Answer Key for the Review Packet for Exam #3


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1 Answer Key for the Review Packet for Eam # Professor Danielle Benedetto Math MaMin Problems. Show that of all rectangles with a given area, the one with the smallest perimeter is a square. Diagram: y Variables: Let =length of the rectangle. Let y =width of the rectangle. Let A =area of rectangle. Let P =perimeter of rectangle. Equations: We know A = y is fied, so that y = A. must be minimized. The commonsense Then the perimeter P = + y = + A boundsdomain of P is { : > }. Minimize: Net P = A. Setting P = we solve for = + A. We take the positive square root here since we are talking about positive lengths. Signtesting the critical number does indeed yield a minimum for the perimeter function. P ց A ր MIN Answer: P Since = A then y = A A = A. As a result, = y and the smallest perimeter occurs when the rectangle is a square.. A rectangle lies in the first quadrant, with one verte at the origin, two sides along the coordinate aes, and the fourth verte on the line + y 6 =. Find the maimum area of the rectangle. Diagram: + y 6 =, y y 6
2 Variables: Let = coordinate of point on the line. Let y = ycoordinate of point on the line. Let A =area of inscribed rectangle. Equations: The verte point, y lies on the line + y 6 =. The fied line + y 6 = can be rewritten as y =. The area of the rectangle is given as A = y = = The commonsenseboundsdomain of A is { : 6}. Maimize: Net, A =. Setting A = we solve for = as the critical number. and must be maimized. Signtesting the critical number does indeed yield a maimum for the area function. A A ր ց MAX Answer: Finally, = = y =. As a result the maimum area is Area=y = = 9 square units.. A farmer wants to use a fence to surround a rectangular field, using an eisting stone wall as one side of the plot. She also wants to divide the field into 5 equal pieces using fence parallel to the sides that are perpendicular to the stone wall see diagram. The farmer must use eactly feet of fence. What is the maimum area possible for this field? Diagram: y y y y y y wall Variables: Let =length of side parallel to wall. Let y =width of sides perpendicular to wall. Let L =length of fence. Let A =area of enclosed field. Equations: We know that the length of fence used L = + 6y = is fied so that = 6y. Then the area A = y = 6yy = y 6y must be maimized. The commonsenseboundsdomain of A is {y : y }. Maimize:
3 Net A = y. Setting A = we solve for y =. Signtesting the critical number does indeed yield a maimum for the area function. A A ր ց MAX Answer: Since y = then = 6 = 6. As a result, the maimum area possible is 6, square feet.. You work for a soup manufacturing company. Your assignment is to design the newest can in the shape of a cylinder. You are given a fied amount of material, 6 cm, to make your can. What are the dimensions of your can which will hold the maimum volume of soup? Diagram: r πr r r h top bottom side of can h Variables: Let r =radius of can. Let h =height of can. Let M =amount of material surface area. Let V =volume of can. Equations: We know that the amount of material used M = πr + πrh = 6 is fied so that 6 πr h =. πr 6 πr Then the volume V = πr h = πr 6 πr = r = r πr must be πr maimized. The commonsenseboundsdomain of V is {r : < r π }. Maimize: Net V = πr. Setting V = yields r = π as the critical number. Signtesting the critical number does indeed yield a maimum for the volume function. V π V ր ց MAX Answer:
4 Since r = 6 π π then h = π π π = 6 π = π. The dimensions of the can with the largest volume are r = π and h = π in cm. A bit of etra work does indeed check that the amount of material used is 6 square cm. M = πr + πrh = π π + π π π = + = A rectangular bo with square base cost $ per square foot for the bottom and $ per square foot for the top and sides. Find the bo of largest volume which can be built for $6. Diagram: y Variables: Let =length of side on base of bo. Let y =height of bo. Let Cost=Cost for amount of material surface area. Let V =volume of bo. Equations: Then the Cost of materials, which is fied is given as Cost = cost of base + cost of top + cost of sides = $ + $ + y$ = + y = 6 6 = y = 6 Then the volume of the bo V = y = = 9 must be maimized. The commonsenseboundsdomain of V is { : < }. Maimize: Net V = 9 9. Setting V = we solve for = as the critical number. Signtesting the critical number does indeed yield a maimum for the volume function. V V ր ց MAX Answer: Since = then y =, each in feet. 6 8 =. As a result, the bo of largest volume will measure
