to be /. approximation becomes P(3) P(2)+P (2) 1 74.9. 98 63 35 = . Then the linear 0 3 3. 35 63.23 kPa. 3. 3. Extend the tangent line at the point (2030,21) to the t axis. Answers will vary based on this approximationwe’ll use t=1900 as our t intercept. The linearization is then / P(t) P(2030)+P (2030)(t2030) 1.

(35) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials. 21 (t2030) 130 21 P(2040) =21+ (20402030) 22.6% 130 21 P(2050) =21+ (20502030) 24.2% 130 21+. These predictions are probably too high since the tangent line lies above the graph at t=2030. N(1990)N(1985) 19.317.0 N(1980)N(1985) 15.017.0 = =0.4 and B= = =0.46 . Then 5 19901985 5 19801985 N(t)N(1985) A+B / N (1985)= lim =0.43 million / year. So t1985 2 t 1985. 4. Let A=. /. N(1984) N(1985)+N (1985)(19841985) 17.0+0.43(1)=16.57 million. N(1995)N(2000) 22.024.9 / N (2000) = =0.58 million / year. 19952000 5 /. N(2006) N(2000)+N (2000)(20062000) 24.9+0.58(6)=28.38 million. 3. /. 2. /. /. 5. f (x)=x f (x)=3x , so f (1)=1 and f (1)=3 . With a=1 , L(x)= f (a)+ f (a)(xa) becomes /. L(x)= f (1)+ f (1)(x1)=1+3(x1)=3x2 . /. /. /. 6. f (x)=ln x f (x)=1/x , so f (1)=0 and f (1)=1 . Thus, L(x)= f (1)+ f (1)(x1)=0+1(x1)=x1. =1 . Thus, 2 / . L(x)= f +f x =01 x =x+ 2 2 2 2 2 1 2/3 1 1/3 / / 3 8. f (x)= x =x f (x)= x , so f (8)=2 and f (8)= . Thus, 3 12 1 1 4 / L(x)= f (8)+ f (8)(x+8)=2+ (x+8)= x . 12 12 3 /. 7. f (x)=cos x f (x)=sin x , so f. 2. =0 and f. /. 9. f (x)= 1x 2.

(36) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials. 1 1 / , so f (0)=1 and f (0)= . Therefore, 2 2 1x 1x = f (x) f (0)+ f /(0)(x0) 1 1 = 1+ (x0)=1 x 2 2 1 1 So 0.9 = 10.1 1 (0.1)=0.95 and 0.99 = 10.01 1 (0.01)=0.995 . 2 2 /. f (x)=. 1/3. 3. /. 10. g(x)= 1+x =(1+x) g (x)=. 1 2/3 (1+x) , 3. 1 1 / 3 . Therefore, 1+x =g(x) g(0)+g (0)(x0)=1+ x . So 3 3 1 1 3 3 3 0.95= 1+ ( 0.05) 1+ ( 0.05) =0.983 , and 1.1 = 1+0.1 1+ (0.1)=1.03 . 3 3 /. so g(0)=1 and g (0)= 3. 1 1 2/3 / , so f (0)=1 and f (0)= . Thus, (1x) 3 3 1 1 3 / 3 f (x) f (0)+ f (0)(x0)=1 x . We need 1x 0.1<1 x< 1x +0.1 , which is true when 3 3 1.204<x<0.706 . 1/3. 3. /. 11. f (x)= 1x =(1x) f (x)=. /. 2. /. /. 12. f (x)=tan x f (x)=sec x , so f (0)=0 and f (0)=1 . Thus, f (x) f (0)+ f (0)(x0)=0+1(x0)=x . We need tan x0.1<x<tan x+0.1 , which is true when 0.63<x<0.63 .. 3.

(37) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials. 4. 1. 13. f (x)=. 4. 5. /. =(1+2x) f (x)=4(1+2x) (2)=. (1+2x). 8. /. 5. , so f (0)=1 and f (0)=8 . Thus,. (1+2x). /. f (x) f (0)+ f (0)(x0)=1+(8)(x0)=18x . 4. 4. We need 1/(1+2x) 0.1<18x<1/(1+2x) +0.1 , which is true when 0.045<x<0.055 .. x. x. /. /. /. 14. f (x)=e f (x)=e , so f (0)=1 and f (0)=1 . Thus, f (x) f (0)+ f (0)(x0)=1+1(x0)=1+x . x. x. We need e 0.1<1+x<e +0.1 , which is true when 0.483<x<0.416.. /. 4. 3. 15. If y= f (x) , then the differential dy is equal to f (x)dx . y=x +5x dy=(4x +5)dx . 16. y=cos x dy=sin x dx= sin xdx 17. y=xln x dy= x 2. 18. y= 1+t dy=. 19. y=. 1 +ln x 1 dx=(1+ln x)dx x. 1 2 1/2 (1+t ) (2t)dt= 2. t 1+t. dt 2. u+1 (u1)(1)(u+1)(1) 2 dy= du= du 2 2 u1 (u1) (u1) 4. 5. 5. 20. y=(1+2r) dy=4(1+2r) 2dr=8(1+2r) dr 2. 21. (a) y=x +2x dy=(2x+2)dx (b) When x=3 and 4.

(38) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials. dx=. 1 , dy=[2(3)+2] 2 x/4. 22. (a) y=e dy=. 1 2. =4 .. 1 x/4 e dx 4 1 0 e 4. (b) When x=0 and dx=0.1 , dy=. (0.1)=0.025 .. 1 5 1/2 (4+5x) 5dx= dx 2 2 4+5x 5 1 1 5 (0.04)= = =0.05 . (b) When x=0 and dx=0.04 , dy= 4 25 20 2 4 23. (a) y= 4+5x dy=. 1. 24. (a) y=1/(x+1) dy=. 2. dx. (x+1). (b) When x=1 and dx=0.01 , dy=. 1 1 1 (0.01)= = =0.0025 . 2 4 100 400 2 1. 2. 25. (a) y=tan x dy=sec xdx 2. 2. (b) When x= /4 and dx=0.1 , dy=[sec ( /4)] (0.1)=( 2 ) (0.1)=0.2 . 26. (a) y=cos x dy=sin xdx (b) When x= /3 and dx=0.05 , dy=sin ( /3)(0.05)=0.5 3 (0.05)=0.025 3 0.043 . 2. 2. 2. 27. y=x , x=1 , x=0.5 y=(1.5) 1 =1.25 . dy=2xdx=2(1)(0.5)=1. 28. y= x , x=1 , x=1 y= 2 1 = 2 1 0.414 1 1 dy= dx= (1)=0.5 2 2 x. 5.

(39) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials. 2. 2. 2. 29. y=6x , x=2 , x=0.4 y=(6(1.6) )(6(2) )=1.44 dy=2xdx=2(2)(0.4)=1.6. 30. y=. 16 16 16 4 , x=4 , x=1 y= = . dy= x 3 4 3. 5. 4. 16 2. dx=. x. 16 2. (1)=1. 4. 4. 31. y= f (x)=x dy=5x dx . When x=2 and dx=0.001 , dy=5(2) (0.001)=0.08 , so 5. (2.001) = f (2.001) f (2)+dy=32+0.08=32.08 . 32. y= f (x)= x dy=. 1. dx . When x=100 and dx=0.2 , dy=. 1 (0.2)=0.01 , so 2 100. 2 x 99.8 = f (99.8) f (100)+dy=100.01=9.99 . 2 2 2/3 33. y= f (x)=x dy= 3 dx . When x=8 and dx=0.06 , dy= 3 (0.06)=0.02 , so 3 x 3 8 2/3. (8.06) = f (8.06) f (8)+dy=4+0.02=4.02 . 2. 2. 34. y= f (x)=1/x dy=(1/x )dx . When x=1000 and dx=2,dy=[1/(1000) ](2)=0.000002 , so 1/1002= f (1002) f (1000)+dy=1/10000.000002=0.000998 6.

