Quiz 6 Practice Problems

Transcription

1 Quiz 6 Practice Problems Practice problems are similar, both in difficulty and in scope, to the type of problems you will see on the quiz. Problems marked with a are for your entertainment and are not essential. SUGGESTED REFERENCE MATERIAL: As you work through the problems listed below, you should reference sections 6.6 and 7. of the recommended textbook (or the equivalent section in your alternative textbook/online resource) and your lecture notes. EXPECTED SKILLS: Be able to use integration to compute the work done by a variable force in moving an object along a straight path from x = a to x = b. Specifically, be able to solve spring problems, lifting problems, and pumping problems. Be able to use integration by parts to evaluate various integrals, including integrands involving products of functions, isolated logarithmic functions, or isolated inverse trigonometric functions.

2 PRACTICE PROBLEMS:. A variable force F (x) is applied in the positive x direction, as shown in the graph below. Find the work done by the force in moving a particle from x = to x = 9. Spring Problems. A force of lb is required to hold a spring stretched 5 in beyond its natural length. How much work is done in stretching the spring from its natural length to 7 in beyond its natural length. 3. A spring has a natural length of ft. A body of lb hanging on the spring stretches it to a total length of.4 ft. (a) Find the spring constant (in pounds per foot). (b) How far beyond its natural length would a body of 5 lb stretch the spring? (c) How much work is required to stretch the spring from its natural length to a length of 3 ft?

3 4. If the work required to stretch a spring foot beyond its natural length is 5 ft-lb, how much work is needed to stretch it 8 inches beyond its natural length? Lifting Problems 5. A cable that weighs 3 lb/ft is used to lift, lb of iron ore up a mineshaft which is 45 ft deep. Find the work done. 6. A 5 lb bucket containing lb of water is hanging at the end of a 3 ft rope which weighs / lb/ft. The other end of the rope is attached to a pulley. Assume that the rope is wound onto the pulley at a rate of 3 ft/s, causing the bucket to be lifted. Find the work done in winding the rope onto the pulley. 7. Repeat problem 6 adding the assumption that water leaks out of the bucket at a rate of 4 lb/s. 8. A bag of sand originally weighing 44 lb is lifted at a constant rate. As it rises, sand also leaks out at a constant rate. At the instant when the bag has been lifted to a height of 8 feet, exactly half of the original amount of sand remains. (Neglect the weight of the bag and the lifting equipment.) (a) Describe the weight of the sand as a function of x, where x is the height of the bag above the ground. (b) Calculate the work done in lifting the sand to the height of 8 ft from the ground. 3

4 Pumping Problems 9. A circular swimming pool has a diameter of meters. The sides are 4 meters high. And, the depth of the water is 3.5 meters. How much work is required to pump all of the water over the side? Recall that water weighs 98 N/m 3.. At Charlie s Chocolate Factory, a tank in the shape of an inverted circular cone having a height of meters and a radius (at the top) of 6 meters is filled with chocolate pudding to a height of meters. In order to sterilize the tank, the factory needs to empty the tank. Find the work required to empty the tank by pumping the chocolate pudding through a hole in the top of the tank. Note: the weight density of chocolate pudding is,78 N/m 3.. The trough pictured below is 5 feet long and 4 feet wide at the top. The ends of the trough are isosceles triangles with a height of 3 feet. If the trough is filled to a height of foot with water, find the work required to pump all of the water over the side. Recall that water weighs 6.4 lb/ft 3. 4

5 . The tank shown below is half full of oil which has a weight-density of 96 N/m 3. Let x = correspond to the bottom of the tank. Set up an integral which represents the work (in joules) required to pump the oil out of the outlet at the top which is one meter above the top of the tank. Do not evaluate your integral. For problems 3-, evaluate the given integral. 3. xe 4x dx 4. x cos x dx 5. x ln x dx ln x dx x 4 sin x dx x sec x dx e x cos 3x dx 5

6 . x 3 cos(x ) dx.. π 3x sin x dx x e x dx 3. Suppose that u and v are differentiable functions of x with and the following functional values. v du = 3 x u(x) v(x) Use this information to compute u dv. 4. Evaluate sin x dx by first making an appropriate substitution and then applying integration by parts. 5. Evaluate (sin x) dx. 6. Find the area of the region which is enclosed by y = ln x, y =, and x = e. 6

7 7. Let R be the region enclosed by the graphs y = ln x, x = e, and the x-axis (as shown below). Find the volume of the solid that results from revolving R around the line y =. 8. Let f be a differentiable function. Use integration by parts to show: f(x) dx = xf(x) xf (x) dx. The Great Pyramid of King Khufu was built of limestone in Egypt over a -year time period from 58 bc to 56 bc. Its base is a square with side length 756 ft and its height when built was 48 ft. (It was the tallest man-made structure in the world for more than 38 years.) The density of the limestone is about 5 lb/ft 3. (a) Estimate the total work done in building the pyramid. (b) If each laborer worked hours a day for years, for 34 days a year, and did ft-lb/h of work in lifting the limestone blocks into place, about 7