5 6. Among all the rectangles with given perimeter P, find the one with the maimum area. Diagram: y Variables: Let =length of the rectangle. Let y =width of the rectangle. Let A =area of rectangle. Let P =perimeter of rectangle. Equations: Let P equal the given perimeter. We know + y = P is fied, so that y = P. P Then the area A = y = = P must be maimized. The commonsenseboundsdomain of A is { : P }. Maimize: Net A = P. Setting A = we solve for = P. Signtesting the critical number does indeed yield a maimum for the area function. P P P ր ց MAX Answer: Since = P then y = P P the rectangle is a square. = P. As a result, = y and the largest area occurs when 7. Consider a cone such that the height is 6 inches high and its base has diameter 6 in. Inside this cone we inscribe a cylinder whose base lies on the base of the cone and whose top intersects the cone in a circle. What is the maimum volume of the cylinder? Diagram: r 6 h r h h Variables: 5
6 Let r =radius of cylinder. Let h =height of the cylinder. Let V =volume of inscribed cylinder. Equations: Using similar triangles, for the cross slice of the cone and cylinder, we see r = 6 h which 6 implies that 6r = 8 h = h = 6 r. Then the volume of the cylinder, given by, V = πr h = πr 6 r = 6πr πr must be maimized. The commonsenseboundsdomain of V is {r : r }. Maimize: Net V = πr 6πr. Setting V = we see 6πr r = and solve for r = or r = as critical numbers. Of course r = will not lead to a maimum since no cylinder eists there. Signtesting the critical number r = does indeed yield a maimum for the volume function. V V ր ց MAX Answer: Since r =, then h =, and the maimum volume V = π = 8π. As a result the maimum volume of the cylinder is 8π cubic inches. 8. Consider the right triangle with sides 6, 8 and. Inside this triangle, we inscribe a rectangle such that one corner of the rectangle is the right angle of the triangle and the opposite corner of the rectangle lies on the hypotenuse. What is the maimum area of the rectangle? Diagram: 6 6 y y 8 Variables: Let = coordinate of point on the triangle line. Let y = ycoordinate of point on the triangle line. Let A =area of inscribed rectangle. Equations: Using similar triangles we find the relationship 6 y = 6 which implies that 8 8y = 8 6 = 8y = 8 6 = y = Another way to think about this is that this picture corresponds to the line with yintercept 6 with slope 6 8 If you draw the triangle with width 6 and height 8 instead, the math all still works out the same in the end. The area A = y = = must be maimized. 6
7 The commonsenseboundsdomain of A is { : 8}. Maimize: Net A = 6. Setting A = we solve for = as the critical number. Signtesting the critical number does indeed yield a maimum for the area function. A A ր ց MAX Answer: Finally, = = y = = As a result the maimum area is Area=y = = square units. 9. A toolshed with a square base and a flat roof is to have volume of 8 cubic feet. If the floor costs $6 per square foot, the roof $ per square foot, and the sides $5 per square foot, determine the dimensions of the most economical shed. Diagram: y Variables: Let =length of base of toolshed. Let y =height of the toolshed. Let V =volume of toolshed. Let Cost=cost of amount of material to make toolshed. Equations: We know the volume of the toolshed is given by V = y = 8 is fied, so that y = 8. Then the Cost of materials, which must be minimized, is given as Cost = cost of floor + cost of top + cost of sides = $6 + $ + y$5 = 8 + y 8 = = 8 + The commonsenseboundsdomain of Cost is { : > }. Minimize: Net Cost = 6 6. Setting Cost = we solve =, = =. Signtesting the critical number does indeed yield a maimum for the area function. 7
8 Cost Costց ր MIN Answer: Since = then y = 8 = 8. As a result, the most economical shed has dimensions 8, each in feet.. A rectangular sheet of metal 8 inches wide and inches long is folded along the center to form a triangular trough. Two etra pieces of metal are attached to the ends of the trough. The trough is filled with water. Diagram: a. How deep should the trough be to maimize the capacity of the trough? b. What is the maimum capacity? h Variables: Let =width of endcap of triangular trough. Let y =heightdepth of trough. Let V =volume of trough. Let A =area of endcap of triangular trough. Equations: We will maimize the volume of the trough, by maimizing the area of the metal pieces on the ends of the trough. By the Pythagorean Thrm., h = 6 = 6 Then the area, which must be maimized, is given as A = base height = h = 6 8
