Section 5. 5. (a Note that 5 y 5, so y. Then c 5( + y + 0y 5 + 0y 5 5 + 0 5 +, 750 dc 0( 75 0, 750 d The critical point occurs at 5. Since dc dc < 0 for 0 < < 5 and 0 d d > for > 5, the critical point corresponds to the minimum cost. The values of and y are 5 ft and y 5 ft. (b The material for the tank costs 5 dollars/sq ft and the ecavation charge is 0 dollars for each square foot of the cross-sectional area of one wall of the hole.. Let be the height in inches of the printed area. Then the wih of the printed area is 50 in. and the overall dimensions are + 8 in. by 50 + in. The amount of paper used is 50 00 A ( ( + 8 + + 8 + in. ( 00 Then A ( 00 and the critical point (for > 0 occurs at 0. Since A ( < 0for0< < 0 and A ( > 0 for > 0, the critical point corresponds to the minimum amount of paper. Using + 8 and 50 + for 0, the overall dimensions are 8 in. high by 9 in. wide.. (a st ( 6t + 96t+ vt ( s ( t t+ 96 At t 0, the velocity is v(0 96 ft/sec. (b The maimum height occurs when v(t 0, when t. The maimum height is s( 56 ft and it occurs at t sec. (c Note that st ( 6t + 96t+ 6( t+ ( t 7, so s 0 at t or t 7. Choosing the positive value, of t, the velocity when s 0 is v(7 8 ft/sec. 5. We assume that a and b are held constant. Then A( θ absinθ and A ( θ abcos θ. The critical point (for 0 < θ < occurs at θ. Since A ( θ > 0 for 0 < θ < and A ( θ < 0 for < θ <, the critical point corresponds to the maimum area. The angle that maimizes the triangle s area is θ or 90 (. 6. Let the can have radius r cm and height h cm. 000 Then r h 000, so h. The area of r material used is 000 A r + rh r +, so r da r 000 r 000r. The dr r critical point occurs at 000 / da r 0 cm. Since 0 dr < / da / for 0 < r < 0 and > 0 for r > 0, dr the critical point corresponds to the least amount of material used and hence the lightest possible can. The dimensions are / r 0 6. 8cm and / h 0 6. 88cm. In Eample, because of the top of the can, the best design is less big around and taller. 7. Note that 000 r h 000, so h. Then r 000 A 8r + rh 8 r +, so r da 6( r 5 6r 000r. The dr r critical point occurs at r 5 5 cm. Since Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
5 Section 5. da da < 0 for 0 < r < 5 and > 0 for r > 5, the dr dr critical point corresponds to the least amount of aluminum used or wasted and hence the most economical can. The dimensions are 0 r 5 cm and h, so the ratio of h to r is 8 to. 8. (a The base measures 0 in. by 5 in, so the volume formula is ( 0 ( 5 V( 5 + 75. 5 5 7 96., which confirms the 6 result in (c. 9. (a The sides of the suitcase will measure in. by 8 in. and will be in. apart, so the volume formula is V( ( ( 8 8 68 + 86. (b We require > 0, < 8, and <. Combining these requirements, the domain is the interval (0, 9. (b We require > 0, < 0, and < 5. Combining these requirements, the domain is the interval (0, 5. (c The maimum volume is approimately 66.0 in when.96 in. (c (d V ( 6+ 86 ( + 6 The critical point is at 7 ±, that is, 9or. 06.. We discard the larger value because it is not in the domain. Since V ( (, which is negative when.9, the critical point corresponds to the maimum volume. The maimum value occurs at 7 9.. (d V ( 6 50+ 75 The critical point occurs when V ( 0, 50 ± ( 50 ( 6( 75 at 6 ( 50 ± 700 5 ± 5 7, 6 that is,.96 or 6.7. We discard the larger value because it is not in the domain. Since V ( 50, which is negative when.96, the critical point corresponds to the maimum volume. The maimum volume occurs when (e The maimum volume is approimately 09.95 in when.9 in. 8 68 + 86 0 8( + 08 0 0 8( ( 5( 0 Since is not in the domain, the possible values of are in. or 5 in. (f The dimensions of the resulting bo are in., ( in., and (8 in. Each of these measurements must be positive, so that gives the domain of (0, 9 Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
Section 5. 55 0. Let be the distance from the point on the shoreline nearest Jane s boat to the point where she lands her boat. Then she needs to row + mi at mph and walk 6 mi at 5 mph. The total amount of time to reach + 6 the village is f( + hours (0 6. Then 5 ( ( f 5 5. + + Solving f ( 0, we have: + 5 5 + 5 ( + 6 ± We discard the negative value of because it is not in the domain. Checking the endpoints and critical point, we have f( 0., f., and f(6.6. Jane should land her boat 087. miles down the shoreline from the point nearest her boat.. If the upper right verte of the rectangle is located at (, cos 0.5 for 0 < <, then the rectangle has wih and height cos 0.5, so the area is A( 8 cos 0.5. Then A ( 8 ( 0.5 sin 0.5 + 8(cos 0.5( sin 0.5 + 8 cos 0.5. Solving A ( graphically for 0 < <, we find that.7. Evaluating and cos 0.5 for.7, the dimensions of the rectangle are approimately. (wih by.6 (height, and the maimum area is approimately 8.98.. Let the radius of the cylinder be r cm, 0 < r < 0. Then the height is Vr ( r 00 rcm. Then V ( r r ( r + ( 00 r ( r 00 r r + r( 00 r 00 r r( 00 r 00 r 00 r and the volume is 00 The critical point for 0 < r < 0 occurs at r 0. Since V ( r > 0 for 0< r < 0 and V ( r > 0 for 0 < r < 0, the critical point corresponds to the maimum volume. The dimensions are 0 000 r 0 8. 6 cm and h. 55 cm, and the volume is 8. 0 cm. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
