Review Problems for Test 3

Transcription

1 Review Problems for Test 3 Math 6-3/6 7 These problems are provided to help you study. The presence of a problem on this handout does not imply that there will be a similar problem on the test. And the absence of a topic does not imply that it won t appear on the test. (. Compute ) d.. Compute ( + ) d. 3. Compute 4. Compute 5. Compute 6. Compute (cos) d. ( ) 7 + d. ( 4 + 5) 4 d. ( + ) 3 d. 7. Compute 8. Compute d. ( + ) /5 e e + d Compute d.. Compute (csc) 5 cotd.. Compute sin (cos + ) d.. Compute ( + ) 3 + d. 3. Compute tan5 d. 4. Compute csc(7 + ) cot(7 + ) d. (3 5. Compute ) d. 6. Compute ( )( + 3) d.

2 ( ) 7. Compute d. 8. Compute (7 5) 43 d. e 3 + e 3 9. Compute e 3 d. e 3 ((sin. Compute 5) 4 6 sin 5 + ) cos5 d. f () lnf(). Compute d. f(). Compute (f()g () + g()f ()) d. 3. Compute (3 + )( 4) d. 4. Compute 5. Compute 6. Compute 7. Compute 8. Compute ( + ) + 3 d d. 7 d. e5 3 sec d. 5 e d. 9. Compute e e (e) d. ((cos7) 3. Compute (sin 7) ) d. 3. Compute 3. Compute (sec ) 3 d. (e 3 e )(4e 5 + e 3 ) d. ( 33. Compute cos( + 4) 5 sin( 3 + ) ) d. 34. (a) Given that d + 5 (5 ) d = (, what is 4 + 5) ( d ) (b) Compute 4 + d + d. (5 ) ( 4 + 5) d? 35. Find functions f() and g() such that the antiderivative of f() g() is not equal to the antiderivative of f() times the antiderivative of g().

3 36. (a) Compute ( ) d. (b) Compute the eact value of 37. Approimate d. sin d using 4 rectangles of equal width and using the midpoint of each subinterval to obtain the rectangles heights. 38. Compute 4 ( + 3) d by writing the integral as the limit of a rectangle sum. 39. The position of a bowl of potato salad at time t is s(t) = t 3 3t. (a) Find the velocity v(t) and the acceleration a(t). (b) When is the velocity equal to? When is the acceleration equal to? 4. (a) Find the total area of the region bounded by y = ( )( 4) and the -ais. (b) Find the area of the region in the first quadrant bounded on the left by y =, on the right by y =, and below by the -ais. (c) Find the area of the region bounded by the graphs of y = and y =. 4. (a) Compute d cos t d 7 t 4 + dt. (b) Compute d d 4. (a) Prove that sin 4.5 t + dt. d.5. + (b) Use the Integral Mean Value Theorem to estimate 3 e 4 d.

4 43. Some values for a function y = f() are shown below. (a) Approimate the rectangle heights. (b) Approimate the rectangle heights. f() f() d using 5 rectangles of equal width and using the left-hand endpoints to obtain f() d using rectangles of equal width and using the right-hand endpoints to obtain 44. Find the point(s) on y = closest to the point (, ). 45. Calvin Butterball wants to fence in two equal-size rectangular pens in his yard for his pet fish. (Calvin does not have much luck with pets, for some reason.) As shown in the picture below, one side of each pen will be bounded by an eisting stone wall (and will therefore not require any fence). wall y pen pen y If Calvin has 3 feet of fence, what should the dimensions of the pens be to maimize the total area? 46. A rectangular bo with a bottom and a top consists of two identical partitions which share a common wall. Each partition has a square bottom. If the total volume of the bo (i.e the sum of the volumes of the two partitions) is 67 cubic inches, what dimensions give the bo which has the smallest total surface area? 47. A cylindrical can with a top and a bottom is to have a volume of 8π cubic inches. The material for the top and bottom costs cents per square inch, while the material for the sides costs 3 cents per square inch. What dimensions yield a can which costs the least? 48. Calvin Butterball sits in his rowboat 9 miles from a long straight shore. Phoebe Small waits in a car at a point on shore 5 miles from the point on the shore closest to Calvin. Calvin rows to a point on the shore, 4

