f(t) 3 + 4t t2 for 0 t 2. If we substitute t =2intothese two inequalities, we get bounds on the position at time 2: 21 f(2) 25.

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1 96 Chapter Three /SOLUTIONS That is, f(t) 3 + 4t + 7 t for t. In the same way, we can show that the lower bound on the acceleration, 5 f (t) leads to: f(t) 3 + 4t + 5 t for t. If we substitute t =intothese two inequalities, we get bounds on the position at time : f() Consider the function f() =h() g(). Since f () =h () g (), we know that f is nondecreasing by the Increasing Function Theorem. This means f() f(b) for a b. However, f(b) =h(b) g(b) =, so f(), which means h() g(). 4. If f () =, then both f () and f (). By the Increasing and Decreasing Function Theorems, f is both nondecreasing and nonincreasing, so f is constant. 5. Let h() =f() g(). Then h () =f () g () = for all in (a, b). Hence, by the Constant Function Theorem, there is a constant C such that h() =C on (a, b). Thus f() =g()+c. 6. We will show f() =Ce by deducing that f()/e is a constant. By the Constant Function Theorem, we need only show the derivative of g() =f()/e is zero. By the quotient rule (since e ), we have Since f () =f(), we simplify and obtain which is what we needed to show. g () = f ()e e f() (e ). g () = f()e e f() = =, (e ) e 7. Apply the Racetrack Principle to the functions f() f(a) and M( a); we can do this since f(a) f(a) =M(a a) and f () M. We conclude that f() f(a) M( a). Similarly, apply the Racetrack Principle to the functions m( a) and f() f(a) to obtain m( a) f() f(a). If we substitute = b into these inequalities we get Now, divide by b a. m(b a) f(b) f(a) M(b a). 8. (a) Since f (), f () is nondecreasing on (a, b). Thus f (c) f () for c < b and f () f (c) for a< c. (b) Let g() =f(c)+f (c)( c) and h() =f(). Then g(c) =f(c) =h(c), and g () =f (c) and h () =f (). If c < b, then g () h (), and if a < c, then g () h (), by (a). By the Racetrack Principle, g() h () for c < b and for a< c, as we wanted. Solutions for Chapter 3 Review Eercises. w = (t + ) 99 (t) = t(t + ) 99.. y = 3 e 3 w. ) (te t t 3. f (t) = d dt 4. g (t) = 4(3 + ( ) t) t / = =e t +te t + t 3/. t(3 + t)

2 SOLUTIONS to Review Problems for Chapter Three Using the quotient rule, h (t) = ( )(4 + t) (4 t) (4 + t) = 8 (4 + t). ( 6. f () = (3 ln ) + = 9 9 ) ln = ln 7. Either notice that f() = +3 + ( + )( + ) can be written as f() = + + giving f () =, or use the quotient rule which gives which reduces to f() = +, f () = ( + )( + 3) ( +3 + ) ( + ) = ( + ) = + + ( + ) ( + ) = ( + ) =. 8. Using the chain rule, g (θ) = (cos θ)e sin θ. 9. Since h(θ) =θ(θ / θ )=θθ / θθ = θ / θ, we have h (θ) = θ / + θ.. f (θ) = sin θ = tan θ. cos θ d. dy ln 3 ln(y3 )= ln(y 3 ) y 3 6y = y ln(y 3 ).. g () = d ( k + k ) = k k + k ln k. d 3. y = dz 4. dθ = 3 sin θ cos θ 5. f (t) = cos(3t + 5) ( sin(3t + 5))3 = 6 cos(3t + 5) sin(3t + 5) 6. M (α) = tan( + 3α) cos ( + 3α) 3 tan( + 3α) =6 cos ( + 3α) 7. s (θ) = d dθ sin (3θ π) = 6 cos(3θ π) sin(3θ π). 8. h ( (t) = e t ). e t t 9. ( ) d sin(5 θ) = cos(5 θ)( )θ sin(5 θ)(θ) dθ θ θ 4 θ cos(5 θ) + sin(5 θ) =. θ 3

3 98 Chapter Three /SOLUTIONS. w (θ) = θ cos θ sin θ sin 3 θ. f (θ) = ( + e θ ) (e θ e θ )( ) = ( + e θ ).. g (w) = d ( ) = w ln + e w dw w + e w ( w + e w ). 3. g () = d ( ) = d Using the quotient rule and the chain rule, h (z) = ( ) / [ ] sin(z) cos(z) cos(z) sin(z)( sin(z)) cos(z) cos (z) ( ) / [ ] cos(z) cos (z) + sin (z) = sin(z) cos (z) = (cos(z)) / (sin(z)) / cos (z) = sin(z) cos3 (z). 5. Using the chain rule and simplifying, q (θ) = (4θ sin (θ)) / (8θ sin(θ)( cos(θ))) = 4θ sin(θ) cos(θ). 4θ sin (θ) 6. Using the product rule and factoring gives dw dz = 4z [ 4 ln() sin(πz)+πcos(πz)]. 7. s () = d (arctan( )) = d + ( ). 8. r (θ) = d ( e (eθ +e θ ) ) = e (eθ +e θ ) ( e θ e θ). dθ 9. Using the chain rule, we get: m (n) = cos(e n ) (e n ) 3. Using the chain rule we get: k (α) =e tan(sin α) (tan(sin α)) = e tan(sin α) cos α. cos (sin α) 3. Here we use the product rule, and then the chain rule, and then the product rule. g (t) = cos( te t )+t(cos te t ) = cos( te t )+t( sin( te t ) ( te t ) ) = cos( te t ) t sin( ( te te t ) t + ) t et ( ) 3. f (r) =e(tan + tan r) e (tan + tan r) = e(tan + tan r) e cos r d 33. d etan = e tan + e tan cos. dy 34. d =e sin (3)+e ( sin(3) cos(3)3) = e sin(3)(sin(3) + 3 cos(3)) 35. g 6 () = + (3 + ) = dy 36. d = (ln )sin cos cos + sin ( sin ) = ( sin (ln ) cos sin )

4 37. h() =a ln e = a, so h () =a. 38. k () =a 39. f (θ) =ke kθ 4. Using the product rule and factoring gives f (t) =e 4kt (cos t 4k sin t). 4. Using the product rule gives 4. d a dθ sin θ = a sin θ 43. Using the chain rule gives f () = 5 ln(a)a Using the quotient rule gives H (t) = ate ct c(at + b)e ct =( cat +at bc)e ct. sin θ cos θ ( sin θ cos θ) =. a sin θ f () = ( )(a + ) ()(a ) (a + ) = 4a (a + ). SOLUTIONS to Review Problems for Chapter Three Using the quotient rule gives w (r) = ar(b + r3 ) 3r (ar ) (b + r 3 ) = abr ar4 (b + r 3 ). 46. Using the quotient rule gives 47. ( ) dy d = + ( ) = Using the chain rule gives r (t) = cos( t k ) sin( t k ) ( k s a + s s a f (s) = +s (a s ) (a + s ) = s(a + s ) s(a s ) (a + s ) 3/ = a s s 3 a s + s 3 (a + s ) 3/ = 3a s s 3 (a + s ) 3/. 49. Since g(w) = 5(a w ), g (w) = (a w ) 3 ( w) = 5. ). w (a w ) 3 dy d = (e + e )(e + e ) (e e )(e e ) (e + e ) = (e + e ) (e e ) = (e + + e ) (e +e ) (e + e ) (e + e ) 4 = (e + e )