(40) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials. . 2. . 35. y= f (x)=tan x dy=sec xdx . When x=45 and dx=1 , 2. . . 2. . . dy=sec 45 ( /180)=( 2 ) ( /180)= /90 , so tan 44 = f (44 ) f (45 )+dy=1 /90 0.965 . 1 1 dx . When x=1 and dx=0.07 , dy= (0.07)=0.07 , so 1 x ln 1.07= f (1.07) f (1)+dy=0+0.07=0.07.. 36. y= f (x)=ln x dy=. /. /. 37. y= f (x)=sec x f (x)=sec xtan x , so f (0)=1 and f (0)=1 0=0 . The linear approximation of f at /. 0 is f (0)+ f (0)(x0)=1+0(x)=1 . Since 0.08 is close to 0 , approximating sec 0.08 with 1 is reasonable. 6. /. 5. 38. If y=x , y =6x and the tangent line approximation at (1,1) has slope 6 . If the change in x is 6. 0.01 , the change in y on the tangent line is 0.06 , and approximating (1.01) with 1.06 is reasonable. /. /. 39. y= f (x)=ln x f (x)=1/x , so f (1)=0 and f (1)=1 . The linear approximation of f at 1 is /. f (1)+ f (1)(x1)=0+1(x1)=x1 . Now f (1.05)=ln 1.05 1.051=0.05 , so the approximation is reasonable. 2. /. /. 40. (a) f (x)=(x1) f (x)=2(x1) , so f (0)=1 and f (0)=2 . /. Thus, f (x) L (x)= f (0)+ f (0)(x0)=12x . 2x. g(x)=e. f /. 2x. g (x)=2e. /. , so g(0)=1 and g (0)=2 . /. Thus, g(x) L (x)=g(0)+g (0)(x0)=12x . g. /. h(x)=1+ln (12x) h (x)=. 2 / , so h(0)=1 and h (0)=2 . 12x. /. Thus, h(x) L (x)=h(0)+h (0)(x0)=12x . h. Notice that L =L =L . This happens because f , g , and h have the same function values and the f. g. h. same derivative values at a=0 .. (b) The linear approximation appears to be the best for the function f since it is closer to f for a larger domain than it is to g and h . The approximation looks worst for h since h moves away from L faster 7.

(41) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials. than f and g do. 3. 2. 2. 41. (a) If x is the edge length, then V =x dV =3x dx . When x=30 and dx=0.1 , dV =3(30) (0.1)=270 3. , so the maximum possible error in computing the volume of the cube is about 270 cm . The relative error is calculated by dividing the change in V , V , by V . We approximate V with dV . 2. dV 3x dx dx 0.1 V = =3 =3 =0.01 . Relative error = 3 V x 30 V x Percentage error = relative error 100%=0.01 100%=1% . 2. (b) S=6x dS=12xdx . When x=30 and dx=0.1 , dS=12(30)(0.1)=36 , so the maximum possible 2. error in computing the surface area of the cube is about 36 cm . S dS 12xdx dx 0.1 = =2 =2 =0.006 . Relative error = 2 S S x 30 6x Percentage error = relative error 100%=0.006 100%=0.6% . 2. 42. (a) A= r dA=2 r dr . When r=24 and dr=0.2 , dA=2 (24)(0.2)=9.6 , so the maximum 2. possible error in the calculated area of the disk is about 9.6 30 cm . A dA 2 rdr 2dr 2(0.2) 0.2 1 = = = = = =0.016 . (b) Relative error = 2 A A r 24 12 60 r Percentage error = relative error 100%=0.016 100%=1.6% . 2. 43. (a) For a sphere of radius r , the circumference is C=2 r and the surface area is S=4 r , so 2 84 2 2 r=C/(2 ) S=4 (C/2 ) =C / dS=(2/ )C dC . When C=84 and dC=0.5 , dS= (84)(0.5)= , 84 dS 84/ 1 2 27 cm . Relative error = 2 = 0.012 so the maximum error is about S 84 84 / 4 3 4 (b) V = r = 3 3 dV =. 1 2. 2. 2. (84) (0.5)=. C 2 1764 . 2. 3. =. C. 3. 6. 2. dV =. 1 2. 2. 2. C dC . When C=84 and dC=0.5 ,. , so the maximum error is about. 1764 . 2. 3. 179 cm . The relative error is. 2. dV 1764/ 1 = = 0.018 . approximately 3 2 V 56 (84) /(6 ) 44. For a hemispherical dome, 8.

(42) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials. 2 3 1 2 V = r dV =2 r dr . When r= (50)=25 m and 3 2 5 5 2 dr=0.05 cm =0.0005 m, dV =2 (25) (0.0005)= 2 , so the amount of paint needed is about 8 8 3. m . 2. 45. (a) V = r h V dV =2 rhdr=2 rhr (b) The error is V dV =[ (r+r)2h r 2h]2 rhr= r 2h+2 rhr+ (r)2h r 2h2 rhr 2. = (r) h 3. dF 4kR dR dR 46. F=kR dF=4kR dR . Thus, the relative change in F is about 4 = =4 4 F R kR times the relative change in R . So a 5% increase in the radius corresponds to a 20% increase in blood flow. 4. 3. dc dx=0dx=0 dx d du (b) d(cu)= (cu)dx=c dx=cdu dx dx d du dv (c) d(u+v)= (u+v)dx= + dx dx dx d dv du (d) d(uv)= (uv)dx= u +v dx dx dx du v u u d u dx (e) d = dx= 2 v dx v v d n n n1 (f) d(x )= (x )dx=nx dx dx 47. (a) dc=. /. du dv dx+ dx=du+dv dx dx dv du dx=u dx+v dx=udv+vdu dx dx du dv dv v dxu dx dx vduudv dx dx dx= = 2 2 v v dx=. /. 48. (a) f (x)=sin x f (x)=cos x , so f (0)=0 and f (0)=1 . Thus, /. f (x) f (0)+ f (0)(x0)=0+1(x0)=x . (b). 9.

(43) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials. We want to know the values of x for which y=x approximates y=sin x with less than a 2% difference; that is, the values of x for which xsin x xsin x <0.02 0.02< <0.02. sin x sin x. {. 0.02sin x<xsin x<0.02sin x 0.02sin x>xsin x>0.02sin x. if sin x>0. if sin x<0. {. 0.98sin x<x<1.02sin x 1.02sin x<x<0.98sin x. if sin x>0 if sin x<0. In the first figure, we see that the graphs are very close to each other near x=0 . Changing the viewing rectangle and using an intersect feature (see the second figure) we find that y=x intersects y=1.02sin x at x 0.344 . By symmetry, they also intersect at x 0.344 (see the third figure.). Converting 0.344 . 180 . radians to degrees, we get 0.344 /. . . 19.7 20 , which verifies the statement. /. 49. (a) The graph shows that f (1)=2 , so L(x)= f (1)+ f (1)(x1)=5+2(x1)=2x+3 . f (0.9) L(0.9)=4.8 and f (1.1) L(1.1)=5.2 . /. (b) From the graph, we see that f (x) is positive and decreasing. This means that the slopes of the tangent lines are positive, but the tangents are becoming less steep. So the tangent lines lie above the curve. Thus, the estimates in part (a) are too large. 10.