9 The commonsenseboundsdomain of A is { : 8}. Maimize: Net, A = 6 = Setting A = implies = so that = 6 which implies = 6 = 6. 6 As a result, = and finally = =. Signtesting the critical number does indeed yield a maimum for the area function. A A ր ց MAX Answer: 6 = Since = maimizes the area, then the corresponding height is h = 8 =. The trough should be inches deep. The maimum capacity is 8 cubic inches because Ma Volume=Ma Area of end panel= = 58 = 8.. A manufacturer wishes to produce rectangular containers with square bottoms and tops, each container having a capacity of 5 cubic inches. The material costs $ per square inch for the sides. If the material used for the top and bottom costs twice as much per square inch as the material for the sides, what dimensions will minimize the cost? Diagram: y Variables: 9
10 Let =length of base of container. Let y =height of container. Let V =volume of container. Let Cost=cost of producing containers. Equations: We know the volume of the toolshed is given by V = y = 5 is fied, so that y = 5. Then the Cost of materials, which must be minimized, is given as Cost = cost of floor + cost of top + cost of sides = $ + $ + y$ = 8 + 8y 5 = = 8 + The commonsenseboundsdomain of Cost is { : > }. Minimize: Net Cost = 6. Setting Cost = we solve = = 5 = = 5. 6 Signtesting the critical number does indeed yield a minimum for the Cost function. Cost Costց 5 ր MIN Answer: Since = 5 then y = 5 =. As a result, the dimension that will minimize the cost are 5 55 in inches.. An outdoor track is to be created in the shape shown and is to have perimeter of yards. Find the dimensions for the track that maimize the area of the rectangular portion of the field enclosed by the track. Diagram: r r Variables: Let r =radius of semicircle ends of track. Let =length of side of rectangular portion of track. Let A =area of rectangular portion of track. Let P =perimeter of track.
11 Equations: We know the perimeter of the track is given by P = + πr = is fied, so that πr = = πr. Then the Area of the rectangular portion of the field, which must be minimized, is given as A = r = r πr = r πr The commonsenseboundsdomain of A is {r : r π }. Maimize: Net, A = πr. Setting A =, we solve for r = π. Signtesting the critical number does indeed yield a maimum for the area function. A π A ր ց MAX Answer: Since r = π, then = π π =. The dimensions that maimize the area of the rectangular portion are = and r = π in yards.. Show that the entire region enclosed by the outdoor track in the previous eample has maimum area if the track is circular. Equations: From the previous problem, we still have = πr. Now the entire area enclosed by the track, which must be maimized, is given by A = πr + r = πr + r πr = πr + r πr = πr + r The commonsenseboundsdomain of A is {r : r π }. Maimize: Net, A = πr. Setting A =, we solve for r = π. Signtesting the critical number does indeed yield a maimum for the area function. A A π ր ց MAX Answer:
12 Since r = π, then = π π =, in yards, resulting in a circular track. Initial Valued Differential Equations. Find a function f that satisfies f = + 5, f = 5 and passes through the point,. Antidifferentiating yields f = C. The initial condition f = 5 yields C = 5 = C = = f = + 5. Antidifferentiating one last time yields f = + 5 +C. The other initial condition with f passing through the point, implies f =. As a result, C = = C = 6 5 = 7. Finally, f = Find a function f that satisfies f = + sin, f = 6 and f =. Antidifferentiating yields f = cos + C. The initial condition f = 6 implies cos + C = 6 = C = 7. Then, f = cos + 7 Antidifferentiating one last time yields f = 6 sin+7+c. The other initial condition with f = implies sin++c = = C =. As a result, f = 6 6. Find a function f that satisfies f =, f = 9 and f = 5. sin Antidifferentiating yields f = 6 + C. Antidifferentiating one last time yields f = + C + C. The first initial condition f = 9 implies C = 9 so f = + C + 9. The second initial condition f = 5 implies 6 + C + 9 = 5 = C = 8 = C = 9. As a result f = Find a function f that satisfies f = + +, f = 8 and f = 5. Antidifferentiating yields f = C. Antidifferentiating one last time yields f = C + C. The first initial condition f = 8 implies C = 8 so f = C +8. The second initial condition f = 5 implies +++C +8 = 5 = C = 7. As a result f = Area and Riemann Sums 8. Evaluate d using Riemann Sums. Here a =, b =, = n = n and i = + i = + i n n.