56 Section 5.. Set r ( c ( :. The only positive critical value is, so profit is maimized at a production level of 000 units. Note that ( r c ( ( < 0 for all positive, so the Second Derivative Test confirms the maimum.. Set r ( c ( : /( + (. We solve this equation graphically to find that 0. 9. The graph of y r ( c( shows a minimum at 0. 9 and a maimum at 55., so profit is maimized at a production level of about,55 units. c ( 5. Set c( 0+ 0. The only d c( positive solution to 0 is 5, so d average cost is minimized at a production level d c( of 5000 units. Note that 0 > d for all positive, so the Second Derivative Test Confirms the minimum. c ( 6. Set c( e. The only positive d c( solution to 0 is ln, so d average cost is minimized at a production level of 000 ln, which is about 69 units. Note d c( that e 0 > for all positive, so d the Second Derivative Test confirms the minimum. 7. Revenue: ( [ 00 ( 50 ] r + 00 Cost: c ( 6000 + Profit: p( r( c( + 68 6000, 50 80 Since p ( + 68 ( 67, the critical point occurs at 67. This value represents the maimum because p (, which is negative for all in the domain. The maimum profit occurs if 67 people go on the tour. 8. (a f ( ( e + e ( e ( The critical point occurs at. Since f ( > 0 for 0 < and f ( < 0 for >, the critical point corresponds to the maimum value of f. The absolute maimum of f occurs at. (b To find the values of b, use grapher techniques to solve e 0. e 0., (c e 0. e 0., and so on. To find the a values of A, calculate (b a ae, using the unrounded values of b. (Use the list features of the grapher in order to keep track of the unrounded values for part (d. a b A 0..7 0. 0..86 0. 0..6 0.6 0..0 0. 0.5.76 0.8 0.6.55 0. 0.7.8 0. 0.8. 0.5 0.9. 0.08.0.00 0.00 [0,.] by [ 0., 0.6] (d Quadratic: A 09. a + 05. a+ 0. [ 0.5,.5] by [ 0., 0.6] Cubic: A 7. a 78. a + 86. a+ 09. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
(b Possible answers: No local etrema: y ; local etrema: y Section 5. 57 [ 0.5,.5] by [ 0., 0.6] Quartic: A 9. a + 596. a 687. a + 7. a+ 0. [ 0.5,.5] by [ 0., 0.6] (e Quadratic: According to the quadratic regression equation, the maimum area occurs at a 0.0 and is approimately 0.. Cubic: According to the cubic regression equation, the maiumu area occurs at a 0. and is approimately 0.5. Quartic: According to the quartic regression equation the maimum area occurs at a 0.0 and is approimately 0.6. 9. (a f ( is a quadratic polynomial, and as such it can have 0,, or zeros. If it has 0 or zeros, then its sign never changes, so f(f( has no local etrema. If f ( has zeros, then its sign changes twice, and f(f( has local etrema at those points. 0. Let be the length in inches of each edge of the square end, and let y be the length of the bo. Then we require + y 08. Since our goal is to maimize volume, we assume + y 08 and so y 08. The volume is V( ( 08 08, where 0 < < 7. Then V 6 ( 8, so the critical point occurs at 8 in. Since V ( > 0 for 0 < < 8 and V ( < 0 for 8 < < 7, the critical point corresponds to the maimum volume. The dimensions of the bo with the largest possible volume are 8 in. by 8 in. by 6 in.. Since + y 6, we know that y 8. In part (a, the radius is and the height is 8, and so the volume is given by r h ( 8 ( 8. In part (b, the radius is and the height is 8, and so the volume is given by r h ( 8. Thus, each problem requires us to find the value of that maimizes f( ( 8 in the interval 0 < < 8, so the two problems have the same answer. To solve either problem, note that f( 8 and so f ( 6 (. The critical point occurs at. Since f ( > 0 for 0 < < and f ( < 0 for < < 8, the critical point corresponds to the maimum value of f (. To maimize the volume in either part (a or (b, let cm and y 6 cm.. Note that h + r and so r h. Then the volume is given by V r h ( h h h h for dv 0< h <, and so h ( h. dh The critical point (for h > 0 occurs at h. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
58 Section 5. dv dv Since > 0 for 0 < h < and 0 dh dh < for < h <, the critical point corresponds to the maimum volume. The cone of greatest volume has radius m, height m, and volume m.. (a We require f( to have a critical point at. Since f ( a, we have a f ( and so our requirement is a that 0. Therefore, a 6. To verify that the critical point corresponds to a local minimum, note that we now have f ( 6 and so. f ( +, so f ( 6, which is positive as epected. So, use a 6. (b We require f ( 0. Since f + a, we have f ( + a, so our requirement is that + a 0. Therefore, a. To verify that is in fact an inflection point, note that we now have f (, which is negative for 0 < < and positive for >. Therefore, the graph of f is concave down in the interval (0, and concave up in the interval (,, So, use a. a f ( a, so the only sign a change in f ( occurs at, where the sign changes from negative to positive. This means there is a local minimum at that point, and there are no local maima. 5. (a Note that f ( + a+ b. We require f ( 0 and f ( 0, which give a + b 0 and 7 + 6a + b 0. Subtracting the first equation from the second, we have + 8a 0 and so a. Substituting into the first equation, we have 9 + b 0, so b 9. Therefore, our equation for f( is f( 9. To verify that we have a local maimum at and a local minimum at, note that f ( 6 9 ( + (, which is positive for <, negative for < <, and positive for >. So, use a and b 9. (b Note that f ( + a+ b and f ( 6+ a. We require f ( 0 and f ( 0, which give 8 + 8a + b 0 and 6 + a 0. By the second equation, a, and so the first equation becomes 8 + b 0. Thus b. To verify that we have a local minimum at, and an inflection point at, note that we now have f ( 6 6. Since f changes sign at and is positive at, the desired conditions are satisfied. So, use a and b. 6. Refer to the illustration in the problem statement. Since + y 9, we have 9 y. Then the volume of the cone is given by V r h ( y + ( 9 y ( y + ( y y + 9y + 7, for < y <. dv Thus ( y 6y + 9 dy ( y + y ( y + ( y, so the critical point in the interval (, is dv y. Since 0 dy > for < y < and dv dy < 0 for < y <, the critical point does correspond to the maimum value, which is V ( cubic units. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