5 then runs down the shore to the car. Then they drive to the shopping mall, where they purchase two rolls of duct tape,a tub of margarine, a copy of Doom 3, a lovely pink tank top, two cans of Jolt Cola, a copy of Mother Earth News, a bowling ball, a metric he key set, an ionic air purifier, three Cinnabons, leather pants, a mm case fan, curly fries, and a bazooka. If Calvin can row at 4 miles per hour and run at 5 miles per hour, at what point on shore should he land in order to minimize his total travel time? 49. Use a calculating device to approimate to at least three decimal places. 5. Write the series in summation form: (a) (b) 3 sin sin sin sin(99 ). 5. (a) Epress the following sum in terms of n: (b) Find the eact value of (n + 5n 7). 5. Use the fact that to evaluate. Compute k= k k. n= n ( 5k + 7k ). k= k k = k k Solutions to the Review Problems for Test 3 ( ) d. ( ) d = C.. Compute ( + ) d. ( + ) d = ( + + ) d = C. 3. Compute (cos) d. (cos) d = (sec) d = tan + C. 5

6 4. Compute ( ) 7 + d. ( ) 7 + d = u 7 du = 8 u8 + C = 8 u = + du = ] d, d = du ( ) C. 5. Compute ( 4 + 5) 4 d. ( 4 + 5) 4 d = u 4 du = 6 u 3 + C = 6 ( 4 + 5) 3 + C. ] u = du du = ( 4) d = ( ) d, d = ( ) 6. Compute ( + ) 3 d. d = ( + ) 3 u 3 du = u + C = ( + ) + C. u = + du = ] d, d = du 7. Compute ( + ) /5 d. u d = du = ( + ) /5 u/5 u = + du = d] (u 4/5 u /5 ) du = 5 9 u9/5 5 4 u4/5 + C = 5 9 ( + )9/5 5 4 ( + )4/5 + C. 8. Compute e e + d. e e + d = du u = ln u + C = ln e + + C. u = e +, du = e d, d = du ] e 6

7 9. Compute d d = du 4 u = 4 ln u + C = 4 ln C. u = 4 + 4, du = 4 3 d, d = du ] 4 3. Compute (csc) 5 cotd. (csc) 5 cotd = u 4 du = 5 u5 + C = 5 (csc )5 + C.. Compute ] du u = csc du = csc cotd, d = csc cot sin (cos + ) d. sin sin (cos + ) d = u du du sin = u = u + C = cos + + C. u = cos +, du = sin d, d = du ] sin. Compute ( + ) 3 + d. ( + ) 3 ((u + d ) + ) 3 udu = (u u + )u /3 du = (u 7/3 u 4/3 + u /3 ) du = u = +, du = d; = u ] 3 u/3 6 7 u7/3 + 3 u4/3 + C = 3 ( + )/3 6 7 ( + )7/3 + 3 ( + )4/3 + C. 3. Compute tan5 d. tan 5 d = sin 5 sin 5 cos5 d = u du du 5 sin 5 = 5 u = cos5, du = 5 sin 5 d, d = u = 5 ln u + C = ln cos5 + C. 5 ] du 5 sin 5 4. Compute csc(7 + ) cot(7 + ) d. csc(7+) cot(7+) d = csc u cotu du 7 = 7 csc u cotudu = 7 csc u+c = 7 csc(7+)+c. 7