13 d = lim n n i= f i = lim n = lim n n f i= n i= = lim n n + i n n + i n n n + n i= = lim n n n + n n i= i n n i i= = lim n n n + nn + n = lim + n n + n n n = lim + + n n 9. Evaluate 5 d using Riemann Sums. Here a =, b =, = n = n and i = + i = + = = i n n.
14 5 d = lim n n i= f i = lim n = lim n n f i= i= = lim n n = lim n i n n n i i 5 n n n 8 n n i= n i= i n n i n n i= i n n i i= 8 nn + n + = lim nn + n n 6 n 8 n n + = lim n 6 n n n + n n n n + n = lim n n n n. Use Riemann Sums to estimate endpoints. = 8 = 8 = + d f + f + f + f = f + f + f + f = = = = = 7 + d using equallength subintervals and right Sand is added to a pile at a rate of + t cubic feet per hour for t 8. Compute the Riemann Sum to estimate 8 + t dt using subintervals and the left endpoint of each subinterval. Finally, what two things does this Riemann Sum approimate?
15 8 + t dt ft t + ft t + ft t + ft t = f + f + f + f6 = = 96 = 9 This Riemann sum approimates at least two things: the area under the curve y = +t from t = to t = 8, that is the definite integral. By the Net Change Thrm, it also approimates the net change of sand from time t = to time t = 8. Recall, Amount Sandt = 8 Amount Sandt =.. Compute 8 rate of change dt = d using three different methods: a using Area interpretations of the definite integral, b Fundamental Theorem of Calculus, and c Riemann Sums. } }{{} a Area= base height = = 9 b d = = 8 = + = 9 c Here a =, b =, = n = n and i = + i = + i n n. 5
16 d = lim n n i= f i = lim n = lim n n f i= n i= = lim n n = lim n 9 n + i n n + i n n n i= i n n i i= 9 nn + = lim n n 9 n n + = lim n n n 9 = lim n + n = 9 Differentiation Answer each of the following questions regarding derivatives:. Suppose that e y + y =. Compute dy d Differentiating implicitly d d ey + y = d d yields ey dy d + y epanding and solving for dy d d d d d d d d d 7 sin t dt = sint dt = d d 7 = y yey e y + t dt = sin cos cos t dt = d d 8. Find g if g = 7 sint dt = sin = sin cos t dt = cos = cos 7t + sint dt. 6 + dy d + y. Finally,
17 9.. g = d d sin 7 7t + sin t dt = d d g = 78 9 cos d tan t t dt = tan + d t t sec = tan + tan 7 7t + sin t dt = 7 + sin = 89 sec t t dt tan sec tan = tan + d t t + sint dt = t + sint dt + sin = d cos t + sin = cos cos + sin = cos + + sin. Find d d a. b. d d t dt using two different methods. t dt = = 6 We used the FTC Part I. d t dt = d d d We used FTC Part II.. Find d d a. b. d d d d t t = d d t + sint dt using two different methods. = 6 t + sint dt = d t + sin t dt = + sin We used the FTC Part I. d t + sint dt = d t d cos t = d cos d cos = sin We used FTC Part II.. Find f if ftdt = Differentiating both sides yields:. Find f if ftdt =. Differentiating both sides yields: d d d d d d ftdt = d. That is, f =. d ftdt = d d or ftdt = d d. That is, f = f =. 7
18 5. Differentiate y = sin e + y = sin e e + cos 6. Differentiate y = e cos e y = e cos e + e e cos sin 7. Differentiate y = e e cose y = e e sine e + cose e e e 8. Differentiate y = e y = e e = e + e 9. Differentiate y = + e e 7 y = e7 e + e e 7 7 e 7 = e + e 5 + 7e 7 + 7e 5 e 7 = e + 9e 5 + 7e 7 e 7. Differentiate y = sine cose y = sine sine e +cose cose e = sine sine e + cose cose e. Differentiate y = e e 7 y = 7e e 6 e e = 7e e 6 e + e... Integration Evaluate each of the following integrals: dz = dz = 7 5z 7 5z 5 u = 7 5z Here du = 5dz π du = dz 5 sec θ dθ = tanθ d π = tan π tan = sin π cos π u du = 5 u + C = 5 7 5z + C = = We can solve this one two different ways. We will detail both: First 8