Section 5. 59 7. (a Note that w + d, so d w. Then we may write S kwd kw( w kw kw for 0 < w <, so ds k kw k( w 8. dw The critical point (for 0 < w < occurs at ds w 8. Since 0 dw > for ds 0< w < and 0 dw < for < w <, the critical point corresponds to the maimum strength. The dimensions are in. wide by 6 in. deep. ds kw ( w ( w + k( w dw ( k w ( w + w ( k w ( w 6 The critical point (for 0 < w < occurs ds at w 6. Since 0 dw > for (b / / ds 0 < w< 6 and < 0 for 6 < w<, the dw critical point corresponds to the maimum stiffness. The dimensions are 6 in. wide by 6 in. deep. (b The graph of S w w is shown. The maimum strength shown in the graph occurs at w 6., 9 which agrees with the answer to part (a. (c [0, ] by [ 000, 8000] / The graph of S w( w is shown. The maimum stiffness shown in the graph occurs at w 6, which agrees with the answer to part (a. (c The graph of S d d is shown. The maimum strength shown in the graph occurs at d 6 9., 8 which agrees with the answer to part (a, and its value is the same as the maimum value found in part (b, as epected. Changing the value of k changes the maimum strength, but not the dimensions of the strongest beam. The graphs for different values of k look the same ecept that the vertical scale is different. 8. (a Note that w + d, so d w. Then we may write / S kwd kw( w, so [0, ] by [ 000, 8000] The graph of S d d is shown. The maimum stiffness shown in the graph occurs at d 6 0. agrees with the answer to part (a, and its value is the same as the maimum value found in part (b, as epected. Changing the value of k changes the maimum stiffness, but not the dimensions of the stiffest beam. The graphs for different values of k look the same ecept that the vertical scale is different. 9. (a vt ( s ( t 0 sint The speed at time t is 0 sin t. The maimum speed is 0 cm/sec and it 5 occurs at t, t, t, and t 7 sec. The position at these times is s 0 cm (rest position, and the Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
60 Section 5. acceleration at ( v ( t 0 cost is 0 cm / sec at these times. (b Since at ( 0 cos t, the greatest magnitude of the acceleration occurs at t 0, t, t, t, and t. At these times, the position of the cart is either s 0cmor s 0cm, and the speed of the cart is 0 cm/sec. di 0. Since sin t+ cos t, the largest magnitude of the current occurs when sin t + cos t 0, or sin t cos t. Squaring both sides gives sin t cos t, and we know that sin t+ cos t, so sin t cos t. Thus the possible values of t are 5,,, and so on. Eliminating etraneous solutions, the solutions of sin t cost are t + k for integers k, and at these times i cost+ sin t. The peak current is amps.. The square of the distance is 9 ( 0 D ( + +, so D ( and the critical point occurs at. Since D ( < 0 for < and D ( > 0for >, the critical point corresponds to the minimum distance. The 5 minimum distance is D (.. Calculus method: The square of the distance from the point (, to (, 6 is given by D ( ( + ( 6 + + 6 8 + + 0 8. Then D ( ( 6 8 6 +. 8 Solving D ( 0, we have: 6 8 6 ( 8 9 8 8 ± We discard as an etraneous solution, leaving. Since D ( < 0 for < < and D ( > 0 for < <, the critical point corresponds to the minimum distance. The minimum distance is D (. Geometry method: The semicircle is centered at the origin and has radius. The distance from the origin to (, is ( +. The shortest distance from the point to the semicircle is the distance along the radius containing the point (,. That distance is.. No. Since f( is a quadratic function and the coefficient of is positive, it has an absolute minimum at the point where f ( 0, and the point is,.. (a Because f( is periodic with period. (b No; since f( is continuous on [0, ], its absolute minimum occurs at a critical point or endpoint. Find the critical points in [0, ]: f ( sin sin 0 sin sin cos 0 (sin ( + cos 0 sin 0orcos 0,, The critical points (and endpoints are (0, 8, (, 0, and (, 8. Thus, f( has an absolute minimum at (, 0 and it is never negative. 5. (a sin t sin t sin t sin tcost (sin t( cos t 0 sin t 0orcost t k, where k is an integer. The masses pass each other whenever t is an integer multiple of seconds. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
Section 5. 6 6. (a (b The vertical distance between the objects is the absolute value of f( sint sin t. Find the critical points in [, 0 ]: f ( cost cost 0 ( cos t cost 0 ( cos t cos t 0 ( cos t+ (cos t 0 cost or cost t,, 0, The critical points (and endpoints are (0, 0,,,,, and (, 0 The distance is greatest when t sec and when t sec. The distance at those times is meters. sin t sin t+ sin t sin tcos + costsin sin t sin t+ cost sin t cost tan t Solving for t, the particles meet at t sec and at t sec. (b The distance between the particles is the absolute value of f( t sin t+ sin t cost sin t. Find the critical points in [0, ]: f ( t sint cost 0 sin t cost tan t 5 The solutions are t and 6 t 6, so 5 the critical points are at and 6, 6,, and the interval endpoints are at 0,, and,. The particles 5 are farthest apart at t sec and at 6 t 6 sec, and the maimum distance between the particles is m. (c We need to maimize f ( t, so we solve f ( t 0. f ( t cost+ sint 0 sin t cost This is the same equation we solved in part (a, so the solutions are t sec and t sec. For the function y f ( t, the critical points occur at, and,, and the interval endpoints are at 0, and,. Thus, f ( t is maimized at t and t. But these are the instants when the particles pass each other, so the graph of y f( t has d corners at these points and f ( t is undefined at these instants. We cannot say that the distance is changing the fastest at any particular instant, but we can say that Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