8 u = 7 +, du = 7 d, d = du ] 7 (3 5. Compute ) d. ( ) d = ln (3 7 7 ) + C. 6. Compute ( )( + 3) d. ( )( + 3) d = ( 4 + 3) d = C. ( ) 7. Compute d. ( ) d = d = ( ) d = 4 + ln + C. 8. Compute (7 5) 43 d. (7 5) 43 d = u 43 du = 7 38 u44 + C = 38 (7 5)44 + C. u = 7 5 du = 7 d, d = du ] 7 e 3 + e 3 9. Compute e 3 d. e 3 e 3 + e 3 e 3 + e 3 du e 3 d = e 3 u 3(e 3 + e 3 ) d = du 3 u = 3 ln u + C = 3 ln e3 e 3 + C. ] u = e 3 e 3, du = (3e 3 + 3e 3 ) d = 3(e 3 + e 3 du ) d, d = 3(e 3 + e 3 ) d ((sin. Compute 5) 4 6 sin 5 + ) cos5 d. ((sin 5) 4 6 sin 5 + ) (u cos5 d = 4 6u + ) du (cos5) 5 cos5 = (u 4 6u + ) du = 5 u = sin 5, du = 5 cos5 d, d = du ] 5 cos5 8

9 ( ) 5 5 u5 3u + u + C = ( ) 5 5 (sin 5)5 3(sin 5) + (sin 5) + C. f () lnf(). Compute d. f() f () lnf() d = f() f () u f() u = lnf(), f() f () du = du = f () f() d, u du = u + C = (lnf()) + C. ] f() d = f () du. Compute (f()g () + g()f ()) d. (f()g () + g()f ()) d = f()g() + C, because the Product Rule says that d d f()g() = f()g () + g()f (). 3. Compute (3 + )( 4) d. (3 + )( 4) d = (3(u + 4) + )u du = (3u + )u du = (3u + u ) du = u = 4, du = d, = u + 4] 3 u + u + C = 3 ( 4) + ( 4) + C. 4. Compute ( + ) + 3 d. ( + ) ((u 3) + ) (u ) u 4u + 4 d = du = du = du = + 3 u u u u = + 3, du = d, = u 3] ( u 3/ 4u / + 4u /) du = 5 u5/ 8 3 u3/ +8u / +C = 5 (+3)5/ 8 3 (+3)3/ +8(+3) / +C. 5. Compute d d = du u ( 3) = du = u u + C = C. 9

10 ] u = du 6 + 5, du = ( 6) d = ( 3) d, d = ( 3) 6. Compute 7 d. e5 7 d = e5 7e 5 d = 7 5 e 5 + C. 7. Compute 3 sec d. 3 sec d = 3 cosd = 3 sin + C. 8. Compute 5 e d. 5 e d = 5 e e 4 d = 5 e 4 e 4 e 4 d = 5 + 7e 4 u ] u = + 7e 4, du = 8e 4 du d, d = 8e ln u + C = 5 8 ln + 7e 4 + C. du 8e 4 = 5 du 8 u = 9. Compute e e (e) d. e e (e) d = e e u du = e u = e, du = e d, d = du ] e e u du = eu + C = e(e) + C. ((cos7) 3. Compute (sin 7) ) d. ((cos7) (sin 7) ) d = cos4 d = In the first step, I used the double angle formula cos u du 4 = 4 u = 4, du = 4 d, d = du ] 4 (cosθ) (sin θ) = cosθ. cos u du = 4 sin u+c = sin 4+C. 4