5 Chapter Three /SOLUTIONS 5. g (u) = aeau a + b 5. Using the quotient and chain rules, we have dy d = (aea + ae a )(e a + e a ) (e a e a )(ae a ae a ) (e a + e a ) = a(ea + e a ) a(e a e a ) (e a + e a ) = a[(ea + + e a ) (e a +e a )] (e a + e a ) 4a = (e a + e a ) 53. Using the quotient rule gives g (t) = g (t) = g (t) = ( k +) (ln(kt) t) (ln(kt)+t) ( k ) kt kt (ln(kt) t) ( +) (ln(kt) t) (ln(kt)+t) ( ) t t (ln(kt) t) ln(kt)/t + ln(kt) t ln(kt)/t + ln(kt)+t (ln(kt) t) g (t) = ln(kt) (ln(kt) t). 54. Using the quotient and chain rules d dz dt = dt (et + t) sin(t) (e t + t) d (sin(t)) dt (sin(t)) ( ) e t d dt (t ) + sin(t) (e t + t) cos(t) d (t) dt = sin (t) = (tet + ) sin(t) (e t + t) cos(t). sin (t) 55. Using the chain rule twice: f (t) = cos e t + d et + = cos e dt t + e t + d dt (et + ) = cos e t + e t + et = e t cos et + e t Using the chain rule twice: g (y) =e e(y3 ) d dy ( ) e (y3 ) =e e(y3 ) e (y 3 ) d dy (y3 ) = 6y e (y 3 ) e e(y3 ). 57. g () = (54 + ). 58. y = g(z) =z 5 +5z 4 z g (z) = 5z 4 + z f (z) = ( ln 3)z + (ln 4)e z. 6. g () = d d ( /3 +3 e) = ln f () = 6(e 4) + (3 + π)e =6e 4 +3 e + πe.

6 SOLUTIONS to Review Problems for Chapter Three 63. f (θ) = θ sin θ + θ cos θ + cos θ θ sin θ cos θ = θ cos θ. 64. dy dθ = (cos(5θ)) ( sin(5θ) 5) + sin(6θ) cos(6θ) 6 = 5 sin(5θ) + sin(6θ) cos(6θ) cos(5θ) 65. r (θ) = d dθ sin[(3θ π) ] = cos[(3θ π) ] (3θ π) 3 = 6(3θ π) cos[(3θ π) ]. 66. Using the product and chain rules, we have dy d = 3( + 5) ()(3 3 ) +( + 5) 3 [(3 3 )(9 )] = 3()( + 5) (3 3 )[(3 3 ) + ( + 5)(3)] =6( + 5) (3 3 )[ ]. 67. Since tan(arctan(kθ)) = kθ, because tangent and arctangent are inverse functions, we have N (θ) =k. 68. Using the product rule gives h (t) =ke kt (sin at + cos bt)+e kt (a cos at b sin bt). 69. f () = d d ( 4 3 )(6 e 3π) = ( 4 6)(6 e 3π) + ( 4 3 )(6e e ). 7. f (t) = 4(sin(t) cos(3t)) 3 [ cos(t) + 3 sin(3t)] 7. Since cos y + sin y =, we have s(y) = = 3 4. Thus s (y) =. 7. f () = ( +6 )( ) + (4 + 3 )( ) =( ) + ( ) = h () = ( + ) ( 3 +4 ) ( + 3 ) (6 ) = = Note: f(z) = (5z) / +5z / +5z / 5z / + 5, so f (z) = 5 (5z) / + 5 z / 5 z 3/ + 5 z 3/ y dy d 8y 4 dy d = (3y 4 ) dy =8y 3 d dy 8y 3 = d 3y Differentiating implicitly on both sides with respect to, a cos(ay) dy dy b sin(b) =y + d d (a cos(ay) ) dy d = y + b sin(b) dy d = y + b sin(b) a cos(ay).

7 Chapter Three /SOLUTIONS 77. We wish to find the slope m = dy/d. To do this, we can implicitly differentiate the given formula in terms of : Thus, at (, ), m = ()/3( ) = / Taking derivatives implicitly, we find +3y =7 +6y dy d = d d (7) = dy d = 6y = 3y. dy dy + cos y + = d d dy d = + cos y So, at the point =3,y =, dy d = ( )(3) + cos = 6 = First, we differentiate with respect to : At =3, we have Our two points, then, are (3, ) and (3, 4). At (3, ), At (3, 4), dy dy + y + y d d = dy ( +y) = y d dy d = 3y + y =4 y +3y 4 = (y )(y + 4) =. y +y. dy d = 3 + () = 5 ; Tangent line: (y ) = ( 3). 5 dy d = ( 4) 3 + ( 4) = 4 5 ; Tangent line: (y + 4) = 4 ( 3). 5 Problems 8. f (t) = 6t 8t +3 and f (t) = t f () = 8 + f (r) = 8 + r =4 r = =3. 8. f () = ( + ) ( + )( ). Hence or.

8 SOLUTIONS to Review Problems for Chapter Three f () = 3( 5) + (3 + 8) = + f () =. 84. (a) Applying the product rule to h() we get h () = t ()s() + t()s () ( ) 3 + = 6. (b) Applying the product rule to h() we get h () = t ()s() + t()s () ( ) + =. (c) Applying the quotient rule to p() we get p () = t ()s() t()s () ( ) =. (s()) Note that since t() is a linear function whose slope looks like from the graph, t () everywhere. To find s (), draw a line tangent to the curve at the point (,s()), and estimate the slope. 85. Since r() =s(t()), the chain rule gives r () =s (t()) t (). Thus, r () = s (t()) t () s () ( ) ( )( ) = 4. Note that since t() is a linear function whose slope looks like from the graph, t () everywhere. To find s (), draw a line tangent to the curve at the point (,s()), and estimate the slope. 86. (a) Applying the chain rule we get h () = s (s()) s () s (3) =. (b) Applying the chain rule we get h () = s (s()) s () s () s () = ( ) =4. To find s (), draw a line tangent to the curve at the point (,s()), and estimate the slope. 87. We need to find all values for such that dy d = s (s()) s () =. This is the case when either s (s()) = or s () =. From the graph we see that s () = when. Also, s (s()) = when s(), which happens when.4or.4. To find s (a), for any a, draw a line tangent to the curve at the point (a, s(a)), and estimate the slope. 88. (a) Applying the product rule we get h ( ) = ( ) t( ) + ( ) t ( ) ( ) 4 + ( ) =. (b) Applying the chain rule we get p ( ) = t (( ) ) ( ) = t () ( ) ( ) = 4. Note that since t() is a linear function whose slope looks like from the graph, t () everywhere. 89. We have r() = s(t()) s(). By the chain rule, r () =s (t()) t (), so Thus the equation of the tangent line is r () = s (t()) t () s () ( ) ( ) = 4. y = 4( ) y = Note that since t() is a linear function whose slope looks like from the graph, t () everywhere. To find s (), draw a line tangent to the curve at the point (,s()), and estimate the slope. 9. We have (f ) (5) = f (f (5)). From the graph of f() we see that f (5) = 3. From the graph of f () we see that f (3) =.36. Thus (f ) (5) = /.36 = We have (f ) () = f (f ()). From the graph of f() we see that f () = 3. From the graph of f () we see that f (3) =.6. Thus (f ) () = /.6 = We have (f ) (5) = f (f (5)). From the graph of f() we see that f (5) = 3. From the graph of f () we see that f (3) =.73. Thus (f ) (5) = /.73 = Since W is proportional to r 3, we have W = kr 3 for some constant k. Thus, dw/dr = k(3r ) = 3kr. Thus, dw/dr is proportional to r.