6 Section 5. near t or t the distance is changing faster than at any other time in the interval. 7. The trapezoid has height (cos θ ft and the trapezoid bases measure ft and ( + sin θft, so the volume is given by V ( θ (cos θ( + + sin θ( 0 0(cos θ( + sin θ. 8. Find the critical points for 0 θ < : V ( θ 0(cos θ(cos θ + 0( + sin θ( sin θ 0 0cos θ 0sinθ 0sin θ 0 0( sin θ 0sinθ 0sin θ 0 0( sin θ + sinθ 0 0( sin θ (sin θ + 0 sinθ or sinθ θ 6 The critical point is at 5 6,. Since V ( θ > 0 for 0 θ < and V ( θ < 0 for 6 < θ < 6, the critical point corresponds to the maimum possible trough volume. The volume is maimized when θ 6. Sketch segment RS as shown, and let y be the length of segment QR. Note that PB 8.5, and so QB (. 85 85.( 85.. Also note that triangles QRS and PQB are similar. QR PQ RS QB y 85. 85.( 85. (a y 85. 85.( 85. 85. y 8. 5 L + y 85. L + 8. 5 ( 85. + 8. 5 L 85. L 8. 5 (b Note that >. 5, and let f( L. Since y, the 8. 5 approimate domain of f is 5.0 8.. 5 Then ( 85. ( 6 ( ( f ( ( 85. ( 8 5 ( 8. 5 For > 5. 0, the critical point occurs at 5 6. 75in., and this corresponds to 8 a minimum value of f( because f ( < 0 for 5. 0 < < 6. 75 and f ( > 0 for > 6. 75. Therefore, the value of that minimizes L is 6. 75 in. (c The minimum value of L is 675 (.. 0 in. 675 (. 85. C M C 9. Since R M M M, we dr have CM M. dm Let f( M CM M. Then f ( M C M, Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
Section 5. 6 C and the critical point for f occurs at M. This value corresponds to a maimum because ( 0 for C C f M > M < and f ( M < 0 for M >. The value of M that maimizes dr C is M dm. 50. The profit is given by P ( ( n( c a+ b( 00 ( c b + ( 00 + c b + ( a 00bc. Then P ( b+ ( 00 + c b b( 00 + c. The critical point occurs at 00 + c c 50 +, and this value corresponds to the maimum profit because c P ( > 0 for < 50 + and P ( < 0 for c > 50 +. c A selling price of 50 + will bring the maimum profit. 55. B; A bh b + h 00 b 00 h h A 00 h 00 h h A 00 h A 0 when h 50 b 00 50 50 A ma 50 50 5 56. E; length height 0 0 5 area A ( 0 5 60 0 57. da ( 60 0 60 0 d ( 0 5( 0. 5. True. This is guaranteed by the Etreme Value Theorem (Section 5.. 5. False; for eample, consider c 0. 5. D; f( ( 60 f ( ( + ( 60 ( + 0 + 0 ( 0 0 or 0 60 60 60 0 ( 60 0 ( 0 ( 0 ( 600( 0, 000 f( at Let P be the foot of the perpendicular from A to the mirror, and Q be the foot of the perpendicular from B to the mirror. Suppose the light strikes the mirror at point R on the way from A to B. Let: a distance from A to P b distance from B to Q c distance from P to Q distance from P to R To minimize the time is to minimize the total distance the light travels going from A to B. The total distance is D ( + a + ( c + b Then 5. B; since f ( is negative, f( is always decreasing, so f(5. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
6 Section 5. ( ( + [ ( ] D c + a ( c + b c + ( + a c b Solving D ( 0 gives the equation Squaring both sides, we have: ( c + a ( c + b ( c + b ( c ( + a ( c + b ( c + ( c a b ( c a b c c+ a c which we will refer to as Equation. + a ( c + b 0 ( a b a c+ a c 0 [( a+ b ac][( a b ac] ac ac or a+ b a b ac Note that the value a b is an etraneous solution because ac cb c c a b a b, so and c could ac not both be positive. The only critical point occurs at. a + b To verify that critical point represents the minimum distance, notice that D is differentiable for all in [0, c] with a single critical point in the interior of the interval. Since D ( 0 c < 0, D must be decreasing c + b c from 0 to the critical point, and since D ( c > 0, D must be increasing from the critical point to c + b c. PR QR We now know that D( is minimized when Equation is true, or, equivalently,. This means that AR BR the two right triangles APR and BQR are similar, which in turn implies that the two angles must be equal. dv 58. ka k d ka a The critical point occurs at k, which represents a maimum value because a a ka negative for all. The maimum value of v is ka k ka k. d v k, which is d Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
Section 5. 65 59. (a v cr0 r cr dv cr0r cr cr ( r0 r dr r0 The critical point occurs at r. (Note that r 0 is not in the domain of v. The critical point represents a maimum d v because cr 0 6cr c( r0 r, dr r0 which is negative when r. (b We graph v ( 05. r r, and observe that the maimum indeed occurs at v 05.. 6. The profit is given by p( r( c( 6 ( 6 + 5 + 6 9, for 0. Then p ( + 9 ( (, so the critical points occur at and. Since p ( < 0 for 0 <, p ( > 0 for < <, and p ( < 0 for >, the relative maima occur at the endpoint 0 and at the critical point. Since p( 0 p( 0, this means that for 0, the function phas ( its absolute maimum value at the points (0, 0 and (, 0. This result can also be obtained graphically, as shown. h 60. (a Since A ( q kmq +, the critical km h point occurs when q, or km q. h This corresponds to the minimum value of A(q because A ( q kmq, which is positive for q > 0. (b The new formula for average weekly cost ( k + bq m hq is Bq ( + cm+ q km hq + bm + cm + q Aq ( + bm Since B(q differs from A(q by a constant, the minimum value of B(q will occur at the same q-value as the minimum value of A(q. The most economical km quantity is again. h 6. The average cost is given by c ( a( 0+ 0, 000. Therefore, a ( 0 and the critical value is 0, which represents the minimum because a (, which is positive for all. The average cost is minimized at a production level of 0 items. 6. (a According to the graph, y ( 0 0. (b According to the graph, y ( L 0. (c y ( 0 0, so d 0. Now y ( a + b + c, so y ( 0 0 implies that c 0. Therefore, y ( a + b and y ( a + b. Then y ( L al + bl H and y ( L al bl 0, so we have two linear equations in the two unknowns a and b. The second equation gives al b. Substituting into the first al equation, we have al + H, or al H H, so a. Therefore, L Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
66 Section 5. b H and the equation for y is L ( H H y +, or L L y( H +. L L a 6. (a The base radius of the cone is r and so the height is a a Therefore, V( r h a. a h a r a. (b To simplify the calculations, we shall consider the volume as a function of r: volume f ( r r a r, where 0 < r < a. d f ( r ( r a r dr r ( r + ( a r ( r a r r + r( a r a r ( a r r a r r( a r a r a The critical point occurs when r, a 6 which gives r a. Then a a a h a r a. a 6 a Using r and h, we may now find the values of r and h for the given values of a 6 5 6 5 when a : r, h ; when a 5 : r, h ; 8 6 8 when a 6: r 6, h ; when a 8 : r, h a 6 (c Since r and a h, r the relationship is. h 65. (a Let 0 represent the fied value of at point P, so that P has coordinates ( 0, a and let m f ( 0 be the slope of line RT. Then the equation of line RT is y m( 0 + a. The y-intercept of this line is m0 a m( 0 0 + a a m0, and the -intercept is the solution of m( 0 + a 0, or. Let O m designate the origin. Then Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