11 3. Compute (sec ) 3 d. (sec ) 3 d = (sec u) 3 u = ) ( 3 = (secu) du = tanu + C = tan + C. ], du = d, d = 3 3 du 3. Compute (e 3 e )(4e 5 + e 3 ) d. I multiply the two terms out, using the rule e a e b = e a+b : (e 3 e )(4e 5 + e 3 ) d = (4e 8 + e 6 8e 4 e ) d = e8 + 6 e6 e 4 e + C. ( 33. Compute cos( + 4) 5 sin( 3 + ) ) d. ( cos( + 4) 5 sin( 3 + ) ) d = cos( + 4) d 5 sin( 3 + ) d = u = + 4, du = d, d = du ; w = 3 +, dw = 3 d, d = dw ] 3 cos u du 5 sin w dw 3 = cos u du 5 sin w dw = 3 sin u cos w + C = sin( + 4) cos(3 + ) + C. 34. (a) Given that d + 5 d (5 ) = (, what is 4 + 5) (5 ) ( 4 + 5) d = C. ( d ) (b) Compute 4 + d + d. (5 ) ( 4 + 5) d? ( d ) 4 + d + d = C. 35. Find functions f() and g() such that the antiderivative of f() g() is not equal to the antiderivative of f() times the antiderivative of g().

12 There are lots of possibilities. For eample, take f() = and g() =. Then f()g() d = d = + C, but Obviously,. f() d = d = + C and g() d = d = + C. 36. (a) Compute ( ) d. Here is the graph of y = ( ) d is the area under the graph and above the -ais, from = to =. This is the shaded region in the picture. It is a triangle with height and base, so its area is (b) Compute the eact value of 7 ( ) d = = d. You can t compute this integral directly using techniques you know now. If y = 7 + 6, then y = 7 + 6, 6 + y = 7, y = 7 + 9, ( 3) + y = 6. (I knew to add 9 to both sides, since ( 6) = 3, then ( 3) = 9. You should have seen this when you took algebra; it s called completing the square.) The equation represents a circle of radius 6 = 4 centered at (3, ). If I go 4 units to the left and right of = 3, I get =, and = 7, which are the limits on the integral. Finally, y = represents the top semicircle of the circle, because always gives the positive square root

13 Putting everything together, I see that the integral represents the area of a semicircle of radius 4. Since the area of a circle of radius r is πr, d = π 4 = 8π. 37. Approimate sin d using 4 rectangles of equal width and using the midpoint of each subinterval to obtain the rectangles heights., ] is broken up into 4 equal subintervals, each of width.5: I m using the midpoint of each subinterval for the rectangle heights. The midpoints are.5,.375,.65,.875. I compute the value of f() = sin at each point, add up the f-values, then multiply the total by the width.5. f() The approimate value of the integral is = The actual value of the integral is Compute 4 ( + 3) d by writing the integral as the limit of a rectangle sum. The width of a typical rectangle is = 4 n I ll use the right-hand endpoints of the subintervals. (You could use the left-hand endpoints or the midpoints; the computation would look different, but the final answer would be the same.) = n. I m going from to 4 in steps of size. The diagram shows that right-hand endpoints: n + n + 4 n + 6 (n ) + + n n n n = 4 The function is f() = + 3. The rectangle sum is ( ( f + ) ( + f + 4 ) ( + f + 6 ) ( + + f + n )) n n n n 3 = n f i= ( + i ) = n

14 n n i= ( ( + i ) ) + 3 = n n n i= ( ) 4i n + 7 = 8 n n i + n i= n 7 = i= Hence, 4 8 n(n + ) + 4(n + ) 7n = + 4. n n n ( 4(n + ) ( + 3) d = lim n n ) + 4 = = The position of a bowl of potato salad at time t is (a) Find the velocity v(t) and the acceleration a(t). s(t) = t 3 3t. v(t) = s (t) = 6t 6t = 6t(t ), a(t) = v (t) = t 6 = (t 5). (b) When is the velocity equal to? When is the acceleration equal to? The velocity is at t = and at t =. The acceleration is at t = (a) Find the total area of the region bounded by y = ( )( 4) and the -ais The curve is positive from to and negative from to 4, so the area is ( )( 4) d + 4 ( )( 4) d. ( )( 4) = , so ( )( 4) = ( ) d = C. Using this in the integrals above, I find that the area is ( )( 4) d ( )( 4) d = 8.