(44) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials. /. 2. /. /. 50. (a) g (x)= x +5 g (2)= 9 =3 . g(1.95) g(2)+g (2)(1.952)=4+3(0.05)=4.15 . /. g(2.05) g(2)+g (2)(2.052)=4+3(0.05)=3.85 . /. 2. /. (b) The formula g (x)= x +5 shows that g (x) is positive and increasing. This means that the slopes of the tangent lines are positive and the tangents are getting steeper. So the tangent lines lie below the graph of g . Hence, the estimates in part (a) are too small.. 11.

(45) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules. 1. Product Rule: 2. 3. y=(x +1)(x +1) /. 2. 2. 3. 4. 2. 4. 4. 2. y =(x +1)(3x )+(x +1)(2x)=3x +3x +2x +2x=5x +3x +2x . 2. 3. 5. 3. 2. /. 4. 2. Multiplying first: y=(x +1)(x +1)=x +x +x +1 y =5x +3x +2x (equivalent). 2. Quotient Rule: F(x)= 1/2. / F (x) =. x. 1. x3x x x. =. x3x. 3/2. 1/2. . x. 9 1/2 3/2 x x3x 2. (. 1 1/2 x 2. ). ( x1/2) 2. 9 1 1/2 3 1 1/2 x x + x x 3x = 2 2 2 2 1 1/2 = = x 3 x x 2 x3x x 1 1/2 1/2 / Simplifying first: F(x)= = x 3x=x 3x F (x)= x 3 (equivalent). 2 x For this problem, simplifying first seems to be the better method. 1/2. x . 2 x. /. 2. 3. By the Product Rule, f (x)=x e f (x)=x x. 1/2 x. d x x d 2 2 x x x (e )+e (x )=x e +e (2x)=xe (x+2) . dx dx 1/2 x. /. 4. By the Product Rule, g(x)= x e =x e g (x)=x (e )+e. 5. By the Quotient Rule, y=. 2. x. e. 2. /. x. y =. d x x d 2 (e )e (x ) dx dx 22. x. 1 1/2 1 1/2 x x = x e (2x+1) . 2 2. x. 2 x. =. x. (x ). x. x. x. x (e )e (2x). x. x. x. x. 4. x. =. xe (x2) x. 4. x. =. e (x2) x. 3. x. e / (1+x)e e (1) e +xe e xe 6. By the Quotient Rule, y= . y = = = 2 2 2 1+x (x+1) (x+1) (1+x) QR 3x1 / (2x+1)(3)(3x1)(2) 6x+36x+2 5 7. g(x)= g (x)= = = 2 2 2 2x+1 (2x+1) (2x+1) (2x+1) 8. f (t)=. QR. 2t 4+t. 2. /. f (t)=. 2. (4+t )(2)(2t)(2t) 22. (4+t ). 2. =. 8+2t 4t 22. (4+t ). 2. =. 82t. 2. 22. (4+t ). 1.

(46) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules. 3. PR. 4. 9. V (x)=(2x +3)(x 2x) /. 3. 3. 4. 2. 6. 3. 6. 2. 3. 4. 3. 6. 3. V (x)=(2x +3)(4x 2)+(x 2x)(6x )=(8x +8x 6)+(6x 12x )=14x 4x 6 2. 3. 5. 2. PR. 10. Y (u)=(u +u )(u 2u ) 2. /. 3. 4. 5. Y (u) =(u +u )(5u 4u)+(u 2u )(2u 3u ) 1. 2. 2. 1. 2. 2. 2. =(5u 4u +5u4u )+(2u 3u+4u +6u )=3u +2u+2u 1. 11. F(y)=. 2. 3. . 4. y. 2. (. 3. 4. (y+5y )= y 3y. 2. PR. ) ( y+5y3) . y. ( 2 4) ( 1+15y2) + ( y+5y3) ( 2y3+12y5) 2 4 2 2 4 2 = ( y +153y 45y ) +(2y +12y 10+60y ). /. F (y) = y 3y. 2. 4. 2. 4. =5+14y +9y or 5+14/y +9/y t. 12. R(t)=(t+e )(3 t )= 1 1/2 t t / R (t) =(t+e )( 2 t )+(3 t )(1+e ) 1 1/2 1 1/2 t t t t 3 t t = t t e +(3+3e t t e )=3+3e t t e e /(2 t ) 2 2 2 13. y=. t. 2. QR. 2. 3t 2t+1. . 2. /. y =. 2. (3t 2t+1)(2t)t (6t2) 2. 2. =. 2. (3t 2t+1) 2. =. 2. 2t[3t 2t+1t(3t1)]. 2. 2t (3t 2t+13t +t) 2. 2. 2t(1t). =. 2. (3t 2t+1) 3. 14. y=. t +t 4. t 2. QR. /. 2. 3. 3. (t 2)(3t +1)(t +t)(4t ) 4. 2. 6. =. 4. 2. t 3t 6t 2 4. 2. (t 2). 2. 6. 4. 4. 2. (t 2) 6. =. 4. (3t +t 6t 2)(4t +4t ). (t 2) 6. =. 2. (3t 2t+1) 4. y =. 2. (3t 2t+1). 4. 2. 4. 2. t +3t +6t +2 (t 2). 2.

(47) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules. r. 2. /. r. 2. r. r. r. 2. 2. 15. y=(r 2r)e = y =(r 2r)(e )+e (2r2)=e (r 2r+2r2)=e (r 2) 16. y=. 1. s. /. s+ke. s. =y =. s. (s+ke )(0)(1)(1+ke ) s2. 1+ke. =. (s+ke ). s. s2. (s+ke ). 3. v 2v v 2 2 1/2 / =v 2 v =v 2v y =2v2 17. y= v. 1 2. v. We can change the form of the answer as follows: 2vv w. 3/2. 5/2. 18. z=w (w+ce )=w +cw. 19. y=. 1 4. /. y =. 2. 3/2 w. 4. x 1 x +1. 2. 3. (x +x +1)(0)1(4x +2x). x +x +1. 20. y=. 5 3/2 w +c 2. /. e z =. 4. 2. 2. /. y =. 1 2 x. =2vv. 1/2. . 3/2. 2v v 1 2v 1 1 =2v = = v v v. 1/2. 3/2. w. w. w e +e . 3 1/2 5 3/2 1 1/2 w w = w + cw e (2w+3) 2 2 2. 2. =. (x +x +1) ( x +1). 1/2. 2x(2x +1) 4. 2. 2. (x +x +1). ( x 1) 2. ( x +1). 1 2 x. =. 1 1 1 1 + + 2 2 x 2 2 x 2. =. ( x +1) 2. 1 2. x ( x +1). 2. x / (x+c/x)(1)x(1c/x ) x+c/xx+c/x 2c/x x 2cx f (x)= = = = 21. f (x)= 2 2 2 2 2 2 x+c/x 2 c 2 x (x +c) (x +c) x +c x+ x 2 x x 22. f (x)=. ax+b / (cx+d)(a)(ax+b)(c) acx+adacxbc adbc f (x)= = = 2 2 2 cx+d (cx+d) (cx+d) (cx+d). 2x / (x+1)(2)(2x)(1) / 1 2 y = = . At (1,1) , y = , and an equation of the tangent line 2 2 x+1 2 (x+1) (x+1) 1 1 1 is y1= (x1) , or y= x+ . 2 2 2 23. y=. 24. y=. x x+1 3.