Section 5. 67 (Area of triangle RST (Area of triangle ORT (-intercept of line RT (y-intercept of line RT m0 a ( a m0 m m0 a m0 a m m m m0 a m m a m 0 m Substituting for 0, f ( for m, and f ( for a, we have (b The domain is the open interval (0, 0. To graph, let y f ( 5+ 5, 00 y f ( NDER ( y, and y y A( y. y The graph of the area function y A( is shown below. f ( ( (. f ( A f [0, 0] by [ 00, 000] The vertical asymptotes at 0 and 0 correspond to horizontal or vertical tangent lines, which do not form triangles. (c Using our epression for the y-intercept of the tangent line, the height of the triangle is a m f( f ( 5+ 00 00 5+ 00 + 00 We may use graphing methods or the analytic method in part (d to find that the minimum value of A( occurs at 8. 66. Substituting this value into the epression above, the height of the triangle is 5. This is times the y-coordinate of the center of the ellipse. (d Part (a remains unchanged. The domain is (0, C. To graph, note that B f( B + B B C C + C and Therefore, we have B B f ( (. C C C C Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
68 Section 5. f( A ( f ( f ( B B B+ C C B C C C C B ( BC + B C C C C B B + ( BC + B C ( C BC C B BC C B( C + + BC C BC( C + C BC C BC( C + C C BC C ( C C BC C C + + + C ( C C A ( ( C ( BC C + C ( C+ C + C ( C C ( BC C + C C + C C + ( C C C ( BC C + C C C C C ( C C ( BC C C ( C + C C( C C C ( + ( BC C C C C C ( C To find the critical points for 0 < < C, we solve: C C C C + C C C C 0 ( C 0 C C The minimum value of A( for 0 < < C occurs at the critical point, or. The corresponding triangle height is Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
a m f( f ( B B B+ C + C C C B B C B+ C + C C C BC C ( C B C B + C + C B B B + + B This shows that the triangle has minimum area when its height is B. Section 5.5 Linearization and Differentials (pp. 7 9... Section 5.5 69 dy ( + ( sin ( + cos ( d ( + sin + sin cos ( + cos ( + sin ( + 0.567 Eploration Appreciating Local Linearity. The graph appears to have either a cusp or a corner at (0,.. ( / y + 0000. + 09. / f ( ( + 0000. (. ( + 0. 000 Since f ( 0 0, the tangent line at (0, has equation y.. The corner becomes smooth and the graph straightens out.. As with any differentiable curve, the graph comes to resemble the tangent line. Quick Review 5.5. dy d cos( + ( + d d cos( + 0. 5. f ( ( ( e + ( e ( e e f (0 The lines passes through (0, and has slope. Its equation is y +. 6. f ( ( ( e + ( e ( e e f ( e ( e e The lines passes through (, e + and has slope e. Its equation is y e( + + ( e +, or y e + e +. 7. (a + 0 (b e + e + 0 e ( e + e + e 0. 68 8. f ( f ( ( Since f( and f (, the graph of g( passes through (, and has slope. Its equation is g( ( + (, or g(. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
70 Section 5.5 f( g ( 0.7.57.7 0.8.688.8 0.9.87.9.0..069...07...00. 9. f ( cos f (. 5 cos 5. Since f(.5 sin.5 and f (. 5 cos., 5 the tangent line passes through (.5, sin.5 and has slope cos.5. Its equation is y (cos 5. ( 5. + sin 5., or approimately y 0. 07+ 0. 89 Section 5.5 Eercises. (a f ( We have f ( 7 and f ( 0. L( f( + f ( ( 7+ 0( 0 (b Since f(. 8.06 and L(. 8, the approimation differs from the true value in absolute value by less than 0.. (a f ( ( + 9 + 9 We have f( 5 and f (. 5 L( f( + f ( ( ( 5 ( + 5 9 + 5 5 (b Since f(.9.90 and L(.9.9, the approimation differs from the true value by less than 0. [0, ] by [ 0.,.] 0. For >, f (, and so f (. Since f( and f (, the tangent line passes through (, and has slope Its equation is. y ( +, or y.. (a f ( We have f( and f ( 0. L ( f( + f ( ( + 0( (b Since f (..009 and L (., the approimation differs from the true value by less than 0.. (a f ( + We have f( 0 0 and f ( 0. L ( f( 0 + f (( 0 0 0+ (b Since f(0. 0.095 and L ( 0. 0. the approimation differs from the true value by less than 0. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
5. (a f ( sec We have f( 0 and f (. L ( f( + f ( ( 0+ ( (b Since f( + 0. 0.00 and L( + 0. 0., the approimation differs from the true value in absolute value by less than 0. 6. (a 7. f ( We have f( 0 and f ( 0. L ( f( 0 + f (( 0 0 + ( ( 0 + (b Since f(0..706 and L(0..7080, the approimation differs from the true value in absolute value by less than 0. f ( k( + k We have f( 0 and f ( 0 k. L ( f( 0 + f (( 0 0 + k ( 0 + k 8. (a 00 00 (. 00 ( + 000. + ( 00( 0. 00.; 00. 00. 0. 0 < 0 (b (c 0. (a (b (c f( [ + ( ] [ + ( ( ] + Section 5.5 7 / f( ( + + / f( ( + / / + / + / + f( + / + + + f( / + / + + + + 6+ (b 9. (a /. 009 ( + 0. 009 + ( 0. 009 00. ; 6 5 009. 00. 9 0 < 0 6 f( ( 6 [ + ( ] + 6( 6. 00 / f ( 00 ( 00 0. 05 f ( 00 0 + 0. 05( 0 00 0. 05. 7 / f ( 7 ( 7 7 f ( 7 + ( 6 7 7 y. 96 7 Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
7 Section 5.5. 000 / f ( 000 ( 000 00 y 0 + ( 000 00 y 0 9. 99 50. 8 / f ( 8 ( 8 8 y 9+ ( 80 8 8 y 9 89. 8 5. (a Since dy, dy ( d. d (b At the given values, dy ( ( 005. 9005 (. 05.. 6. (a Since dy ( + ( ( (, d ( + ( + dy d. ( + (b At the given values, ( dy ( 0. [ + ( ] 8 ( 0. 5 00.. 7. (a Since dy ( + (ln ( ln +, d dy ( ln + d. (b At the given values, dy [ ( ln( + ]( 00. 00 (. 00. 8. (a Since dy ( ( ( ( d + + + (, dy d. (b At the given values, ( 0 dy ( 0. 0.. ( 0 dy 9. (a Since sin e cos, d sin dy (cos e d. (b At the given values, sin dy (cos ( e ( 0. ( ( ( 0. 0.. 0. (a Since dy csc cot d csc cot, csc dy cot d. (b At the given values, dy csc cot ( 0. 0. csc cot 0. 0555 Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