15 (b) Find the area of the region in the first quadrant bounded on the left by y =, on the right by y =, and below by the -ais. y = - y = - / I ll find the area using horizontal rectangles; vertical rectangles would require two integrals. The left-hand curve is y = or = y. The right-hand curve is y = The region etends from y = to y =. The area is ( ) y y dy = ( ) or = y. y dy = 3 ( ).764. (c) Find the area of the region bounded by the graphs of y = and y = Solve simultaneously: =, =, ( )( + ) =, = or =. The top curve is y = and the bottom curve is y =. The area is ( ( ) ( ) ) d = 9 = (a) Compute d cos t d 7 t 4 + dt. 5

16 Using the second form of the Fundamental Theorem of Calculus, I get d cos t cos d 7 t 4 dt = (b) Compute d d sin 4 t + dt. In this situation, I m differentiating with respect to, but the top limit in the integral is sin they don t match. In order to apply the second form of the Fundamental Theorem, I need to use the Chain Rule to make them match : d d sin 4 ( ) d(sin ) d t + dt = d d(sin ) sin 4 t + dt = (cos) (sin ) (a) Prove that.5 d.5. + Since +, it follows that +. Hence,.5 d.5.5 d, or d. (b) Use the Integral Mean Value Theorem to estimate e 4 d. If f() = e 4, then f () = 4 3 e 4. f () = for =, and f () is defined for all. f() e The maimum value of f() on is e and the minimum value is. Thus, e 4 e. The Integral Mean Value Theorem says that e 4 d = ( )f(c) = e c4, where c. The ma-min result I derived above shows that e c4 e. Hence, e 4 d e. 6

17 43. Some values for a function y = f() are shown below. (a) Approimate the rectangle heights. = 5 (b) Approimate f() f() d using 5 rectangles of equal width and using the left-hand endpoints to obtain =.4, so the approimation is the rectangle heights. = (.4)( ) = f() d using rectangles of equal width and using the right-hand endpoints to obtain =., so the approimation is (.)( ) = Find the point(s) on y = closest to the point (, ). Draw a picture. Label the things that are relevant to the problem. y (, ) (, ) Write down an epression for the thing you re trying to maimize or minimize. 7

18 The distance from (, ) to (, ) is d = ( ) + ( ) = ( ) +. A distance is smallest eactly when its square is smallest. So we can work with the square of the distance instead: D = ( ) +. This has the advantage of removing the square root and making it easier to differentiate. Look at the etreme cases to determine any endpoints. can be as small as, but it can be arbitrarily large. That is, <. Therefore, dd d = ( ) + = for =. dd d = ( ) +, d D d =. D 3 4 The minimum does not occur at =. To show that = is an absolute min, notice that d D d = >. Therefore, = is a local min. It is the only critical point, so it( is an absolute ) min. The closest point to (, ) on the curve y = is the point,. 45. Calvin Butterball wants to fence in two equal-size rectangular pens in his yard for his pet fish. (Calvin does not have much luck with pets, for some reason.) As shown in the picture below, one side of each pen will be bounded by an eisting stone wall (and will therefore not require any fence). wall y pen pen y If Calvin has 3 feet of fence, what should the dimensions of the pens be to maimize the total area? The area is A = y. The amount of fence is 3 = + 3y, so = 3 3y. Therefore, A = (3 3y)y = 3y 3y. The endpoints are y = and y = (i.e. y ). The derivative is da = 3 6y. dy 8

19 da dy = for y = 5. y 5 A 75 y = 5 gives the absolute ma; = (3 3y) = A rectangular bo with a bottom and a top consists of two identical partitions which share a common wall. Each partition has a square bottom. If the total volume of the bo (i.e the sum of the volumes of the two partitions) is 67 cubic inches, what dimensions give the bo which has the smallest total surface area? y Suppose the square base of a partition has sides of length, and let y be the height of the bo. The total surface area is A = (bottom and top) + (back and front) + (sides and middle) = 4 + 4y + 3y = 4 + 7y. The volume is Plug this into A and simplify: 67 = y, so y = 67 = 336. A = = The only restriction on is >. The derivatives are A = 8 95, A = A is defined for all >. Set A = and solve: The corresponding value for y is y = =, 8 = 95, 3 = 744, = 4. A (4) = = 4 >. Hence, = 4 is a local min. Since it s the only critical point, it s an absolute min. 9