9 4 Chapter Three /SOLUTIONS 94. Taking the values of f, f, g, and g from the table we get: (a) h(4) = f(g(4)) = f(3) =. (b) h (4) = f (g(4))g (4) = f (3) =. (c) h(4) = g(f(4)) = g(4) = 3. (d) h (4) = g (f(4))f (4) = g (4) 3 = 3. (e) h (4) = (f(4)g (4) g(4)f (4)) /f (4) = 5/6. (f) h (4) = f(4)g (4) + g(4)f (4) = (a) H () = r ()s() + r()s () = =. (b) H () = r () r() = 4 = 4. (c) H () = r (s())s () = r () 3, but we don t know r (). (d) H () = s (r())r () = s (4)r () = (a) f() = 4g() f () = 4g () f () = () 4( 4) = = (b) f() = g() f () = g() g () (g()) f () = g() g () (g()) = 3 ( 4) (3) = 9 (c) f() = g() f () = g()+ g () f () = ()(3) + () ( 4) = 6 = 4 (d) f() = (g()) f () = g() g () f () = (3)( 4) = 4 (e) f() = sin(g()) f () = sin(g()) + cos(g()) g () f () = sin(g()) + cos(g()) g () = sin 3 + cos(3) ( 4) = sin 3 8 cos 3 (f) f() = ln(g()) f () = ln(g()) + ( g () g() ) f () = () ln 3 + () ( 4 ) 3 = 4 ln (a) f() = 4g() f() = 4 4(3) = 8 f () = Thus, we have a point (, 8) and slope m =. This gives (b) f() = g() 8 = () + b b = 48, so y = 48. f() = 3 f () = 9 Thus, we have point (, ) and slope m =. This gives =( 9 )() + b b = 3 9 = 6 9, so

10 SOLUTIONS to Review Problems for Chapter Three 5 y = (c) f() = g() f() = 4 g() = 4(3) = f () = 4 Thus, we have point (, ) and slope m = 4. This gives = ( 4) + b b =, so y = 4 +. (d) f() = (g()) f() = (g()) = (3) =9 f () = 4 Thus, we have point (, 9) and slope m = 4. This gives 9 = ( 4) + b b = 57, so y = (e) f() = sin(g()) f() = sin(g()) = sin 3 f () = sin 3 8 cos 3 We will use a decimal approimation for f() and f (), so the point (, sin 3) (,.8) and m 8.6. Thus, (f) f() = ln g() f() = 4 ln g() = 4 ln f () = 4 ln = (8.6) + b b = 5.84, so y = Thus, we have point (, 4.39) and slope m =.94. This gives 4.39 = (.94) + b b =6.7, so y = When we zoom in on the origin, we find that two functions are not defined there. The other functions all look like straight lines through the origin. The only way we can tell them apart is their slope. The following functions all have slope and are therefore indistinguishable: sin tan, cos, sin, and. + cos These functions all have slope at the origin, and are thus indistinguishable: sin arcsin,, arctan, +sin e,, and. + + Now, sin and ln both are undefined at the origin, so they are distinguishable from the other functions. In addition, while sin has a slope that approaches zero near the origin, ln becomes vertical near the origin, so they are distinguishable from each other. Finally, + is the only function defined at the origin and with a vertical tangent there, so it is distinguishable from the others. 99. It makes sense to define the angle between two curves to be the angle between their tangent lines. (The tangent lines are the best linear approimations to the curves). See Figure 3.5. The functions sin and cos are equal at = π 4. For f () = sin, f ( π 4 ) = cos( π 4 )= For f () = cos, f ( π 4 )= sin( π 4 )=.

11 6 Chapter Three /SOLUTIONS Using the point ( π, ) for each tangent line we get y = + ( π ) and y = + ( + π ), respectively ( ) + π 4 y β α y = y = sin + ( ) π 4 ( ) + π 4 π 8 y π 4 α ( ) π 4 π 4 y = cos y = + ( ) + π 4 ( ) π 4 Figure 3.5 Figure 3.6 There are two possibilities of how to define the angle between the tangent lines, indicated by α and β above. The choice is arbitrary, so we will solve for both. To find the angle, α, we consider the triangle formed by these two lines and the y-ais. See Figure 3.6. ( ) π/8 tan α = = π/4 α =.6548 radians α =.3 radians, or 7.5. Now let us solve for β, the other possible measure of the angle between the two tangent lines. Since α and β are supplementary, β = π.3 =.99 radians, or The curves meet when + = +, that is when ( ) = so that =or =. Let Then y () = + and y () = +. y = and y = +. At =, y =, y = so that y y = and the curves are perpendicular. At =, y =, y = so that y y = and the curves are perpendicular.. The curves meet when 3 /3=, that is when =. So the roots of this equation give us the - coordinates of the intersection point. By numerical methods, we see there is one solution near =.3. See Figure 3.7. Let y () = 3 and y () =. 3 So we have y = and y =. However, y () = +, so if the curves are to be perpendicular when they cross, then y must be. Since y =, y = only at = ± which is not the point of intersection. The curves are therefore not perpendicular when they cross. 5 5 y y = Figure 3.7

12 . Differentiating gives dy = ln + b. d To find the point at which the graph crosses the -ais, set y =and solve for : Since >, we have = ln b =(ln b). ln b = = e b. At the point (e b, ), the slope is dy d = ln ( e b) + b = b + b =. Thus the equation of the tangent line is y = ( e b ) y = e b. SOLUTIONS to Review Problems for Chapter Three 7 3. Using the definition of cosh and sinh, we have cosh = e + e and sinh 3 = e3 e 3. Therefore cosh() lim sinh() = lim e + e e 3 e 3 e (e 4 + ) = lim e (e 5 e ) e 4 + = lim e 5 e =. 4. Using the definition of sinh we have sinh = e e. Therefore lim e sinh() = lim e e e = lim e 4 =. 5. Using the definition of cosh and sinh, we have cosh = e + e and sinh = e e. Therefore sinh( ) lim cosh( ) = lim e e e + e e ( e ) = lim e ( + e ) = lim e +e =.