19 d = = = d + + d + + d = = 8 + = = Second, we could use symmetry d [ = d + [ = + ] d ] [ = ] + 9 [ = ] [ = + 8 ] [ ] = = 5. d 9
20 = d + = + d = = 7 6 = 7 = d + + = d = + + d = C = C u + u + 7 = u du u + u 5 + 7u du = u 7 u u + C = u 7 u + C π u d = sin d d + π π = sin d + sin d π π π = cos + cos d = = cos π cos + cosπ cos π = + + = π 6 cos + 6 sin d = 6 Here π u= u= + u du = 6 u u= = = u= = + 6 = = 8 u = + 6 sin { du = 6 cos d = = u = and 6 du = cos d = π 6 = u = + 6 sin π 6 = + 6 =
21 5. π π sin d = u= π u= π sinu du = u= π cos u u= π = cos π + cos π = = Here π sin 6 Here π 8 u = { du = d = π and = u = π du = d = π = u = π cos 6 u= d = 6 u= u du = u = u= u= u = sin 6 du = 6 cos { = = u = 6 d 6du = cos and = π = u = 6 d tan sec d = Here π π 6 Here π 9 6 Here 5 5 u = tan du = sec d du = sec d cos π t dt = t sin t π 6 u = sin t du = cos t dt t du = cos t t dt u= u= and sincos + sin d = u= u= u du = u u= = u= 8 = 8 { = = u = = π 8 = u = tan π = cos t dt = t sin t u= and = π 6 = u = sin = π = u = sin u du = u u du = u u= u= = π π u= u= 6 = sin π 6 = = sin π = = = + = u = + sin { du = 6 sin cos d = = u = and du = sin cos d = π = u = + sin π = + = 5 sin d = why? w cos w + 9 dw = w cos w dw + sinu + 9w + C = sin w + 9w + C u = w Here du = w dw du = w dw 9 dw = cos u du + 9 dw =
22 y 9y siny + 6y 58. dy = y 9 siny + 6y dy = y y + 9 cos y 6y + C cos 59. d = cos u du = sinu + C = sin + C u = Here du = d du = d 6. t sin dt = sinu du = cos u + C = cos t t + C u = t Here du = dt t du = dt t u u du = { w = Here u dw = du u d = w dw = w + C = + C u d + = + + = 5 π = cos sin d π 5π cos sin d + π π = sin + cos + cos sin d = π sin cos d + 5π 5π π + sin + cos + cos sin d = sin π + cos π sin + cos + cos 5π sin 5π cos π sin π + sinπ + cos π sin 5π + cos 5π = = = t 5 + t 6 + t dt = ON EXAM # Here π 5π = + u du = ln u + C = ln t6 + t + C NOTE: THIS WILL NOT BE u = t 6 + t du = 6t 5 + dt du = t5 + dt + d = u u du = u 5 u du = 6 u6 5 u5 +C = C
23 66. { u = + = = u Here du = d 7 cos5 5 sin7 d = sin cos7 + C Here u = 5 du = 5d and 5 du = d 7 cos u du 5 7 w = 7 dw = 7d dw = d 67. d = u du = 6 6 u = Here du = 6d 6du = d 68. d = u C = C u = = = u Here du = d du = d u C = d = u9 7 + Here 8 u u = = = u + du = d du = d 7 + d = sinw dw = 7 5 sinu cos w + C = u + C = 9 u + C = 9 + C u du = 9 u 5 7 du = C d = C e d = e u du = eu + C = e + C u = Here du = d du = d u u du = u 9 5 u5 + u u 7 du = u9 7 + u 7 +C = 7. e d = 6 u= 7 u= e u du = 6 eu u= 7 u= = 6 e 7 e = 6e 6e 7
24 Here 6 u = du = 6d and du = d { = = u = = = u = 7 e 7 d = e u du = 7 7 eu + C = 7 e + C { u = Here du = d e e d = u du = u + C = e + C { u = e Here du = e d e 7 d = e 7 7 Here u = e 7 du = 7e 7 d 7 du = e7 d du = 6 6 u 7 u + C = u + C = e C e e e d = e u du = eu + C = ee + C u = e Here du = e d du = e d Area between Curves 77. Compute the area bounded by y = and y =. You can solve this problem two different ways. I will detail both. First we will use symmetry to integrate one side and then double the value to capture the entire area. Note that y = and y = intersect when = = = or when = or = ±.