Section 5.5 7. (a y+ y 0 y( + y + dy ( + ( ( ( Since d ( + ( +. (a.. 5. 6. d dy ( +. 00. (b At the given values, dy 00.. ( 0+ y y dy d dy yd dy( + ( y d y dy d + (b At the given values, and y from the original equation, ( dy ( 0. 05 0. 075 + dy d dy d dy d dy e d + 5 5 5 dy ( 5e + 5 d d du tan u d + u d u du d dy d + 6 d a (ln a a d 7 dy ( 8 ln 8+ 8 d, (b Since f ( +, f ( 0. Therefore, df d ( 0. 0.. (c f df 0. 0. 00. 8. (a f f(. f( 0. 0 0. (b Since f (, f (. Therefore, df d ( 0. 0.. (c f df 0. 0. 00. 0 9. (a f f( 055. f( 05. (b Since f (, f ( 05.. Therefore, df d ( 005. 0. 5 (c f df + 5 55 0. (a f f(. 0 f(. 00600 0. 00600 (b Since f (, f (. Therefore, df d ( 0. 0 0. 0. (c f df 0. 00600 0. 0 0. 000600 dv. Note that r, dv r dr. When r dr changes from a to a + dr, the change in volume is approimately a dr. When a 0 and dr 0.05, V ( 0 ( 0. 05 0 cm. ds. Note that 8r, so ds 8rdr. When r dr changes from a to a + dr, the change in surface area is approimately 8 adr. When a 0 and dr 0.05, S 8( 0( 005. cm. 7. (a f f( 0. f( 0 0. 0 0. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
7 Section 5.5 dv. Note that, so dv d. d When changes from a to a + d, the change in volume is approimately a d. When a 0 and d 0.05, V ( 0 ( 0. 05 5 cm. ds. Note that, so ds d. d When changes from a to a + d, the change in surface area is approimately a d. When a 0 and d 0.05, S ( 0( 0. 05 6 cm. dv 5. Note that rh, so dv rh dr. dr When r changes from a to a + dr, the change in volume is approimately ah dr. When a 0 and dr 0.05, V ( 0 h( 005. h cm. ds 6. Note that r, so ds rdh. When h dh changes from a to a + dh, the change in lateral surface area is approimately r dh. When a 0 and dh 0.05, 7. 8. 9. 0. S r( 005. 0. r cm. A r da r dr da ( 0( 0. 6. in V r dv r dr dv ((. 8 0 in V s dv s ds dv 5 ( ( 0. 5 cm A da s s ds 0 05 87 cm da ( (... (a Note that f ( 0 cos 0. L ( f( 0 + f (( 0 0 + + (b f( 0. L( 0.. (c The actual value is less than.. This is because the derivative is decreasing over the interval [0, 0.], which means that the graph of f ( is concave down and lies below its linearization in this interval. da. (a Note that A r and r, so dr da r dr. When r changes from a to a + dr, the change in area is approimately a dr. Substituting for a and 0.0 for dr, the change in area is approimately ( ( 0. 0 0. 08 0. 5 (b da A 008. 00. %. Let A cross section area, C circumference, C dd andd diameter. Then D, so dc and dd dc. Also, D C C, so da C A dc C and da dc. When C increases from 0 in. to 0 + in. the diameter increases by dd ( 0. 666 in. and the area increases by approimately 0 da ( 0 in.. Let edge length and V volume. Then V, and so dv d. With 0 cm and d 0.0 0. cm, we have V 0 000 cm and dv (0 (0. 0 cm, so the percentage error in the volume measurement is dv 0 approimately 00. %. V 000 Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
Section 5.5 75 5. Let side length and A area. Then A da and, so da d. We want d da 00. A, which gives d 0. 0, or d 00.. The side length should be measured with an error of no more than %. 6. (a Note that V r h 0r. 5D, where D is the interior diameter of the dv tank. Then 5 D, so dv 5 D dd. dd We want dv 00. V, which gives 5 DdD 0. 0(. 5 D, or dd 0. 005D. The interior diameter should be measured with an error of no more than 0.5%. (b Now we let D represent the eterior diameter of the tank, and we assume that the paint coverage rate (number of square feet covered per gallon of paint is known precisely. Then, to determine the amount of paint within 5%, we need to calculate the lateral surface area S with an error of no more than 5%. Note that ds S rh 0D, so 0 and dd ds 0. We want ds 005. S, which gives 0 dd 0. 05( 0D, or dd 0.5D. The eterior diameter should be measured with an error of no more than 5%. 7. Note that V r h, where h is constant. Then dv rh. The percent change is given by dr dv rhdr dr 0. % r 0. %. V r h r r dv 8. Note that h, so dv h dh. We dh want dv 00. V, which gives 00. h hdh 0. 0( h, or dh. The height should be measured with an error of no more than %. 9. If dc dr and dc inch, then 8 dr inch. 6 Since V r, we have Sphere Type r dv r dr r 6. The volume error in each case is simply r in. True Radius Tape error Radius Error Orange in. 8 in. in. 6 Melon in. 8 in. in. 6 Beach Ball Volume Error in in 7 in. 8 in. in.. 6. 5 in 50. If dc dr and Sphere Type dc inch, then 8 dr inch. 6 Since A r, we have r da 8rdr 8r 6. The surface area error in each case is simply r in. Orange Melon Beach Ball 5. We have True Radius in. in. 7 in dw dg Tape Error Radius Error 8 in. in. 6 8 in. in. 6 8 in. in. 6 Volume Error in in 5in., so. bg dw bg dg dw moon b(. 5 dg Then 7. 87. dw earth b( dg 5. The ratio is about 7.87 to. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
76 Section 5.5 5. (a Note that T L g, so dt dg and L g dt L g dg. (b Note that dt and dg have opposite signs. Thus, if g increases, T decreases and the clock speeds up. (c L g dg dt ( 00 ( 980 dg 0. 00 dg 0. 9765 Since 0. 9765, 980 0. 9765 979. 05. dg g 5. Let f( +. Then f ( + and f( n n + n n+ n n. f ( n n + Note that f is cubic and f is always positive, so there is eactly one solution. We choose 0. 0 075. 0. 686065 5 0. 6896 6 0. 6878 7 0. 6878 Solution: 0. 688. 5. Let f( +. Then f ( + and f( n n + n n+ n n f ( n n + The graph of y f( shows that f( 0 has two solutions.. 5.. 55. 6596. 56. 607. 5669. 605 5. 5669 5. 605 Solution:.567,.605 55. Let f( + sin. Then f ( cosand f ( n n+ n f ( n n n + sin n n n cos n The graph of y f( shows that f( 0 has two solutions 0. 0. 85699 0. 8695 0. 8669 5 0. 8669. 96598. 965695. 965690 5. 965690 Solutions: 0.867,.96569 56. Let f(. Then f ( and f( n n n+ n n. f ( n n Note that f( 0 clearly has two solutions, namely ±. We use Newton s method to find the decimal equivalents. 5.. 78. 9798. 89858 5. 8907 6. 8907 Solutions: ±. 8907 57. True; a look at the graph reveals the problem. The graph decreases after toward a horizontal asymptote of y 0, so the -intercepts of the tangent lines keep getting bigger without approaching a zero. 58. False; by the product rule, duv ( udv+ vdu. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