13 8 Chapter Three /SOLUTIONS 6. (a) Since f() =, we have f () = (/) /. So f(4) = 4 = and f (4) = (/)4 / =/4, and the tangent line approimation is See Figure 3.8. (b) For =4., the true value is whereas the approimation is f() f(4) + f (4)( 4) + ( 4). 4 f(4.) = 4. = , f(4.) + (4. 4) =.5. 4 Thus, the approimation differs from the true value by about.5. (c) For = 6 the true value is: f(6) = 6 = 4 whereas the approimation is f(6) + (6 4) = 5. 4 Thus, the approimation differs from the true value by. (d) The tangent line is a good approimation to the graph near =4, but not necessarily far away. Of course, there s no reason to epect that the curve will look like the tangent line if we go too far away, and usually it does not. (See Figure 3.8.) The problem is that we have traveled too far from the place where the curve looks like a line with slope /4. 5 Tangent line at =4 4 f() = 4 6 Figure 3.8: Local linearization: Approimating f() = by its tangent line at =4 7. (a) From the figure, we see a =. The point with =lies on both the line and the curve. Since we have Since the slope of the line is 3, we have (b) We use the line to approimate the function, so y = =, f(a) =. f (a) = 3. f(.) 3(.) + 7 =.7. This is an underestimate, because the line is beneath the curve for >. Similarly, f(.98) 3(.98) + 7 =.6. This is an overestimate because the line is above the curve for <. The approimation f(.98).6 is likely to be more accurate because.98 is closer to than. is. Since the graph of f() appears to bend away from the line at approimately the same rate on either side of =, in this eample, the error is larger for points farther from =.

14 SOLUTIONS to Review Problems for Chapter Three 9 8. Differentiating gives Rate of change of price = dv dt = 75(.35)t ln.35.5(.35) t dollar/yr. dg 9. (a) dr = GM d ( ) = GM d ( ) r = GM( )r 3 = GM. dr r dr r 3 dg (b) is the rate of change of acceleration due to the pull of gravity. The further away from the center of the earth, the dr weaker the pull of gravity is. So g is decreasing and therefore its derivative, dg, is negative. dr (c) By part (a), dg dr r=64 = GM r 3 r=64 (d) It is reasonable to assume that g is a constant near the surface of the earth. = (6.67 )(6 4 ) (64) (a) If the distance s(t) = e t, then the velocity, v(t), is given by ( ) ( ) v(t) =s (t) = e t = e )( t. (b) Observing the differentiation in (a), we note that s (t) =v(t) = Substituting s(t) for e t, we obtain s (t) = s(t). ( ) e t = s(t). = e t. dy = 7.5(.57) sin(.57t) = 3.8 sin(.57t) dt (a) When t =6, we have dy = 3.8 sin(.57 6) =.38 meters/hour. So the tide is falling at.38 meters/hour. dt (b) When t =9, we have dy = 3.8 sin(.57 9) = 3.76 meters/hour. So the tide is rising at 3.76 meters/hour. dt (c) When t =, we have dy = 3.8 sin(.57 ) =.75 meters/hour. So the tide is rising at.75 meters/hour. dt (d) When t = 8, we have dy = 3.8 sin(.57 8) =. meters/hour. So the tide is falling at. meters/hour. dt. Since we re given that the instantaneous rate of change of T at t = 3 is, we want to choose a and b so that the derivative of T agrees with this value. Differentiating, T (t) =ab e bt. Then we have We also know that at t = 3, T =, so =T (3) = abe 3b or e 3b = ab = T (3) = ae 3b or e 3b = 8 a Thus 8 a = e 3b = ab, so b = =.5 and a = (a) Differentiating, we see (b) We have v = dy = πωy sin(πωt) dt a = dv dt = 4π ω y cos(πωt). y = y cos(πωt) v = πωy sin(πωt) a = 4π ω y cos(πωt).

15 Chapter Three /SOLUTIONS So Amplitude of y is y, Amplitude of v is πωy =πω y, Amplitude of a is 4π ω y =4π ω y. The amplitudes are different (provided πω ). The periods of the three functions are all the same, namely /ω. (c) Looking at the answer to part (a), we see So we see that d y dt = a = 4π ω (y cos(πωt)) = 4π ω y. d y dt +4π ω y =. 4. (a) Since lim e.t =, we see that lim =. Thus, in the long run, close to,, t t + 5e.t people will have had the disease. This can be seen in Figure 3.9. (b) The rate at which people fall sick is given by the first derivative N (t). N (t) N, where t =day. t N (t) = 5,, e.t ( + 5e.t ) = 5,, e.t + 5,,e.t + 4 In Figure 3.3, we see that the maimum value of N (t) is approimately 5,. Therefore the maimum number of people to fall sick on any given day is 5,. Thus there are no days on which a quarter million or more get sick. N,, N(t) dn dt 5,, 5,, 5, N (t) Figure 3.9 t t 5 5 Figure (a) We solve for t to find the time it takes for the population to reach billion. P (t) = 6.7e kt = e kt = 6.7 t = ln(/6.7) k =.448 k Thus the time is f(k) =.448 years. k (b) The time to reach billion with a growth rate of.% is f(.) =.448. years = 33.4 years.

16 SOLUTIONS to Review Problems for Chapter Three (c) We have f (k) =.448 k f (.) = 78. Thus, for growth rates, k, near.% we have Time for world population to reach billion = f(k) f(.) + f (.)(k.) f(k) = (k.) years (d) True time to reach billion when the growth rate is.% is f(.) =.448/. = 4. years. The linear approimation gives Approimate time = (..) = 39. years. 6. (a) Suppose Then g = f(r) = GM r. f (r) = GM. r 3 So f(r + r) f(r) GM ( r). r 3 Since f(r + r) f(r) = g, and g = GM/r, we have g GM r 3 r ( r) = g r. (b) The negative sign tells us that the acceleration due to gravity decreases as the distance from the center of the earth increases. (c) The fractional change in g is given by g g r r. So, since r =4.35 km and r = 64 km, we have ( ) g 4.35 g =.35 =.35% (a) The statement f() = 43 tells us that when the price is $ per gallon, 43 gallons of gas are sold. (b) Since f() = 43, we have f (43) =. Thus, 43 gallons are sold when the price is $ per gallon. (c) The statement f () = 5 tells us that if the price increases from $ per gallon, the sales decrease at a rate of 5 gallons per $ increase in price. (d) The units of (f ) (43) are dollars per gallon. We have (f ) (43) = f (f (43)) = f () = 5 =.8. Thus, when 43 gallons are already sold, sales decrease at the rate of one gallon per price increase of.8 dollars. In others words, an additional gallon is sold if the price drops by.8 dollars. 8. Since f() =, we have f () =, so (f ) () = f (f ()) = f (). Therefore (f ) ()f () =. Option (b) is wrong. 9. Since f() is decreasing, its inverse function f () is also decreasing. Thus (f ) () for all. Option (b) is incorrect.