(48) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules. (x+1) /. y =. 1. x (1). 2 x. =. 2. (x+1)(2x). 1x. =. 2. /. . At (4,0.4) , y =. 2. (x+1) 2 x (x+1) 2 x (x+1) equation of the tangent line is y0.4=0.03(x4) , or y=0.03x+0.52 . x. x. /. x. x. /. 3 =0.03 , and an 100. 0. 25. y=2xe y =2(x e +e 1)=2e (x+1) . At (0, 0) , y =2e (0+1)=2 1 1=2 , and an equation of the tangent line is y0=2(x0) , or y=2x . x. x. x. x. e / x e e 1 e (x1) / 26. y= y = . At (1, e) , y =0 , and an equation of the tangent line is = 2 2 x x x ye=0(x1) , or y=e . 1. 27. (a) y= f (x)=. 2. /. 2. f (x)=. (1+x )(0)1(2x) 22. 2x 22. . So the slope of the tangent line at the. (1+x ) (1+x ) 1 1 1 1 / 2 is f (1)= = and its equation is y = (x+1) or y= x+1 . 2 2 2 2 2 2 1+x. point. =. 1 1, 2. (b) x. 28. (a) y= f (x)=. 2. /. 2. f (x)=. 1+x. (1+x )1x(2x) 22. 2. =. (1+x ). /. point (3,0.3) is f (3)=. 1x. 22. . So the slope of the tangent line at the. (1+x ). 8 and its equation is y0.3=0.08(x3) or y=0.08x+0.54 . 100. (b) 29. (a) f (x)=. e x. x 3. /. f (x)=. 3. x. x. 32. (x ). 2 x. 2. x (e )e (3x ). =. x e (x3) x. 6. x. =. e (x3) x. 4. 4.

(49) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules. (b) /. f =0 when f has a horizontal tangent line, f when f is increasing. 30. f (x)=. x. 2. (x 1)1x(2x). /. 2. x 1. f (x)=. 2. 2. /. is negative when f is decreasing, and f. 2. 2. (x 1). is positive. 2. x 1. =. /. =. 2. (x 1). x +1 2. 2. Notice that the slopes of all tangents to. (x 1). /. f are negative and f (x)<0 always.. /. /. 31. We are given that f (5)=1 , f (5)=6 , g(5)=3 , and g (5)=2 . /. /. /. (a) ( fg) (5)= f (5)g (5)+g(5) f (5)=(1)(2)+(3)(6)=218=16 (b). (c). f g. /. g f. /. /. (5)=. /. g(5) f (5) f (5)g (5) 2. =. (3)(6)(1)(2) 2. (3). [g(5)] /. (5)=. =. /. f (5)g (5)g(5) f (5) 2. =. (1)(2)(3)(6) 2. [ f (5)]. 20 9. =20. (1) /. /. 32. We are given that f (3)=4 , g(3)=2 , f (3)=6 , and g (3)=5 . /. /. /. (a) ( f +g ) (3)= f (3)+g (3)=6+5=1 /. /. /. (b) ( fg ) (3)= f (3)g (3)+g(3) f (3)=(4)(5)+(2)(6)=2012=8 (c). f g. /. /. (3)=. /. g(3) f (3) f (3)g (3) 2. [g(3)]. =. (2)(6)(4)(5) 2. (2). =. 32 =8 4. (d). 5.

(50) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules. /. f f g. /. (3) =. /. /. [ f (3)g(3)] f (3) f (3)[ f (3)g (3)] 2. [ f (3)g(3)] (42)(6)4(65) 12+44 = = =8 2 2 (42) 2 x. x. /. x. /. x. /. 33. f (x)=e g(x) f (x)=e g (x)+g(x)e =e g (x)+g(x) . /. 0. /. f (0)=e g (0)+g(0) =1(5+2)=7 d 34. dx. h(x) x. /. =. xh (x)h(x) 1 2. x. h(x) x. d dx. /. =. 2h (2)h(2) 2. x=2. 2. =. 2(3)(4) 10 = =2.5 4 4. 35. (a) From the graphs of f and g , we obtain the following values: f (1)=2 since the point (1,2) is on /. the graph of f ; g(1)=1 since the point (1,1) is on the graph of g ; f (1)=2 since the slope of the line 40 / segment between (0,0) and (2,4) is =2 ; g (1)=1 since the slope of the line segment between 20 04 / / / (2,4) and (2,0) is =1 . Now u(x)= f (x)g(x) , so u (1)= f (1)g (1)+g(1) f (1)=2 (1)+1 2=0 . 2(2) 8 1 2 / / 2 3 3 2 / g(5) f (5) f (5)g (5) 3 3 (b) v(x)= f (x)/g(x) , so v (5)= = = = 2 2 4 3 2 [g(5)] /. /. /. 2 3 +2 0= . 4 2 1 2 1 5 4 3. 36. (a) P(x)=F(x)G(x) , so P (2)=F(2)G (2)+G(2)F (2)=3 /. /. (b) Q(x)=F(x)/G(x) , so Q (7)=. /. G(7)F (7)F(7)G (7) 2. [G(7)] /. /. =. 2. 1. =. 1 10 43 + = 4 3 12. /. 37. (a) y=xg(x) y =xg (x)+g(x) 1=xg (x)+g(x) /. /. x / g(x) 1xg (x) g(x)xg (x) (b) y= y = = 2 2 g(x) [g(x)] [g(x)] /. /. g(x) / xg (x)g(x) 1 xg (x)g(x) (c) y= y = = 2 2 x (x) x 38. (a) 6.

(51) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules. 2. /. 2. /. /. x f (x) f (x)(2x). y=x f (x) y =x f (x)+ f (x)(2x) (b) y=. f (x) 2. y =. 2. /. 22. x. /. =. xf (x)2 f (x) x. (x ). 2. 2. 3. /. x / f (x)(2x)x f (x) y = (c) y= 2 f (x) [ f (x)] 1+xf (x) (d) y= x 1 / x[xf (x)+ f (x)][1+xf (x)] / 2 x y = 2 ( x) 1 1/2 1 1/2 3/2 / 1/2 x f (x)+x f (x) x x f (x) 1/2 2 / 2 2 2x xf (x)+2x f (x)1 = = 1/2 3/2 x 2x 2x 39. If P(t) denotes the population at time t and A(t) the average annual income, then T (t)=P(t)A(t) is /. /. /. the total personal income. The rate at which T (t) is rising is given by T (t)=P(t)A (t)+A(t)P (t) /. /. /. T (1999) =P(1999)A (1999)+A(1999)P (1999)=( 961,400 )( $1400 / yr )+( $30,593 )( 9200 / yr ) =$ 1,345,960,000 / yr +$ 281,455,600 / yr =$ 1,627,415,600 / yr So the total personal income was rising by about $ 1.627 billion per year in 1999. /. The term P(t)A (t) $ 1.346 billion represents the portion of the rate of change of total income due to /. the existing population’s increasing income. The term A(t)P (t) $ 281 million represents the portion of the rate of change of total income due to increasing population. 40. (a) f (20)=10 , 000 means that when the price of the fabric is $20/ yard, 10 , 000 yards will be sold. /. f (20)=350 means that as the price of the fabric increases past $20/ yard, the amount of fabric which will be sold is decreasing at a rate of 350 yards per (dollar per yard). /. /. /. /. (b) R( p)= pf ( p) R ( p)= pf ( p)+ f ( p) 1 R (20)=20 f (20)+ f (20) 1=20(350)+10,000=3000. This means that as the price of the fabric increases past $20/ yard, the total revenue is increasing at $3000/($/yard). Note that the Product Rule indicates that we will lose $7000/($/yard) due to selling less fabric, but that that loss is more than made up for by the additional revenue due to the increase in price. 7.