Section 5.5 77 59. B; f( e f ( e L ( e + e( L ( e 60. D; y tan dy (sec d (sec 05. 05. f( + 6. D; f ( n n + n+ n n ( + ( ( + 8 ( 6. A; f( ; 6 f ( 6 ( 6 8 97 66 + ( 66 6 8 The percentage error is 66 97 / 00. %. 66 6. If f ( 0, we have f( 0. Therefore, f ( f (, and all later approimations are also equal to. 6. If h, then f ( and h h h h h h. If h, then / h f ( and h h h h+ h h / h 65. Note that f ( and so f( n n+ n f ( n / n n n / n n n. For, we have,, 8, n and 5 6; n. The approimations alternate in sign and rapidly get farther and farther away from the zero at the origin. [ 0, 0] by [, ] 66. (a i. Qa ( f( a implies that b0 f( a. ii. Since Q ( b+ b( a, Q ( a f ( a implies that b f ( a. iii. Since Q ( b, Q ( a f ( a f ( a implies that b In summary, f ( a b0 f( a, b f ( a, and b. (b f( ( f ( ( ( ( f ( ( ( ( Since f( 0, f ( 0, and f ( 0, the coefficients are b0, b, and b. The quadratic approimation is Q ( + +. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
78 Section 5.5 (c (d As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical. g ( g ( g ( Since g(, g (, and g (, the coefficients are bo, b, and b. The quadratic approimation is Q ( ( + (. (f The linearization of any differentiable function u( at a is L( u( a + u ( a( a b0 + b( a, where b0 and bare the coefficients of the constant and linear terms of the quadratic approimation. Thus, the linearization for f( at 0 is + ; the linearization for g( at is ( or ; and the linearization for h( at 0 is +. 67. Finding a zero of sin by Newton s method would use the recursive formula sin( n n+ n n tan n, and that is cos( n eactly what the calculator would be doing. Any zero of sin would be a multiple of. 68. Just multiply the corresponding derivative formulas by d. (e As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical / h ( ( + / h ( ( + / h ( ( + Since h( 0, h ( 0, and h ( 0, the coefficients are b0, b, and b 8. The quadratic approimation is Q ( +. 8 (a Since d ( c 0, d ( c 0. d d du (b Since ( cu c, d( cu c du. d d (c Since d ( u+ v du + dv, d d d d(u + v du + dv. (d Since d ( u v u dv + v du, d d d duv ( udv+ vdu. du d d u v u (e Since d v v u vdu udv d. v v (f Since n d u d n nu n du ( nu du. n du, d dv d, As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
Section 5.6 79 69. First, note that d tan sec, and d sec ( 0, so the linearization of tan at 0 is y 0 ( 0, or more simply y. tan sin / cos lim lim sin lim 0 cos sin lim lim 0cos 0 ((. 0 0 70. g(a c, so if E(a 0, then g(a f(a and c f(a. Then E( f( g( f( f(a m( a. E ( f( fa ( Thus, m. a a f lim ( f ( a f ( a, so a a E lim ( f ( a m. a a 7. E ( Therefore, if the limit of a m f ( a and g( L(. is zero, then f ( + cos + We have f( 0 and f ( 0 L( f( 0 + f (( 0 0 + The linearization is the sum of the two individual linearizations, which are for sin and + for +. 7. The equation for the tangent is y f( n f ( n( n. Set y 0 and solve for. 0 f( n f ( n( n f( n f ( n f ( n n f ( n f ( n n f( n f( n n (If f ( n 0 f ( n + The value of is the net approimation. n Section 5.6 Related Rates (pp. 50 59 Eploration The Sliding Ladder. Here the -ais represents the ground and the y-ais represents the wall. The curve (, y gives the position of the bottom of the ladder (distance from the wall at any time t in 0 t 5. The curve (, y gives the position of the top of the ladder at any time in 0 t 5.. 0 t 5. This is a snapshot at t.. The top of the ladder is moving down the y-ais and the bottom of the ladder is moving to the right on the -ais. The end of the ladder is accelerating. Both aes are hidden from view. 6. 7. dy T 0 ( T y (. 5ft/sec. The negative number means the y-side of the right triangle is decreasing in length. lim y ( t, the speed of the top of t 5 the ladder is infinite as it hits the ground. 8. Since Quick Review 5.6.. D ( 7 0 + ( 0 5 9 + 5 7 D ( b 0 + ( 0 a a + b. Use implicit differentiation. d d ( y + y ( + y d d dy dy dy + y( + y ( + d d d dy ( + y y d dy y d + y Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
80 Section 5.6. Use implicit differentiation. d d ( sin y ( y d d dy dy ( (cos y + (sin y( y( d d dy ( + cos y y sin y d dy y sin y d + cos y dy y + sin y d + cos y 5. Use implicit differentiation. d d tan y d d dy sec y d dy d sec y dy cos y d 6. Use implicit differentiation. d d ln( + y ( d d dy + + y d dy + ( + y d dy + y d 7. Using A(, we create the parametric equations + at and y + bt, which determine a line passing through A at t 0. We determine a and b so that the line passes through B(, at t. Since + a, we have a 6, and since + b, we have b. Thus, one parametrization for the line segment is + 6t, y t, 0 t. (Other answers are possible. 8. Using A(0,, we create the parametric equations 0 + at and y + bt, which determine a line passing through A at t 0. We now determine a and b so that the line passes through B(5, 0 at t. Since 5 0 + a, we have a 5, and since 0 + b, we have b. Thus, one parametrization for the line segment is 5t, y + t, 0 t. (Other answers are possible. 9. One possible answer: t 0. One possible answer: t Section 5.6 Eercises da da dr da dr. Since, we have r. dr ds ds dr ds dr. Since, we have 8 r. dr dv dv dh. (a Since, we have dh dv dh r. dv dv dr (b Since, we have dr dv dr rh. (c. (a dv d d r h ( r h dv dh ( dr r + h r dv dh dr r + rh dp d ( RI dp d dr R I + I dp di dr R I I + dp di dr RI + I dp (b If P is constant, we have 0, which di dr means RI + I 0, or dr R di P di. I I Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