17 Chapter Three /SOLUTIONS. (a) If y = ln, then y = y = and so (b) If y = e, then so that (c) If y = e cos, then Combining these results we get y = 3 y = 3 4 y (n) =( ) n+ (n )! n. y = e + e y = e +e y = e +3e y (n) = e + ne. y = e (cos sin ) y = e sin y = e ( cos sin ) y (4) = 4e cos y (5) = e ( 4 cos + 4 sin ) y (6) =8e sin. y (n) =( 4) (n )/4 e (cos sin ), n =4m +, m =,,, 3,... y (n) = ( 4) (n )/4 e sin, n =4m +, m =,,, 3,... y (n) = ( 4) (n 3)/4 e (cos + sin ), n =4m +3, m =,,, 3,... y (n) =( 4) (n/4) e cos, n =4m, m =,, 3,..... (a) We multiply through by h = f g and cancel as follows: ( f f f f f + g g fg + g g f + g g = h h ) fg = h h fg fg = h h h f g + g f = h, which is the product rule. (b) We start with the product rule, multiply through by /(fg) and cancel as follows: which is the additive rule shown in part (a). f g + g f = h (f g + g f) fg = h fg (f g) fg +(g f) fg = h fg f f + g g = h h,

18 . This problem can be solved by using either the quotient rule or the fact that f f = d d (ln f) and g g = d (ln g). d SOLUTIONS to Review Problems for Chapter Three 3 We use the second method. The relative rate of change of f/g is (f/g) /(f/g), so ( ) (f/g) = d f f/g d ln = d d d f (ln f ln g) = (ln f) (ln g) = g d d d f g g. Thus, the relative rate of change of f/g is the difference between the relative rates of change of f and of g. CAS Challenge Problems 3. (a) Answers from different computer algebra systems may be in different forms. One form is: (b) Both the answers in part (a) follow the general rule: (c) Applying this rule to g(), we get d d ( + ) = ( + ) +( + ) ln( + ) d d (sin ) = cos (sin ) + (sin ) ln(sin ) d d f() = f ()(f()) +(f()) ln(f()). d d (ln ) = (/)(ln ) + (ln ) ln(ln ) = (ln ) + (ln ) ln(ln ). This agrees with the answer given by the computer algebra system. (d) We can write f() =e ln(f()). So (f()) =(e ln(f()) ) = e ln(f()). Therefore, using the chain rule and the product rule, d d (f()) = d d ( ln(f())) e ln(f()) = (ln(f()) + d ) d ln(f()) e ln(f()) ( = ln(f()) + f ) () (f()) = ln(f()) (f()) + f ()(f()) f() 4. (a) A CAS gives f () =. (b) By the chain rule, = f ()(f()) +(f()) ln(f()).. f () = cos(arcsin ) Now cos t = ± sin t. Furthermore, if π/ t π/ then cos t, so we take the positive square root and get cos t = sin t. Since π/ arcsin π/ for all in the domain of arcsin, we have cos(arcsin ) = (sin(arcsin )) =, so d d sin(arcsin()) = =. (c) Since sin(arcsin()) =, its derivative is. 5. (a) A CAS gives g (r) =. (b) Using the product rule, g (r) = d dr ( r ) 4 r + r d dr (4r )= ln r 4 r + r ln 4 4 r = ln 4 r 4 r + ln 4 r 4 r =( ln 4 + ln 4) r 4 r = r 4 r =. (c) By the laws of eponents, 4 r = ( ) r = r, so r 4 r = r r = =. Therefore, its derivative is zero.

19 4 Chapter Three /SOLUTIONS 6. (a) A CAS gives h (t) = (b) By the chain rule h (t) = d dt ( ) t t + ( d t dt t t t ) = t + t t t t (t ) t t = (t ) t + = t t t t t t + t t =. (c) The epression inside the first logarithm is (/t) = (t )/t. Using the property log A + log B = log(ab), we get ( ln ) ( t ) ( t ) ( t ) + ln = ln + ln t t t t ( ) t t = ln = ln =. t t Thus h(t) =, so h (t) = also. CHECK YOUR UNDERSTANDING. True. Since d( n )/d = n n, the derivative of a power function is a power function, so the derivative of a polynomial is a polynomial.. False, since 3. True, since cos θ and therefore cos θ are periodic, and 4. False. Since ( ) d π = d ( ) π = π 3 = π d d. 3 d d ln( )= = d (tan θ) = dθ cos θ. and d d ln( )= d ( ) = d, we see that the second derivative of ln( ) is negative for >. Thus, the graph is concave down. 5. True. Since f () is the limit the function f must be defined for all. f f( + h) f() () = lim, h h 6. True. The slope of f()+g() at =is the sum of the derivatives, f () + g () = = False. The product rule gives Differentiating this and using the product rule again, we get (fg) = fg + f g. (fg) = f g + fg + f g + f g = fg +f g + f g. Thus, the right hand side is not equal to fg + f g in general. 8. True. If f() is periodic with period c, then f( + c) =f() for all. By the definition of the derivative, we have and f f( + h) f() () = lim h h f f( + c + h) f( + c) ( + c) = lim. h h Since f is periodic, for any h, we have f( + h) f() f( + c + h) f( + c) =. h h Taking the limit as h, we get that f () =f ( + c), so f is periodic with the same period as f().

20 CHECK YOUR UNDERSTANDING 5 9. True; differentiating the equation with respect to, we get y dy d + y + dy d =. Solving for dy/d, we get that dy d = y y +. Thus dy/d eists where y +. Now if y + =, then = y. Substituting for in the original equation, y + y =, we get y y =. This simplifies to y + =, which has no solutions. Thus dy/d eists everywhere.. True. We have tanh = (sinh ) / cosh =(e e )/(e + e ). Replacing by in this epression gives (e e )/(e + e )= tanh.. False. The second, fourth and all even derivatives of sinh are all sinh.. True. The definitions of sinh and cosh give sinh + cosh = e e + e + e = e = e. 3. False. Since (sinh ) = cosh >, the function sinh is increasing everywhere so can never repeat any of its values. 4. False. Since (sinh ) = sinh cosh and ( sinh cosh ) = sinh + cosh >, the function sinh is concave up everywhere. 5. False. If f() =, then f() is not differentiable at =and f () does not eist at =. 6. False. If f() = ln, then f () = /, which is decreasing for >. 7. False; the fourth derivative of cos t+c, where C is any constant, is indeed cos t. But any function of the form cos t+p(t), where p(t) is a polynomial of degree less than or equal to 3, also has its fourth derivative equal to cos t. So cos t + t will work. 8. False; For eample, the inverse function of f() = 3 is /3, and the derivative of /3 is (/3) /3, which is not /f () = /(3 ). 9. False; for eample, if both f() and g() are constant functions, such as f() = 6,g() =, then (fg) () =, and f () = and g () =.. True; looking at the statement from the other direction, if both f() and g() are differentiable at =, then so is their quotient, f()/g(), as long as it is defined there, which requires that g(). So the only way in which f()/g() can be defined but not differentiable at =is if either f() or g(), or both, is not differentiable there.. False; for eample, if both f and g are constant functions, then the derivative of f(g()) is zero, as is the derivative of f(). Another eample is f() = and g() = +.. True. Since f () > and g () > for all, we have f ()+g () > for all, which means that f()+g() is concave up. 3. False. Let f() = and g() =. Let h() =f()g(). Then h () =. Since h () <, clearly h is not concave up for all. 4. False. Let f() = and g() =. Then f() g() =, which is concave up for all. 5. False. Let f() =e and g() =. Let h() =f(g()) = e. Then h () = e and h () = ( + 4 )e. Since h () <, clearly h is not concave up for all. 6. (a) False. Only if k = f (a) is L the local linearization of f. (b) False. Since f(a) =L(a) for any k, we have lim a(f() L()) = f(a) L(a) =, but only if k = f (a) is L the local linearization of f. 7. (a) This is not a countereample. Although the product rule says that (fg) = f g + fg, that does not rule out the possibility that also (fg) = f g. In fact, if f and g are both constant functions, then both f g + fg and f g are zero, so they are equal to each other. (b) This is not a countereample. In fact, it agrees with the product rule: ( d d ) d (f()) = d () f()+ d d f() =f()+f () =f ()+f().