20 47. A cylindrical can with a top and a bottom is to have a volume of 8π cubic inches. The material for the top and bottom costs cents per square inch, while the material for the sides costs 3 cents per square inch. What dimensions yield a can which costs the least? r h Let r be the radius of the can, and let h be the height. The total cost is the cost of the sides plus the cost of the top and bottom: The volume is 8π, so Plug h = 8 into C and simplify: r C = (3)(πrh) + ()(πr ) = 6πrh + πr. 8π = πr h, and h = 8 r. C = 6πr 8 r + πr = 8π + πr. r The only restriction on r is that r >. Since I don t have two endpoints, I ll use the Second Derivative Test. Differentiate: C = 8π r + 4πr, C = 6π r 3 + 4π. Find the critical points: This gives h = 8 9 C = 8π r + 4πr =, 4πr = 8π r, r 3 = 7, r = 3. =. Now C (3) = 6π 7 + 4π = π >. Hence, r = 3 is a local min. Since it s the only critical point, it must give an absolute min. 48. Calvin Butterball sits in his rowboat 9 miles from a long straight shore. Phoebe Small waits in a car at a point on shore 5 miles from the point on the shore closest to Calvin. Calvin rows to a point on the shore, then runs down the shore to the car. Then they drive to the shopping mall, where they purchase two rolls of duct tape,a tub of margarine, a copy of Doom 3, a lovely pink tank top, two cans of Jolt Cola, a copy of Mother Earth News, a bowling ball, a metric he key set, an ionic air purifier, three Cinnabons, leather pants, a mm case fan, curly fries, and a bazooka.

21 If Calvin can row at 4 miles per hour and run at 5 miles per hour, at what point on shore should he land in order to minimize his total travel time? Phoebe 9 X + 8 Calvin Let be the distance from the point on shore closest to Calvin to the point where he lands. The distance that he rows is + 8, and the distance that he runs is 5. Since the time elapsed is equal to the distance divided by the speed, his total travel time is T = (The first term is his rowing distance divided by his rowing speed, and the second term is his running distance divided by his running speed.) The endpoints are = (where he rows to the nearest point on the shore) and = 5 (where he rows directly to Phoebe). Differentiate: T = 4 ( + 8) / () 5 = Find the critical points by setting T = : = = 5 5 = = 6( + 8) 5 = = 96 = 44 = (I omitted = because it isn t in the interval 5.) Test the critical point and the end points: 5 T He should row to a point miles from the point closest to shore to minimize his travel time.

22 49. Use a calculating device to approimate to at least three decimal places = n 3 + n n= 5. Write the series in summation form: (a) = (b) 3 sin sin sin sin(99 ). n= n + 5 n. 3 sin sin sin sin(99 ) = n sin(n ). n=3 5. (a) Epress the following sum in terms of n: Use the formulas Therefore, n ( 5k + 7k ) = k= k= n ( 5k + 7k ). k= n c = nc k= n k = k= n k = k= k= (b) Find the eact value of (n + 5n 7). (n +5n 7) = n= n= n= n +5 n(n + ) n(n + )(n + ) 6 n n n 5 k + 7 k = n n= n n= k= 7 = ()()() 6. 5n(n + ) + 7n(n + )(n + ) ()() 7 =

23 5. Use the fact that to evaluate k= k k. ( ) ( + ) + 3 k= k k = k k k k = ( 3 4 k= ) + + ( k ) = k ( 999 ) = = 999. The earth keeps some vibrations going There in your heart, and that is you. - Edgar Lee Masters c 8 by Bruce Ikenaga 3