Section 5.6 8 5. 6. ds d + y + z ds d ( y z y z + + + + ds d dy dz y z + + y z + + ds dy d + y + z dz + y + z da d absinθ da da db d b sinθ + a sinθ + ab sinθ da da db dθ bsinθ asinθ abcosθ + + dv 7. (a Since V is increasing at the rate of volt/sec, volt/sec. (b Since I is decreasing at the rate of di amp/sec, amp/sec. (c Differentiating both sides of V IR, we have dv I dr + R di. (d Note thatv IR gives R, so R 6 ohms. Now substitute the known values into the equation in (c. dr + 6 dr dr ohms/sec R is changing at the rate of ohms/sec. Since this value is positive, R is increasing. 8. Step : r radius of plate A area of plate Step : dr At the instant in question, 0. 0 cm/sec, r 50 cm. Step : da We want to find. Step : A r Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
8 Section 5.6 Step 5: da dr r Step 6: da ( 50( 0. 0 cm /sec At the instant in question, the area is increasing at the rate of cm /sec. 9. Step : l length of rectangle w wih of rectangle A area of rectangle P perimeter of rectangle D length of a diagonal of the rectangle Step : dl At the instant in question, cm/sec, dw cm/sec, l cm, and w 5 cm. Step : da dp dd We want to find,, and. Steps, 5, and 6: (a A lw da dw dl l + w da ( ( + ( 5( cm /sec The rate of change of the area is cm /sec. (b P l+ w dp dl dw + dp ( + ( 0 cm/sec The rate of change of the perimeter is 0 cm/sec. (c D l + w dd dl dw l w + l + w l dl + wdw l + w dd ( ( + ( ( + 5 cm/sec 5 The rate of change of the length of the diameter is cm/sec. (d The area is increasing, because its derivative is positive. The perimeter is not changing, because its derivative is zero. The diagonal length is decreasing, because its derivative is negative. 0. Step :, y, z edge lengths of the bo V volume of the bo S surface area of the bo s diagonal length of the bo Step : At the instant in question, d dy dz m/sec, m/sec, m/ sec, m, y m, and z m. Step : We want to find dv, ds, and ds. Steps, 5, and 6: (a V yz dv dz dy d y + z + yz dv ( ( ( + ( ( ( + ( ( ( m /sec The rate of change of the volume is m /sec. (b S ( y+ z+ yz ds dy d dz d dz dy y z y z + + + + + ds [( ( + ( ( + ( ( + (( + (( + (( ] 0 m /sec The rate of change of the surface area is 0 m /sec. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
Section 5.6 8 (c s + y + z ds d dy dz + y + z + y + z dy d dy + y + z dz + y + z The rate of change of the diagonal length is 0 m/sec.. Step : r radius of spherical balloon S surface area of spherical balloon V volume of spherical balloon Step : dv At the instant in question, 00 ft /min and r 5 ft. Step : We want to find the values of dr ds and. Steps, 5, and 6: (a (b V r dv dr r dr 00 ( 5 dr ft/min The radius is increasing at the rate of ft/min. S r ds dr 8 r ds 8 (( 5 ds 0 ft /min The surface area is increasing at the rate of 0 ft /min.. Step : r radius of spherical droplet S surface area of spherical droplet V volume of spherical droplet Step : No numerical information is given. Step : We want to show that dr is constant. Step : S r, V r, dv ks for some constant k Steps 5 and 6: Differentiating V r, we have dv dr r. dv Substituting ks for and S for r, we dr dr have ks S, or k.. Step : s (diagonal distance from antenna to airplane horizontal distance from antenna to airplane Step : At the instant in question, ds s 0 mi and 00 mph. Step : d We want to find. Step : + 9 s or s 9 Step 5: d ds s ds s s 9 s 9 Step 6: d 0 ( 00 0 9 000 mph 5 0. 08 mph The speed of the airplane is about 0.08 mph.. Step : s length of kite string horizontal distance from Inge to kite Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
8 Section 5.6 Step : d At the instant in question, 5 ft/sec and s 500 ft Step : ds We want to find. Step : + 00 s Step 5: d ds d ds s or s Step 6: At the instant in question, since + 00 s, we have s 00 500 00 00. Thus ( 00( 5 ds ds ( 500, so, so ds 0 ft/sec. Inge must let the string out at the rate of 0 ft/sec. 5. Step : The cylinder shown represents the shape of the hole. r radius of cylinder V volume of cylinder Step : At the instant in question, dr 000in.. in./min and (since the min 000 diameter is.800 in., r.900 in. Step : dv We want to find. Step : V r ( 6 6r Step 5: dv dr r Step 6: dv (. 900 000 9 500 0. 0076 0. 09 in /min. The volume is increasing at the rate of approimately 0.09 in /min. 6. Step : r base radius of cone h height of cone V volume of cone Step : At the instant in question, h m and dv 0 m /min. Step : We want to find dh dr and. Step : Since the height is of the base diameter, we 8 have h 8 ( r or r h. We also have h 6 V r h h h 7. We will 6 h use the equations V and 7 r h. Step 5 and 6: (a dv 6 h dh 9 6 ( dh 0 9 dh 5 5 m/ min cm/ min 8 The height is changing at the rate of 5. 9 cm/ min. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.
Section 5.6 85 (b Using the results from Step and part (a, we have dr dh 5 75 cm/ min. 8 The radius is changing at the rate of 75. 9cm/ min. 8 7. Step : 6 m h 5 m r r radius of top surface of water h depth of water in reservoir V volume of water in reservoir Step : At the instant in question, dv 50 m /min and h 5 m. Step : dh dr We want to find and. Step : h 6 Note that r 5 by similar cones, so r 75. h. Then V 75 875 r h (. h h. h Steps 5 and 6: dv dh (a SinceV 8. 75h, 56. 5h. dh Thus 50 56. 5 ( 5, and dh 8 so m/min cm/min. 5 9 The water level is falling by. 9 cm/min. dh (Since < 0, the rate at which the water level is falling is positive. (b Since r 75. h, dr dh 80 75. cm/min. The rate of change of the radius of the water s surface 80 is 89. cm/min. 8. (a Step : y depth of water in bowl V volume of water in bowl Step : At the instant in question, dv 6 m /min and y 8 m. Step : dy We want to find the value of. Step : V y ( 9 y or V y y Step 5: dv dy ( 6y y Step 6: dy 6 6( 8 ( 8 dy 6 dy 0. 06 m/min 5 or. 6 cm/min 6 (b Since r + ( y, r 69 ( y 6y y. (c Step : y depth of water r radius of water surface V volume of water in bowl Step : At the instant in question, dv 6 m /min, y 8 m, and therefore (from part (a dy m/min. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.