21 6 Chapter Three /SOLUTIONS (c) This is not a countereample. Although the product rule says that d d (f() )= d d f() f() =f ()f()+f()f () = f()f (), it could be true that f () =, so that the derivative is also just f(). In fact, f() = is an eample where this happens. (d) This would be a countereample. If f (a) =g (a) =, then d d (f()g()) = f (a)g(a)+f(a)g (a) =. =a So fg cannot have positive slope at = a. Of course such a countereample could not eist, since the product rule is true. 8. True, by the Increasing Function Theorem, Theorem??. 9. False. For eample, let f() = +5, and g() = 3. Then f () g () for all, but f() >g(). 3. False. For eample, let f() = 3 + and g() = False. For eample, if f() =, then f () for all, but f( ) =, so f( ) >. 3. The function f() = is continuous on [, ], but there is no number c, with < c <, such that that is, the slope of f() = is never. 33. Let f be defined by f (c) = ( ) = ; { if < f() = 9 if = Then f is differentiable on (, ) and f () = for all in (, ). Thus there is no c in (, ) such that f (c) = f() f() = 9. The reason that this function does not satisfy the conclusion of the Mean Value Theorem is that it is not continuous at =. 34. Let f be defined by { if < f() = / if =. Then f is not continuous at =, but f is differentiable on (, ) and f () = for < <. Thus, c =/4 satisfies f f() f() (c) = = ( ), since f = 4 4 =. PROJECTS FOR CHAPTER THREE. Let r = i/. (For eample if i = 5%, r =.5.) Then the balance, $B, after t years is given by B = P ( + r) t, where $P is the original deposit. If we are doubling our money, then B =P, so we wish to solve for t in the equation P = P ( + r) t. This is equivalent to = ( + r) t. Taking natural logarithms of both sides and solving for t yields ln = t ln( + r), ln t = ln( + r).

22 PROJECTS FOR CHAPTER THREE 7 We now approimate ln( + r) near r =. Let f(r) = ln( + r). Then f (r) = /( + r). Thus, f() = and f () =, so f(r) f() + f ()r becomes ln( + r) r. Therefore, t = ln ln( + r) ln ln = 7 r i i, as claimed. We epect this approimation to hold for small values of i; it turns out that values of i up to give good enough answers for most everyday purposes.. (a) (i) Set f() = sin, so f () = cos. Guess =3. Then =3 sin 3 cos = sin cos , which is correct to one billionth! (ii) Newton s method uses the tangent line at =3, i.e. y sin 3 = cos(3)( 3). Around =3, however, sin is almost linear, since the second derivative sin (π) =. Thus using the tangent line to get an approimate value for the root gives us a very good approimation. f() = sin tangent line 3 (iii) For f() = sin, we have f(3) =.4 f(4) =.7568, so there is a root in [3, 4]. We now continue bisecting: [3, 3.5] : f(3.5) =.3578 (bisection ) [3, 3.5] : f(3.5) =.89 (bisection ) [3.5, 3.5] : f(3.5) =.659 (bisection 3) [3.5, 3.875] : f(3.875) =.4584 (bisection 4) We continue this process; after bisections, we know the root lies between 3.4 and 3.46, which still is not as good an approimation as what we get from Newton s method in just two steps. (b) (i) We have f() = sin 3 and f () = cos 3. Using =.94, =.94 sin(.94) 3 (.94) cos(.94) ,

23 8 Chapter Three /SOLUTIONS Using =.95, Now using =.96, =4.74 sin(4.74) 3 (4.74) cos(4.74) 3 3 =.433 sin(.43) 3 (.43) cos(.43) 3 4 =.499 sin(.5) 3 (.5) cos(.5) 3 5 =.496 sin(.496) 3 (.496) cos(.496) 3 =.95 sin(.95) 3 (.95) cos(.95) 3 =4.643 sin(4.643) 3 (4.643) cos(4.643) 3 3 =.98 sin(.98) 3 (.98) cos(.98) 3 4 = sin( 3.996) 3 ( 3.996) cos( 3.996) 3 5 =.43 sin(.43) 3 (.43) cos(.43) 3 6 =.5 sin(.5) 3 (.5) cos(.5) 3 =.96 sin(.96) 3 (.96) cos(.96) 3 =4.584 sin(4.584) 3 (4.584) cos(4.584) 3 3 =.5 sin(.59) 3 (.59) cos(.59) 3 4 =.3 sin(.7) 3 (.7) cos(.7) 3.43,.5,.496, ,.98, 3.996,.43,.5, ,.59,.7,.9, 5 =.543 sin(.9) 3 (.9) cos(.9), 3 (ii) Starting with.94 and.95 yields the same value, but the two paths to get to the root are very different. Starting with.96 leads to a different root. Our starting points were near the maimum value of f. Consequently, a small change in makes a large change in.

24 Solutions for Chapter 4 Review SOLUTIONS to Review Problems for Chapter Four 33 Eercises. See Figure Local and global ma Local ma f() Local ma Local min Local and global min Local ma f() Local and global min Local and global ma Critical point (not ma or min) Figure 4.53 Figure See Figure (a) We wish to investigate the behavior of f() = 3 3 on the interval 3. We find: f () = 3 6 =3( ) f () = 6 6 = 6( ) (b) The critical points of f are =and =since f () = at those points. Using the second derivative test, we find that =is a local maimum since f () = and f () = 6 <, and that =is a local minimum since f () = and f () = 6 >. (c) There is an inflection point at =since f changes sign at =. (d) At the critical points, f() = and f() = 4. At the endpoints: f( ) = 4,f(3) =. So the global maima are f() = and f(3) =, while the global minima are f( ) = 4 and f() = 4. (e) See Figure incr. decreasing incr. concave down concave up 3 4 π increasing concave down concave up π Figure 4.55 Figure (a) First we find f and f ; f () = + cos and f () = sin. (b) The critical point of f is = π, since f (π) =. (c) Since f changes sign at = π, it means that = π is an inflection point. (d) Evaluating f at the critical point and endpoints, we find f() =, f(π) = π, f(π) = π,. Therefore, the global maimum is f(π) = π, and the global minimum is f() =. Note that = π is not a local maimum or minimum of f, and that the second derivative test is inconclusive here. (e) See Figure (a) First we find f and f : f () = e sin + e cos f () =e sin e cos e cos e sin = e cos

25 33 Chapter Four /SOLUTIONS (b) The critical points are = π/4, 5π/4, since f () = here. (c) The inflection points are = π/, 3π/, since f changes sign at these points. (d) At the endpoints, f() =, f(π) =. So we have f(π/4) = (e π/4 )( /) as the global maimum; f(5π/4) = e 5π/4 ( /) as the global minimum. (e) See Figure conc. down incr. π π concave up decreasing 3π π conc. down increasing. 3.5 decr incr concave up Figure 4.57 Figure (a) We first find f and f : f () = = ( ) f () = = ( 5) (b) Critical point: =. (c) There are no inflection points, since f does not change sign on the interval (d) At the endpoints, f(.).948 and f(3.5).959. So, the global minimum is f() and the global maimum is f(3.5).959. (e) See Figure The polynomial f() behaves like 3 as goes to. Therefore, lim f() = and lim f() =. We have f () = = 6( )( ), which is zero when =or =. Also, f () = 8 = 6( 3), which is zero when =3/. For <3/, f () < ; for >3/, f () >. Thus =3/isan inflection point. The critical points are =and =, and f() = 6, f() = 5. By the second derivative test, f () = 6 <, so =isalocal maimum; f () = 6 >, so =isalocal minimum. Now we can draw the diagrams below. y > y < y > increasing = decreasing = increasing y < y > concave down =3/ concave up The graph of f() = is shown in Figure It has no global maimum or minimum.