(52) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules. x / (x+1)(1)x(1) 1 , then f (x)= = . When x=a , the equation of the tangent 2 2 x+1 (x+1) (x+1) a a 1 1 line is y = (xa) . This line passes through (1,2) when 2 = (1a) 2 2 a+1 a+1 (a+1) (a+1) 41. If y= f (x)=. 2. 2. 2. 2. 2(a+1) a(a+1)=1a 2a +4a+2a a1+a=0a +4a+1=0 . 2. 4 4 4(1)(1) 4 12 = =2 3 , The quadratic formula gives the roots of this equation as a= 2(1) 2 so there are two such tangent lines. Since 2 3. f (2 3 ) =. 2 3 +1. =. 2 3 1 3. . 1 3 1 3. 2 2 3 3 3 1 3 1 3 = = , 13 2 2 1 3 the lines touch the curve at A 2+ 3, (0.27,0.37) and 2 1+ 3 (3.73,1.37) . B 2 3, 2 x1 / (x+1)(1)(x1)(1) 2 42. y= . If the tangent intersects the curve when x=a , y = = 2 2 x+1 (x+1) (x+1) 1 2 then its slope is 2/(a+1) . But if the tangent is parallel to x2y=2 , that is, y= x1 , then its slope is 2 1 1 2 2 . Thus, = (a+1) =4 a+1= 2 a=1 or 3 . When a=1 , y=0 and the equation of the 2 2 2 (a+1) 1 1 1 tangent is y0= (x1) or y= x . 2 2 2 1 1 7 When a=3 , y=2 and the equation of the tangent is y2= (x+3) or y= x+ . 2 2 2 =. /. /. /. /. /. /. /. /. /. 43. (a) ( fgh) =[( fg)h] =( fg) h+( fg)h =( f g+ fg )h+( fg)h = f gh+ fg h+ fgh. / 8.

(53) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules. (b) Putting f =g=h in part (a) , we have d 3 / / / / / 2 / [ f (x)] = ( fff ) = f ff + ff f + fff =3 fff =3[ f (x)] f (x). dx. (c). d 3x d x 3 x2 x 2x x 3x (e )= (e ) =3(e ) e =3e e =3e dx dx. 44. (a) d dx. d d (1)1 [g(x)] dx dx. g(x). 1 g(x). =. [g(x)] /. =. g(x) 01 g (x) 2. /. =. [g(x)]. (b) y=. 1 4. [Quotient Rule]. 2. 2. 2. x +x +1. 2. 2. (x +x +1) 1 x. n. =. g (x) 2. [g(x)] 2. 4x +2x 4. /. [g(x)] 3. /. y =. d n d (c) (x )= dx dx. 0g (x). or. 2x(2x +1) 4. 2. 2. (x +x +1). n /. =. (x ). n2. (x ). [by the Reciprocal Rule] =. nx. n1. 2n. =nx. n12n. =nx. n1. x. 9.

(54) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences. 2. /. 1. (a) s= f (t)=t 10t+12 v(t)= f (t)=2t10 (b) v(3)=2(3)10=4 ft / s (c) The particle is at rest when v(t)=02t10=0t=5 s. (d) The particle is moving in the positive direction when v(t)>02t10>02t>10t>5 . (e) Since the particle is moving in the positive direction and in the negative direction, we need to calculate the distance traveled in the intervals [0,5] and [5,8] separately. | f (5) f (0)|=|1312|=25 ft and | f (8) f (5)|=|{4(13)|=9 ft. The total distance traveled during the first 8 s is 25+9=34 ft.. (f) 3. 2. /. 2. 2. (a) s= f (t)=t 9t +15t+10 v(t)= f (t)=3t 18t+15=3(t1)(t5) (b) v(3)=3(2)(2)=12 ft / s (c) v(t)=0t=1 s or 5 s (d) v(t)>00 t<1 or t>5 (e) | f (1) f (0)|=|1710|=7 , | f (5) f (1)|=|1517|=32 , and | f (8) f (5)|=|66(15)|=81 . Total distance =7+32+81=120 ft.. (f) 3. 2. /. 2. 3. (a) s= f (t)=t 12t +36t v(t)= f (t)=3t 24t+36 (b) v(3)=2772+36=9 ft / s 2. (c) The particle is at rest when v(t)=0 . 3t 24t+36=0 3(t2)(t6)=0 t=2 s or 6 s. (d) The particle is moving in the positive direction when v(t)>0 . 3(t2)(t6)>00 t<2 or t>6 . (e) Since the particle is moving in the positive direction and in the negative direction, we need to calculate the distance traveled in the intervals (0,2) , (2,6) , and [6,8] separately. | f (2) f (0)|=|320|=32 . | f (6) f (2)|=|032|=32 . | f (8) f (6)|=|320|=32 . The total distance is 32+32+32=96 ft. (f). 1.

(55) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences. 4. /. 3. 4. (a) s= f (t)=t 4t+1 v(t)= f (t)=4t 4 3. (b) v(3)=4(3) 4=104 ft / s 3. 2. (c) It is at rest when v(t)=4(t 1)=4(t1)(t +t+1)=0t=1 s. 3. (d) It moves in the positive direction when 4(t 1)>0t>1 . (e) Distance in positive direction =| f (8) f (1)|=|4065(2)|=4067 ft Distance in negative direction =| f (1) f (0)|=|21|=3 ft Total distance traveled =4067+3=4070 ft. (f) t. 5. (a) s=. /. 2. v(t)=s (t)=. 2. (t +1)(1)t(2t) 2. t +1. (t +1) 2. (b) v(3)=. 1(3) 2. 2. 2. (3 +1). =. 19 2. 10. =. =. 1t 2. 2 2. (t +1). 8 2 ft / s = 100 25 2. (c) It is at rest when v=01t =0t=1 s [ t 1 since t 0 ]. 2. 2. (d) It moves in the positive direction when v>01t >0t <10 t<1 . 1 1 (e) Distance in positive direction =|s(1)s(0)|= 0 = ft 2 2 8 1 49 Distance in negative direction =|s(8)s(1)|= ft = 65 2 130 1 49 57 Total distance traveled = + ft = 2 130 65. (f) 2.

(56) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences. 2. 5/2. 3/2. 1/2. 6. (a) s= t (3t 35t+90)=3t 35t +90t 15 3/2 105 1/2 15 / 1/2 15 1/2 2 v(t)=s (t)= t t +45t = t (t 7t+6)= (t1)(t6) 2 2 2 2 t 15 (b) v(3)= (2)(3)=15 3 ft / s 2 3 (c) It is at rest when v=0t=1 s or 6 s. (d) It moves in the positive direction when v>0(t1)(t6)>00 t<1 or t>6. (e) Distance in positive direction =|s(1)s(0)|+|s(8)s(6)|=|580|+|4 2 (12 6 ) =58+4 2 +12 6 93.05 ft Distance in negative direction =|s(6)s(1)|=|12 6 58|=58+12 6 87.39 ft Total distance traveled =58+4 2 +12 6 +58+12 6 =116+4 2 +24 6 180.44 ft. (f) 3. 2. /. 2. 2. 7. s(t)=t 4.5t 7t v(t)=s (t)=3t 9t7=5 3t 9t12=0 3(t4)(t+1)=0 t=4 or 1 . Since t 0 , the particle reaches a velocity of 5 m / s at t=4 s. ds =5+6t , so v(2)=5+6(2)=17 m / s. dt (b) v(t)=35 5+6t=35 6t=30 t=5 s. 2. 8. (a) s=5t+3t v(t)=. dh =101.66t , so v(3)=101.66(3)=5.02 m / s. dt 10 17 2 2 3.54 or 8.51 . (b) h=25 10t0.83t =25 0.83t 10t+25=0 t= 1.66 The value t =(10 17 )/1.66 corresponds to the time it takes for the stone to rise 25 m and 2. 9. (a) h=10t0.83t v(t)=. 1. t =(10+ 17 )/1.66 corresponds to the time when the stone is 25 m high on the way down. Thus, 2. v(t )=101.66[(10 17 )/1.66]= 17 4.12 m / s. 1. /. 10. (a) At maximum height the velocity of the ball is 0 ft / s. v(t)=s (t)=8032t=0 32t=80t=. 5 . 2. So the maximum height is 3.