25 [ ] Area = top bottom d [ ] = d [ ] = = [8 ] = 8 If you don t use symmetry we see that Area = = top bottom d + d + = + d = = + = 8 top bottom d 78. Compute the area bounded by y = and y = 8. You can solve this problem two different ways. I will detail both. First we will use symmetry to integrate one side and then double the value to capture the entire area. Note that y = and y = 8 intersect for > when = 8 = + 8 = = + = 5
26 or when = or =. We will ignore = here since we are considering the side with >. [ ] Area = top bottom d [ ] = 8 d [ ] = + 8 d [ ] = + 8 = [ 8 ] + 6 [ = 8 ] [ 6 = 8 ] = [ ] 8 = 56 If you don t use symmetry we see that 6
27 Area = top bottom d + = = 8 d d + top bottom d 8 d + 8 d = = = = 6 + = = Compute the area bounded by y =, y = +, =, and =. Note that these two curves intersect when = + = + = = + = = = or =. 7
28 Area = = = top bottom d + top bottom d + d + + d + + d = + = + d = = = = 6 Position, Velocity, Acceleration 8. A ball is thrown upward with a speed of 8 ft/sec from the edge of a cliff ft above the ground. Find its height above the ground t seconds later. When does it reach its maimum height? When does it hit the ground? Note v = 8 ft sec, s = ft. at = vt = t + v = vt = t + 8 st = 6t + 8t + s = st = 6t + 8t + Ma height occurs when vt =. That is, t + 8 = or when t = seconds. The ball hits the ground when st = 6t + 8t + = or when 6t 8t 9 = so that 6t 9t + =. Then t = 9 or t =. We will ignore the negative time here. The ball hits the ground after 9 seconds. 8. The skid marks made by an automobile indicate that its brakes were fully applied for a distance of 9 ft before it came to a stop. Suppose that it is known that the car in question has a constant deceleration of ft/sec under the conditions of the skid. Suppose also that 8
29 the car was travelling at 6 ft/sec when the brakes were first applied. How long did it take for the car to come to a complete stop? Note v = 6 ft sec, s = ft. s = s stop = 9 v = 6 t stop =? at = vt = t + v = vt = t + 6 st = t + 6t + s = st = t + 8t + You can solve this two ways. First, the car stops when vt =. Set vt = t + 6 = and solve for the stopping time. That is, t = seconds. Second, the car stops when st = 9. Set st = t + 6t = 9 and solve for the stopping time. We see that t 6t + 9 = implies t t = or the car stops when t = seconds. 8. Suppose that a bolt was fired vertically upward from a powerful crossbow at ground level, and that it struck the ground 8 seconds later. If air resistance may be neglected, find the initial velocity of the bolt and the maimum height it reached. Note v =? ft sec, s = ft, t impact = 8sec. at = vt = t + v st = 6t + v t + s = st = 6t + v t + = st = 6t + v t First we use the impact information, s8 =. s8 = 68 + v 8 = = 8v = 68 = v = 68 8 = 68 = 768 ft sec, As a result vt = t and the ma height occurs when vt = = t = 768 = seconds. Finally the ma height is s = = = = 96 feet. 8. Jack throws a baseball straight downward from the top of a tall building. The initial speed of the ball is 5 feet per second. It hits the ground with a speed of 5 feet per second. How tall is the building? Note v = 5 ft sec, s =?ft, v impact = 5 ft sec? 9
30 at = vt = t + v = vt = t 5 st = 6t 5t + s The ball hits the ground when vt = t 5 = 5 or when t = 5 5 = 8 which is when t impact = seconds. Finally, we solve s = for s. The ball hits the ground when s = or when 56 + s = which is when s = 56 feet. As a result, the building is 56 feet tall. 8. A ball is dropped from the top of the building 576 feet high. With what velocity should a second ball be thrown straight downward seconds later so that the two balls hit the ground simultaneously? Note v = ft sec, s = 576ft. 576 Ball has the following motion equations: at = vt = t + v = vt = t st = 6t + s = st = 6t The first ball hits the ground when st = 6t +576 = or when 6t = 576 which is when t = or when t impact = 6 seconds. For the second ball to be thrown seconds later and hit the ground at the same time as the first ball, the second ball must travel just seconds before hitting the ground. Ball has the following motion equations: at = vt = t + v st = 6t + v t + s = st = 6t + v t We solve s = to find the second ball s initial velocity. Set 6 + v = = v = = = v =. Finally, the second ball must have an initial velocity of feet per second, or we can say the second ball must be thrown straight down with a speed of feet per second. 85. A particle starts from rest at the point = and moves along the ais with acceleration function at = t. Find its resulting position function. Note s =, v =. at = t vt = 6t + v = vt = 6t st = t + s = st = t + As a result, the resulting position function is given as st = t The skid marks made by an automobile indicate that its brakes were fully applied for a distance of 6 ft before it came to a stop. Suppose that it is known that the car in question