26 SOLUTIONS to Review Problems for Chapter Four 333 f() = f() = 4 + Figure 4.59 Figure If we divide the denominator and numerator of f() by we have since Using the quotient rule we get lim ± 4 + = lim ± lim ± =. 4 + =4 f () = ( + )8 4 () 8 = ( + ) ( + ), which is zero when =, positive when >, and negative when <. Thus f() has a local minimum when =, with f() =. Because f () = 8/( + ), the quotient rule implies that f () = ( + ) 8 8[( + )] ( + ) 4 = ( + ) 3 = 8( 3 ) ( + ) 3. The denominator is always positive, so f () = when = ± /3, positive when /3 < < /3, and negative when > /3 or < /3. This gives the diagram y < decreasing = y > increasing y < concave down = /3 y > concave up = /3 y < concave down and the graph of f looks Figure 4.6. with inflection points = ± /3, a global minimum at =, and no local or global maima (since f() never equals 4). 9. As, e, so e. Thus lim e =. As,, since e grows much more quickly than. Thus lim e e =. Using the product rule, f () =e e = ( )e, which is zero when =, negative when >, and positive when <. Thus f() = /e =/e is a local maimum.

27 334 Chapter Four /SOLUTIONS Again, using the product rule, f () = e e + e = e e =( )e, which is zero when =, positive when >, and negative when <, giving an inflection point at (, e ). With the above, we have the following diagram: y > increasing = y < decreasing y < concave down = y > concave up The graph of f is shown in Figure 4.6. and f() has one global maimum at /e and no local or global minima. f() =e 3 h(z) = z +4z / z Figure 4.6 Figure 4.6. Since f() = sin() is continuous and the interval π/ is closed, there must be a global maimum and minimum. The possible candidates are critical points in the interval and endpoints. Since there are no points where f () is undefined, we solve f () = to find all the critical points: f () = cos() =, so cos() = /. Therefore = π/3, 5π/3,.... Thus the only critical point in the interval is = π/6. We then compare the values of f at the critical points and the endpoints: f() =, f(π/6) = π/6 3/ =.34, f(π/) = π/. Thus the global maimum is π/ at = π/ and the global minimum is.34 at = π/6.. Since f() = is continuous and the interval 3 4 is closed, there must be a global maimum and minimum. The possible candidates are critical points in the interval and endpoints. The derivative f is not defined at =. To find the other critical points we solve f () =. For > we have f() =, so f () = =. Thus, =is the only critical point for < < 4. For <, we have: f() = +, so f () = + =. Thus = is the only critical point for 3 < <. We then compare the values of f at the critical points and the endpoints: f( 3) = 3, f( ) =, f() =, f() =. f(4) = 8. Thus the global maimum is 8 at =4and the global minimum is at = and =.

28 SOLUTIONS to Review Problems for Chapter Four 335. We rewrite h(z) as h(z) =z +4z. Differentiating gives so the critical points satisfy h (z) = z +8z, z +8z = z =8z 8z 3 = z 3 = 8 z =. Since h is negative for < z < / and h is positive for z>/, there is a local minimum at z =/. Since h(z) as z + and as z, the local minimum at z =/ is a global minimum; there is no global maimum. See Figure 4.6. Thus, the global minimum is h(/) = Since g(t) is always decreasing for t, we epect it to a global maimum at t =but no global minimum. At t =, we have g() =, and as t, we have g(t). Alternatively, rewriting as g(t) = (t 3 + ) and differentiating using the chain rule gives g (t) = (t 3 + ) 3t. Since 3t =when t =, there is a critical point at t =, and g decreases for all t>. See Figure / f() = ( ) + g(t) = t 3 + Figure 4.63 Figure We begin by rewriting f(): f() = Differentiating using the chain rule gives so the critical points satisfy ( ) + = (( ) + ) =( + 3). f () = ( + 3) ( ) = ( + 3) = = = =. ( + 3), Since f is positive for < and f is negative for >, there is a local maimum at =. Since f() as and as, the local maimum at = is a global maimum; there is no global minimum. See Figure Thus, the global maimum is f() = /.

29 336 Chapter Four /SOLUTIONS 5. lim f() = +, and lim f() =. There are no asymptotes. f () = = 3( + 3)( ). Critical points are = 3, =. f () = 6( + ). 3 f + + f f Thus, = is an inflection point. f( 3) = is a local maimum; f() = is a local minimum. There are no global maima or minima. See Figure f() = Figure lim + f() = +, and lim f() =. There are no asymptotes. f () = =5 ( 9) = 5 ( + 3)( 3). The critical points are =, = ±3. f changes sign at 3 and 3 but not at. f () = 3 9 = ( 9). f changes sign at, ±3/. So, inflection points are at =, = ±3/. 3 3/ 3/ 3 f + + f f Thus, f( 3) is a local maimum; f(3) is a local minimum. There are no global maima or minima f() =

30 7. lim f() = +, and lim f() = Hence, =isavertical asymptote. f () = =, so =istheonlycritical point. f () =, which can never be zero. So there are no inflection points. SOLUTIONS to Review Problems for Chapter Four 337 f() = ln f + f f Thus, f() is a local and global minimum. 8. lim + f() = +, lim f() =. y =isthehorizontal asymptote. f () = e 5 +5 e 5 = e 5 (5 + ). Thus, = and =are the critical points. 5 So, = ± 5 are inflection points. f () = e 5 +e e e 5 = e 5 (5 + + ) f f f f() = e 5.4 So, f( ) is a local maimum; f() is a local and global minimum Since lim f() = lim + f() =, y = is a horizontal asymptote. f () = e. So, =istheonlycritical point. f () = (e + ( )e ) = e ( ) = e ( )( + ).