(57) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences. s. 5 2. =80. 5 2. 16. 2. 5 2. 2. =200100=100 ft.. 2. 2. (b) s(t)=80t16t =96 16t 80t+96=0 16(t 5t+6)=0 16(t3)(t2)=0 . So the ball has a height of 96 ft on the way up at t=2 and on the way down at t=3 . At these times the velocities are v(2)=8032(2)=16 ft / s and v(3)=8032(3)=16 ft / s, respectively. 2. /. /. 2. 11. (a) A(x)=x A (x)=2x . A (15)=30 mm / mm is the rate at which the area is increasing with respect to the side length as x reaches 15 mm. 1 1 (4x)= P(x) . The figure suggests that if x is small, 2 2 then the change in the area of the square is approximately half of its perimeter ( 2 of the 4 sides) /. (b) The perimeter is P(x)=4x , so A (x)=2x=. 2. times x . From the figure, A=2x( x)+( x) . If x is small, then A 2x( x) and so A/ x 2x .. dV dV 2 =3x . dx dx increasing as x increases past 3 mm. 3. 12. (a) V (x)=x . 2. 3. =3(3) =27 mm / mm is the rate at which the volume is. x=3. 1 2 1 (6x )= S(x) . The figure suggests that if x is 2 2 small, then the change in the volume of the cube is approximately half of its surface area (the area of 2. /. 2. (b) The surface area is S(x)=6x , so V (x)=3x =. 2. 2. 3. 3 of the 6 faces) times x . From the figure, V =3x ( x)+3x( x) +( x) . If x is small, then 2. 2. V 3x ( x) and so V / x 3x .. 4.

(58) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences. 13. (a) (i) A(3) A(2) 9 4 = =5 32 1 (ii) A(2.5) A(2) 6.25 4 = =4.5 2.52 0.5 (iii) A(2.1) A(2) 4.41 4 = =4.1 2.12 0.1 2. /. /. (b) A(r)= r A (r)=2 r , so A (2)=4 . /. (c) The circumference is C(r)=2 r=A (r) . The figure suggests that if r is small, then the change in the area of the circle (a ring around the outside) is approximately equal to its circumference times r . Straightening out this ring gives us a shape that is approximately rectangular with length 2 r and 2. 2. 2. width r , so A 2 r(r) . Algebraically, A=A(r+r) A(r)= (r+r) r =2 r(r)+ (r) . So we see that if r is small, then A 2 r(r) and therefore, A/r 2 r .. /. 2. 14. (a) A (1)=7200 cm / s /. 2. /. 2. (b) A (3)=21 , 600 cm / s (c) A (5)=36 , 000 cm / s /. 2. 15. (a) S (1)=8 ft / ft /. 2. /. 2. (b) S (2)=16 ft / ft (c) S (3)=24 ft / ft 5.

(59) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences. 16. (a) (a) 4 4 (512) (125) V (8)V (5) 3 3 3 = =172 m / m 85 3 (b). 4 4 (512) (125) 3 V (8)V (5) 3 3 = =172 m / m 3 85. (c). 4 4 (216) (125) 3 V (6)V (5) 3 3 = =121.3 m / m 1 65. (d). 4 4 (216) (125) 3 V (6)V (5) 3 3 = =121.3 m / m 1 65. (e). 4 3 4 3 (5.1) (5) V (5.1)V (5) 3 3 3 = =102.013 m / m 0.1 5.15. (f). 4 3 4 3 (5.1) (5) V (5.1)V (5) 3 3 3 = =102.013 m / m 0.1 5.15 /. 2. /. 3. (b) V (r)=4 r , so V (5)=100 m / m. 4 3 / 2 (c) V (r)= r V (r)=4 r =S(r) . By analogy with Exercise 13(c) , we can say that the change in 3 the volume of the spherical shell, V , is approximately equal to its thickness, r , times the surface 2. 2. area of the inner sphere. Thus, V 4 r (r) and so V /r 4 r . 17. (a) (1)=6 kg / m (b) (2)=12 kg / m (c) (3)=18 kg / m 5 =218.75 gal / min 40 10 / (b) V (10)=250 1 =187.5 gal / min 40 20 / (c) V (20)=250 1 =125 gal / min 40 (d) /. 18. (a) V (5)=250. 1. 6.

(60) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences. /. V (40)=250. 1. /. 40 40. =0 gal / min 2. 19. (a) Q (0.5)=3(0.5) 4(0.5)+6=4.75 A /. 2. (b) Q (1)=3(1) 4(1)+6=5 A dF 3 2GmM , which is the rate of change of the =2(GmM)r = 2 3 dr r r force with respect to the distance between the bodies. The minus sign indicates that as the distance r between the bodies increases, the magnitude of the force F exerted by the body of mass m on the body of mass M is decreasing. / / 3 2GmM (b) Given F (20 , 000)=2 , find F (10 , 000) . 2= GmM=20 , 000 . 3 20,000 20. (a) F=. /. GmM. F (10,000)=. 2. =(GmM)r . (. 3. 2 20,000. 3. ) =2 23=16 N / km. 10,000. 21. (a) To find the rate of change of volume with respect to pressure, we first solve for V in terms of P. C dV C . PV =C V = = 2 P dP P (b) From the formula for dV /dP in part (a), we see that as P increases, the absolute value of dV /dP decreases. Thus, the volume is decreasing more rapidly at the beginning. 1 C C 1 1 dV C (c) = = = = = 2 V (PV )P CP P V dP P 22. (a). .. C(6)C(2) 0.02950.0570 = 62 4 (i). .. =0.006875 ( moles/L ) / min C(4)C(2) 0.04080.0570 = 42 2 =0.008 ( moles/L ) / min. 7.

(61) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences. .. C(2)C(0) 0.05700.0800 = 20 2. (iii). =0.0115 ( moles/L ) / min (b) Slope =. C 0.077 0.01 ( moles/L ) / min t 7.8. 1860 1750 110 2070 1860 210 = =11 , m = = =21 , 1 1920 1910 2 1930 1920 10 10 (m +m )2=(11+21)/2=16 million / year 23. (a) 1920: m = 1. 2. 4450 3710 740 5280 4450 830 = =74 , m = = =83 , 1 1980 1970 2 1990 1980 10 10 (m +m )2=(74+83)/2=78.5 million / year 1980: m = 1. 2. 3. 2. 3. 2. (b) P(t)=at +bt +ct+d (in millions of people), where a 0.0012937063 , b 7.061421911 , c 12,822.97902 , and d 7,743,770.396 . /. 2. (c) P(t)=at +bt +ct+d P (t)=3at +2bt+c (in millions of people per year) (d) /. 2. P (1920) =3(0.0012937063)(1920) +2(7.061421911)(1920)+12,822.97902 14.48 million / year / P (1980) 75.29 million / year (smaller, but close) /. (e) P (1985) 81.62 million / year, so the rate of growth in 1985 was about 81.62 million / year. 4. 3. 6. 2. 24. (a) A(t)=at +bt +ct +dt+e , where a=5.8275058275396

(62) 10 , b=0.0460458430461 , c=136.43277039706 , d=179 , 661.02676871 , and e=88 , 717 , 597.060767 . 4. 3. 2. /. 3. 2. (b) A(t)=at +bt +ct +dt+e A (t)=4at +3bt +2ct+d (c) 8.