31 has a constant deceleration of ft/sec under the conditions of the skid. How fast was the car travelling when its brakes were first applied? Note v =? ft sec, s = ft, s stop = 6ft. s = s stop = 6 v =? v stop = at = vt = t + v st = t + v t + s = st = t + v t + You can solve this in two parts. First, the car stops when vt =. Set vt = t + v = and solve for the stopping time in terms of the initial velocity v. That is, t stop = v seconds. v Second, the car stops when st stop = 6 or s = 6. Set and solve for the unknown initial velocity. We see that v v + v = 6 v + v v = 6 v + v v + v v = 6 = 6 = 6 v = 6 v = 6 = 6 v = 8 Finally, the initial velocity v = 8 feet per second. Displacement Total Distance Net Change 87. Suppose that the velocity of a moving particle is vt = t t + feet per second. Find both the displacement and total distance it travels between time t = and t = seconds. Displacement= t t + dt = t t + t = + = 55 + = = 9 = 7 The displacement in question is 7. Note that if you sketch the parabola y = t t +, it passes below the ais between = and = 8.
32 Total Distance= t t + dt 8 = t t + dt + t t + dt + t t + dt 8 = t t + t t t t + t t + t 8 = = = 7 98 = = 6 The total distance in question is Supppose that water is pumped into an initially empty tank. The rate of water flow into the tank at time t in seconds is 5 t liters per second. How much water flows into the tank during the first seconds? Using the Net Change Theorem, the amount of water in the tank between time t = and time t = is W W = rate of water flow dt. Using the FTC, we simply integrate 5 t dt = 5t t = 5 9 = 5 5 = 5. The amount of water that flows into the tank during the first seconds is 5 liters. Curve Sketching 89. Use curve sketching techniques to present a detailed sketch for f = e. Domain: f has domain,. Symmetry: f is an even function since f = f = symmetry about yais. Vertical asymptotes:none. Horizontal asymptotes: There are horizontal asymptotes for this f at y = since lim f = and lim f =. First Derivative Information: Here f = e. The critical points occur where f is undefined never here or zero =. Recall that the eponential is never zero. As a result, = is the critical number. Using sign testing/analysis for f, f f ր ց local ma
33 Therefore, f is increasing on, and decreasing on,. There is a local ma at the point,. Second Derivative Information Net, f = e + e = e. Possible inflection points occur when f is undefined never here or zero = ±. Using sign testing/analysis for f, f f infl. infl. point point Therefore, f is concave up on, and,, whereas f is concave down on,. There are inflection points at ±, e. Piece the first and second derivative information together f ր ր ց ց f infl. pt. local ma infl. pt. Sketch:
34 9. Use curve sketching techniques to present a detailed sketch for f = + e. Domain: f has domain,. Vertical asymptotes:none. Horizontal asymptotes: There is horizontal asymptote for this f at y = since lim f = will see why in Math and lim f =. First Derivative Information: Here f = + e + e + = e = e. The critical points occur where f is undefined never here or zero = and =. Recall that the eponential is never zero. As a result, = is the critical number. Using sign testing/analysis for f, f f ր ց ր local local ma min Therefore, f is increasing on, and,, and decreasing on,. There is a local ma at the point, e and local min at the point, e. Second Derivative Information Net, f = e e = e + 7. Possible inflection points occur when f is undefined never here or zero = 7± 5. Using sign testing/analysis for f, f f infl. infl. point point Therefore, f is concave down on, 7 5 and 7+ 5,, whereas f is concave down on 7 5, There are inflection points at, e 7 = 7 5, and, e 7+ = 7+ 5, 6 5. Piece the first and second derivative information together f ր ց ց ր ր f local ma infl. local pt. min infl. pt.
35 Sketch: 5
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1949_07_ch07_p561599.qd 7/5/06 1:39 PM Page 561 7 Polynomial and Rational Functions 7.1 Polynomial Functions 7. Graphing Polynomial Functions 7.3 Comple Numbers 7.4 Graphing Rational Functions 7.5 Equations
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