31 338 Chapter Four /SOLUTIONS Thus, = ±/ are inflection points. Table 4. / / f f + + f Thus, f() = is a local and global maimum. f() =e. lim + f() = lim f() =. Thus, y =isahorizontal asymptote. Since +is never, there are no vertical asymptotes. f () = ( + ) () = ( + ) ( + ). So, =istheonlycritical point. So, = ± 3 are inflection points. Table 4. f () = ( + ) ( + ) ( + ) 4 = ( + 4 ) ( + ) 3 = ( 3 ) ( + ) f f f Thus, f() = is a local and global minimum. A graph of f() can be found in Figure y = f() = + Figure 4.66

32 SOLUTIONS to Review Problems for Chapter Four 339. We see from the parametric equations that the particle moves along a line. It suffices to plot two points: at t =, the particle is at point (4, ), and at t =, the particle is at point (, 5). Since decreases as t increases, the motion is right to left and the curve is shown in Figure Alternately, we can solve the first equation for t, giving t = ( 4)/, and substitute this into the second equation to get ( ) ( 4) y =4 + = +9. The line is y = y y t = 4 t = 4 6 Figure 4.67 Figure The graph is a circle centered at the origin with radius. The equation is + y = ( sin t) + ( cos t) =4. The particle is at the point (, ) when t =, and motion is clockwise. See Figure (a) We set the derivative equal to zero and solve for to find critical points: We see that there are three critical points: f () = 4 3 4a = 4( a) =. Critical points: =, = a, = a. To find possible inflection points, we set the second derivative equal to zero and solve for : There are two possible inflection points: f () = 4a =. Possible inflection points: = a 3, = a 3. To see if these are inflection points, we determine whether concavity changes by evaluating f at values on either side of each of the potential inflection points. We see that so f is concave up to the left of = a/3. Also, a f ( 3 ) = (4a ) 4a = 6a 4a = a >, 3 f () = 4a <, so f is concave down between = a/3 and = a/3. Finally, we see that f ( a 3 ) = (4a ) 4a = 6a 4a = a >, 3 so f is concave up to the right of = a/3. Since f() changes concavity at = a/3 and = a/3, both points are inflection points.

33 34 Chapter Four /SOLUTIONS (b) The only positive critical point is at = a, so to have a critical point at =, we substitute: Since the critical point is at the point (, 5), we have = a = a a =4. f() = 5 4 (4) + b = b =5 b =. The function is f() = (c) We have seen that a =4, so the inflection points are at = 4/3 and = 4/3. 4. (a) The function f() is defined for. We set the derivative equal to zero and solve for to find critical points: f () = a / = a = = a = a 4. Notice that f is undefined at =so there are two critical points: =and = a /4. (b) We want the critical point = a /4 to occur at =5, so we have: 5= a 4 = a a = ±. Since a is positive, we use the positive square root. The second derivative, f () = 4 a 3/ = 3/ 4 is positive for all >, so the function is concave up and =5gives a local minimum. See Figure y 5 Figure We have If =5, then dm dt dm dt = ( ) d dt. = [3(5 ) +.4(5 3 )](.) =.5 gm/hr.

34 6. We have df dt = K(a y ) dy (a + y ) 5/ dt. (a) When y =, we have df dt = dy Ka 3 dt. So, df /dt is positive and F is increasing. (b) When y = a/, we have df dt = K(a a ) dy (a +(a /)) 5/ dt =. So, df /dt =and F is not changing. (c) When y =a, we have df dt = K(a 8a ) (a +4a ) 5/ dy dt = 7Ka (5a ) 5/ dy dt. 7. We have So So, df /dt is negative and F is decreasing. da dt θ=π/ = da dt = (3 + 4 cos θ + cos(θ)) dθ dt. SOLUTIONS to Review Problems for Chapter Four 34 ( ( ) ) π cos + cos π.3 =.6 cm /min. Problems 8. (a) f () (b) f () changes sign at, 3, and 5. (c) f () has local etrema at and The local maima and minima of f correspond to places where f is zero and changes sign or, possibly, to the endpoints of intervals in the domain of f. The points at which f changes concavity correspond to local maima and minima of f. The change of sign of f, from positive to negative corresponds to a maimum of f and change of sign of f from negative to positive corresponds to a minimum of f. 3. The function f has critical points at =,=3,=5. By the first derivative test, since f is positive to the left of =and negative to the right, =is a local maimum. Since f is negative to the left of =3and positive to the right, =3is a local minimum. Since f does not change sign at =5, this point is neither a local maimum nor a local minimum. 3. The critical points of f occur where f is zero. These two points are indicated in the figure below. f () f has a local min. f has crit. pt. Neither ma or min Note that the point labeled as a local minimum of f is not a critical point of f.

35 34 Chapter Four /SOLUTIONS 3. (a) The function f is a local maimum where f () = and f > to the left, f < to the right. This occurs at the point 3. (b) The function f is a local minimum where f () = and f < to the left, f > to the right. This occurs at the points and 5. (c) The graph of f is climbing fastest where f is a maimum, which is at the point. (d) The graph of f is falling most steeply where f is the most negative, which is at the point. 33. Since the 3 term has coefficient of, the cubic polynomial is of the form y = 3 + a + b + c. We now find a, b, and c. Differentiating gives dy d =3 +a + b. The derivative is at local maima and minima, so dy = 3() +a() + b = 3 + a + b = d = dy d =3 = 3(3) +a(3) + b = 7 + 6a + b = Subtracting the first equation from the second and solving for a and b gives Since the y-intercept is 5, the cubic is 4 + 4a = so a = 6 y = b = 3 ( 6) = 9. Since the coefficient of 3 is positive, =is the maimum and =3is the minimum. See Figure 4.7. To confirm that =gives a maimum and =3gives a minimum, we calculate At =, d y = 6 <, so we have a maimum. d At =3, d y =6>, so we have a minimum. d d y =6 +a =6. d y y (, 4) (, 4) Figure 4.7: Graph of y = Figure 4.7: Graph of y = Since the graph of the quartic polynomial is symmetric about the y-ais, the quartic must have only even powers and be of the form y = a 4 + b + c. The y-intercept is 3, so c =3. Differentiating gives dy d =4a3 +b.

36 Since there is a maimum at (, 4), we have dy/d =if =, so 4a() 3 +b() = 4a +b = so b = a. The fact that dy/d =if = gives us the same relationship We also know that y =4if = ±, so Solving for a and b gives 4a b = so b = a. a() 4 + b() + 3 = a + b + 3 = 4 so a + b =. a a = so a = and b =. Finding d y/d so that we can check that = ± are maima, not minima, we see d y d = a +b = +4. Thus d y = 8 < for = ±, so = ± are maima. See Figure 4.7. d 35. Differentiating y = a b ln, we have Since the maimum occurs at = e, we know that Since a and (e ) b for all b, we have Since ln(e ) =, the equation becomes dy d = abb ln + a b = ab (b ln + ). a(e ) b (b ln(e ) + ) =. b ln(e ) + =. b + = Thus y = a / ln. When = e, we know y =6e, so b =. y = a(e ) / ln e = ae () = 6e SOLUTIONS to Review Problems for Chapter Four 343 a =3. Thus y =3 / ln. To check that = e gives a local maimum, we differentiate twice dy d = 3 3/ ln +3 / = 3 3/ ln +3 3/, d y d = 9 4 5/ ln 3 3/ 3 3 5/ = 9 4 5/ ln 6 5/ = 3 4 5/ (3 ln 8). At = e, since ln(e ) =, we have a maimum because See Figure 4.7. d y d = 3 4 (e ) 5/ ( 3 ln(e ) 8 ) = 3 4 e 5 (3 8) <. y (e, 6e ) 6 Figure 4.7: Graph of y =3 / ln