(63) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences. /. A (1990) 0.0833 years of age per year. (d) 2. a kt rate of reaction 25. (a) C = akt+1 2. 2. 2. 2. d C (akt+1)(a k)(a kt)(ak) a k(akt+1akt) ak = = = = 2 2 2 dt (akt+1) (akt+1) (akt+1) 2. 2. 2. a kt a kt+aa kt a = = (b) If x= C , then ax=a . akt+1 akt+1 akt+1 2. So k(ax) =k. a akt+1. 2. =. 2. ak 2. (akt+1). =. d C dt. =. dx . dt 2. 26. (a) After an hour the population is n(1)=3 500 ; after two hours it is n(2)=3(3 500)=3 500 ; after 2. 3. 4. three hours, n(3)=3(3 500)=3 500 ; after four hours, n(4)=3 500 . From this pattern, we see that the t. t. population after t hours is n(t)=3 500=500 3 . d x x t (3 ) (1.10)3 . Thus, for n(t)=500 3 , (b) From (5) in Section 3.1, we have dx dn d t dn t 6 =500 (3 ) 500(1.10)3 500(1.10)3 400 , 950 bacteria / hour. dt dt dt t=6 27. (a) Using v=. P 2 2 (R r with R=0.01 , l=3 , P=3000 , and =0.027 , we have v as a function of r : 4l. 3000 2 2 (0.01 r ) . v(0)=0.925 cm / s, v(0.005)=0.694 cm / s, v(0.01)=0 . 4(0.027)3 P P Pr 2 2 / (R r ) v (r)= (2r)= (b) v(r)= . When l=3 , P=3000 , and =0.027 , we have 4l 4l 2l. v(r)=. 9.

(64) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences. 3000r / / / . v (0)=0 , v (0.005)=92.592(cm/s)/ cm, and v (0.01)=185.185(cm/s)/ cm. 2(0.027)3 (c) The velocity is greatest where r=0 (at the center) and the velocity is changing most where r=R=0.01 cm (at the edge). /. v (r)=. 28. (a) (a). 1 2L. T = . 1 2. T . L . 1 2L. T = . 1 2. T . L . f=. 1 2L. T = . 1 2L . T. f=. 1 2L. T = . 1 2L . T. f=. 1 2L. T = . T 2L. . 1 2L. T = . T 2L. . f= (b) f= (c). (d). (e). (f) f=. (b) (i). 1. 1. df = dL. 1 2. T . L =. df = dL. 1 2. T . L =. 2. 2. df 1 = dT 2. 1 2L . T. 1/2. . df 1 = dT 2. 1 2L . T. 1/2. . df 1 = d 2. T 2L. . df 1 = d 2. T 2L. . 1 2L. . . 2. T . 2L. 1/2. 1/2. 2. T . 1. 1/2. 1 4L T. 1/2. 1 4L T. =. =. T. 3/2. =. 4L. 3/2. T. 3/2. =. 4L. 3/2. df <0 and L is decreasing f is increasing higher note dL. (ii). df >0 and T is increasing f is increasing higher note dT. (iii). df <0 and is increasing f is decreasing lower note d 2. 3. /. 2. 29. (a) C(x)=2000+3x+0.01x +0.0002x C (x)=3+0.02x+0.0006x /. /. (b) C (100)=3+0.02(100)+0.0006(10 , 000)=3+2+6=$11/ pair. C (100) is the rate at which the cost is increasing as the 100 th pair of jeans is produced. It predicts the cost of the 101 st pair. (c) The cost of manufacturing the 101 st pair of jeans is 10.

(65) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences. C(101)C(100) = (2000+303+102.01+206.0602)(2000+300+100+200) = 11.0702 $11.07 2. 3. /. 2. /. 30. (a) C(x)=84+0.16x0.0006x +0.000003x C (x)=0.160.0012x+0.000009x C (100)=0.13 . This is the rate at which the cost is increasing as the 100 th item is produced. (b) C(101)C(100)=97.1303029997 $0.13 . /. /. p(x) / / xp (x) p(x) 1 xp (x) p(x) 31. (a) A(x)= . A (x)>0 A(x) is increasing; that is, A (x)= = 2 2 x x x the average productivity increases as the size of the workforce increases. p(x) / / / (b) p (x) is greater than the average productivity p (x)>A(x) p (x)> x /. /. /. xp (x)> p(x) xp (x) p(x)>0. xp (x) p(x) 2. /. >0 A (x)>0 .. x 32. (a). (. 1+4x dR = S= dx =. 9.6x. 0.4. 0.6. ) ( 9.6x0.6) ( 40+24x0.4) ( 1.6x0.6) ( 1+4x0.4) 2. +38.4x. 0.2. 64x. 0.6. 0.4 2. ( 1+4x ). 38.4x. 0.2. =. 54.4x. 0.6. ( 1+4x0.4) 2. (b) At low levels of brightness, R is quite large and is quickly decreasing, that is, S is negative with large absolute value. This is to be expected: at low levels of brightness, the eye is more sensitive to slight changes than it is at higher levels of brightness. PV PV 1 = = (PV ) . Using the Product Rule, we have nR (10)(0.0821) 0.821 dT 1 1 / / = [P(t)V (t)+V (t)P (t)]= [(8)(0.15)+(10)(0.10)] 0.2436 K / min. dt 0.821 0.821. 33. PV =nRT T =. 11.

(66) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences. 34. (a) If dP/dt=0 , the population is stable (it is constant). dP P P P (b) =0 P=r 1 P =1 =1 P=P 0 c dt P r P P r c. 0. c. c. 0. If P =10 , 000 , r =5%=0.05 , and =4%=0.04 , then P=10 , 000 c. 0. (c) If =0.05 , then P=10 , 000. 1. 5 5. 1. 1. r. .. 0. 4 5. =2000 .. =0 . There is no stable population.. 35. (a) If the populations are stable, then the growth rates are neither positive nor negative; that is, dC dW =0 and =0 . dt dt (b) ‘‘The caribou go extinct’’ means that the population is zero, or mathematically, C=0 . dC dW (c) We have the equations =aCbCW and =cW +dCW . Let dC/dt=dW /dt=0 , a=0.05 , dt dt b=0.001 , c=0.05 , and d=0.0001 to obtain 0.05C0.001CW =0 (1) and 0 .05W +0.0001CW =0 (2) . Adding 10 times (2) to (1) eliminates the CW terms and gives us 0.05C0.5W =0C=10W . Substituting C=10W into (1) results in 2. 2. 0.05(10W )0.001(10W )W =0 0.5W 0.01W =0 50W W =0 W (50W )=0 W =0 or 50 . Since C=10W , C=0 or 500 . Thus, the population pairs (C,W) that lead to stable populations are (0,0) and (500,50) . So it is possible for the two species to live in harmony.. 12.

(67) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions. /. 1. f (x)=x3sin x f (x)=13cos x /. 2. f (x)=xsin x f (x)=x cos x+(sin x) 1=xcos x+sin x /. 2. 3. y=sin x+10tan x y =cos x+10sec x /. 4. y=2 x+5cos x y =2 xcot x5sin x 3. /. 3. 2. 2. 3. 2. 5. g(t)=t cos t g (t)=t (sin t)+(cos t) 3t =3t cos tt sin t or t (3cos ttsin t) /. 2. 6. g(t)=4sec t+tan t g (t)=4sec ttan t+sec t . 7. h( )=csc +e cot . /. . 2. . 2. h ( )=csc cot +e (csc )+(cot )e =csc cot +e (cot csc ) u. /. u. u. u. 8. y=e (cos u+cu) y =e (sin u+c)+(cos u+cu)e =e (cos usin u+cu+c) 9. y=. x / (cos x)(1)(x)(sin x) cos x+xsin x y = = 2 2 cos x cos x (cos x) 1+sin x x+cos x. 10. y= /. y =. (x+cos x)(cos x)(1+sin x)(1sin x) 2. (x+cos x) 2. =. 2. (x+cos x). =. 2. 2. (x+cos x). 2. xcos x+cos x(cos x). 2. xcos x+cos x(1sin x). =. xcos x 2. (x+cos x). sec 1+sec (1+sec )(sec tan )(sec )(sec tan ). 11. f ( )= /. f ( ). 12. y=. =. 2. (1+sec ). =. (sec tan )[(1+sec )sec ] 2. (1+sec ). =. sec tan 2. (1+sec ). tan x1 sec x 1.

(68) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions. (. 2. 2. ). 2. dy sec xsec x(tan x1)sec xtan x sec x sec xtan x+tan x 1+tan x = = = 2 2 dx sec x sec x sec x /. Another method: Simplify y first: y=sin xcos x y =cos x+sin x . 13. y=. sin x 2. /. y =. 2. x cos x(sin x)(2x). ( x2) 2. x. =. x(xcos x2sin x) x. 4. =. xcos x2sin x x. 3. 14. y= ( +cot ) /. 2. 2. 2. y = (1csc )+( +cot )( cot )= (1csc cot cot ) 2. 2. 2. = (cot cot cot ). 2. {1+cot =csc }. 2. = ( cot 2cot )= cot ( +2cot ) /. 2. 2. 2. 15. y=sec tan y =sec (sec )+tan (sec tan )=sec (sec +tan ) 2. 2. Using the identity 1+tan =sec , we can write alternative forms of the answer as 2. 2. sec (1+2tan ) or sec (2sec 1) /. /. /. /. 16. Recall that if y= fgh , then y = f gh+ fg h+ fgh . y=xsin xcos x dy 2 2 =sin xcos x+xcos xcos x+xsin x(sin x)=sin xcos x+xcos xxsin x dx d d ( csc(x) ) = dx dx. 1 sin x. =. d d ( sec x ) = dx dx. 1 cos x. =. d d 19. ( cot x ) = dx dx. cos x sin x. 17.. 18.. (sin x)(0)1(cos x) 2. =. sin x. 2. =. sin x. (cos x)(0)1(sin x) 2. cos x. =. cos x. =. sin x 2. cos x. (sin x)(sin x)(cos x)(cos x) 2. sin x. =. 1 cos x = xcot x sin x sin x. 1 sin x =sec xtan x cos x cos x 2. =. 2. sin x+cos x 2. =. sin x. 1. 2. 2. = x. sin x. 20. f (x)=cos x f (x+h) f (x) cos (x+h)cos x cos xcos hsin xsin hcos x / = lim = lim = lim f (x) h h h h 0 h 0 h 0 2.

(69) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions. = lim. cos x. h 0. cos h1 sin h sin x h h. =cos x lim h 0. cos h1 sin h sin x lim h h 0 h. =(cos x)(0)(sin x)(1)=sin x /. 2. 21. y=tan x y =sec x the slope of the tangent line at equation of the tangent line is y1=2 x. x. /. x 4 x. ,1 4. or y=2x+1 . 2. is sec. 2. 4. 2. =( 2 ) =2 and an. x. 22. y=e cos x y =e (sin x)+(cos x)e =e (cos xsin x) the slope of the tangent line at (0, 1) is 0. e (cos 0sin 0)=1(10)=1 and an equation is y1=1(x0) or y=x+1 . /. /. 23. y=x+cos x y =1sin x . At ( 0,1 ) , y =1 , and an equation of the tangent line is y1=1(x0) , or y=x+1 . 1 / / cos xsin x 10 [ Reciprocal Rule]. At ( 0,1 ) , y = y = =1 , and an 2 2 sin x+cos x (sin x+cos x) (0+1) equation of the tangent line is y1=1(x0) , or y=x+1 . 24. y=. /. 25. (a) y=xcos x y =x(sin x)+cos x(1)=cos xxsin x . So the slope of the tangent at the point ( , ) is cos sin =1 (0)=1 , and an equation is y+ =(x ) or y=x .. (b) /. 26. (a) y=sec x2cos x y =sec xtan x+2sin x 3 the slope of the tangent line at ,1 is sec tan +2sin =2 3 +2 =3 3 and an 2 3 3 3 3 equation is y1=3 3 x or y=3 3 x+1 3 . 3 (b). 3.

(70) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions. /. 2. 27. (a) f (x)=2x+cot x f (x)=2csc x. (b) /. Notice that f (x)=0 when f has a horizontal tangent. /. f is positive when f is increasing and f negative when the graph of f is steep.. /. /. /. is negative when f is decreasing. Also, f (x) is large. 28. (a) f (x)= x sin x f (x)= x cos x+(sin x). 1 1/2 sin x x = x cos x+ 2 2 x. (b) /. Notice that f (x)=0 when f has a horizontal tangent. f. /. is positive when f is increasing and f. /. is negative when f is decreasing. /. 29. f (x)=x+2sin x has a horizontal tangent when f (x)=0 1+2cos x=0cos x=. 1 2. 2 2 4 4 +2 n or +2 n , where n is an integer. Note that and are units from . 3 3 3 3 3 This allows us to write the solutions in the more compact equivalent form (2n+1) , n an 3 integer. x=. 2. 2. cos x / (2+sin x)(sin x)cos xcos x 2sin xsin xcos x 2sin x1 30. y= y = = = =0 when 2 2 2 2+sin x (2+sin x) (2+sin x) (2+sin x) 4.

(71) Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions. 1 11 7 x= +2 n or x= +2 n , n an integer. So y= 2 6 6 11 1 +2 n, the points on the curve with horizontal tangents are: , 6 3 n an integer.. 2sin x1=0sin x=. 31. (a) x(t)=8sin t v(t)=x 2 (b) The mass at time t= 3 2 2 v =8cos =8 3 3. /. 1 1 or y= and 3 3 7 1 +2 n, , 6 3. (t)=8cos t. has position x 1 2. 2 3. =4 . Since v. =8sin 2 3. 2 =8 3. 3 2. =4 3 and velocity. <0 , the particle is moving to the left.. 32. (a) s(t)=2cos t+3sin t v(t)=2sin t+3cos t. (b) (c) s=0 t 2.55 . So the mass passes through the equilibrium position for the first time when 2. t 2.55 s. (d) v=0 t 0.98 , s(t ) 3.61 cm. So the mass travels a maximum of about 3.6 cm (upward and 1. 1. downward) from its equilibrium position. (e) The speed |v| is greatest when s=0 ; that is, when t=t +n , n a positive integer. 2. 33. From the diagram we can see that sin =x/10 x=10sin . We want to find the rate of change of x with respect to ; that is, dx/d . Taking the derivative of the above expression, dx/d =10(cos ) . dx. 1 So when = , =10cos =10 =5 ft/rad. 3 d 3 2 34. (a) F=. dF ( sin +cos )(0)W ( cos sin ) W (sin cos ) W = = 2 2 d sin +cos ( sin +cos ) ( sin +cos ). (b) 5.