8 how many laborers were needed to construct the pyramid? Find an antiderivative of f(x) = x that is not defined piecewise. Compute the integral ( + x )e x dx. 8

9 . Here W = 9 SOLUTIONS F (x) dx and corresponds to the area under the trapezoid, which is h(b + b ) = (6 lb)(9 ft + 5 ft) = 4 ft-lb.. Recall from Hooke s Law that F (x) = kx where k is the yet-to-bedetermined spring constant. We are given that F (5) =, and so 5k = k = 4 lb/in. The desired work done is thus 7 4x dx = x 7 = (7) in-lb = 98 in-lb = 49 6 ft-lb. 7 F (x) dx = 3. (a) The lb body stretches the spring.4 ft beyond its natural length, and so F (.4) =. Hooke s Law dictates that.4k = k = /.4 lb/ft = 5 lb/ft. (b) Using Hooke s Law, we need 5 = 5x x = ft. (c) If the spring stretches to 3 ft, then it is ft beyond its natural length. Thus W = F (x) dx = 5x dx = 5 x = 5 ft-lb. 4. Based on what is given and Hooke s Law, we see that, if k is the spring constant, then 5 = F (x) dx = kx dx = k x = k k = 3 lb/ft. Since 8 inches is /3 feet, the desired work is 3/4 3x dx = 5x /3 = 5 ( ) ft-lb = 3 3 ft-lb. 3/4 F (x) dx = 5. Let x be the distance from the bottom of the shaft. Notice F (x) = F ore (x) + F cable (x). Certainly F ore (x) =. Now, when the ore is x feet from the bottom of the mineshaft, there is 45 x feet of rope left to pull, which weighs 3(45 x) lb. Hence F (x) = + 3(45 x) = 9

10 x, and so the work done is F (x) dx = (35 3x) dx = [35x 3 ] 45 x = (35)(45) 3 (45) = 753, 75 ft-lb. 6. Let x be the distance that the water has been lifted. Notice F (x) = F bucket (x)+f water (x)+f rope (x). Certainly F bucket (x) = 5 and F water (x) =. Now, when the bucket has been lifted x feet, there is 3 x feet of rope left to pull, which weighs (3 x) lb. Hence F (x) = 5++ (3 x) = x, and so the work done is F (x) dx = (3 ) x dx = [3x 4 ] 3 x = (3)(3) 4 (3) = 675 ft-lb. 7. The only thing that changes here is F water (x). Since the rope is wound onto the pulley at a rate of 3 ft/s, when the bucket has been lifted x feet, x/3 seconds have elapsed, and so x 3 4 = x pounds of water have leaked from the bucket. Thus F water (x) = x, and so F (x) = 5 + x + 5 x = 3 7 x. The work done is now 3 3 F (x) dx = (3 7 ) x dx = [3x 74 ] 3 x = (3)(3) 7 4 (3) = 75 ft-lb. 8. (a) Let F (x) be the weight of the sand as a function of x, and let c be the rate at which sand leaks from the bag (in lb/ft, which is allowable since the bag is lifted at a constant rate). Then F (x) = 44 cx. We are given that F (8) = 7 so that 7 = 44 8c 8c = 7 c = 4. Therefore F (x) = 44 4x. (b) W = 8 F (x) dx = (8) = 944 ft-lb. 8 (44 4x) dx = [44x x ] 8 = (44)(8)

11 9. Consider a thin circular disk of water of thickness dx that is x meters from the bottom of the pool. Note that such a disk has to move a distance of 4 x meters. Let dw be the work done to move this disk of water. Then dw = h(x) df where h(x) = 4 x and df is the weight of the disk. Then df = ρ dv where ρ is the density of water and dv is the volume of the disk. As usual, dv = A(x) dx where A(x) is the crosssectional area. Observe that the cross-sections are circles, and since the pool is presumably cylindrical, we have that A(x) = π(5) = 5π since the radius of the pool is 5 meters (half of the diameter). Hence W = h(x) df = 3.5 (4 x)ρ dv = 3.5 (98)(5π)(4 x) dx J. (4 x)(98) A(x) dx =. Consider a thin circular disk of pudding of thickness dx that is x meters from the bottom of the tank. Note that such a disk has to move a distance of x meters. Let dw be the work done to move this disk of pudding. Then dw = h(x) df where h(x) = x and df is the weight of the disk. Then df = ρ dv where ρ is the density of chocolate pudding and dv is the volume of the disk. As usual, dv = A(x) dx, where A(x) is the cross-sectional area. Observe that the cross-sections are circles so that A(x) = π[r(x)] where r(x) is the radius of the circular cross-section at a height of x meters. We can compute r(x) from similar triangles as follows: r(x) x = 6 = 3 5 r(x) = 3 5 x. ( ) 3 (Draw a picture to see why.) Therefore A(x) = π 5 x = 9 5 πx. Hence W = h(x) df = ( x)ρ dv = ( x)(,78)a(x) dx =

12 ( ) 9 (,78) 5 πx ( x) dx,484,3 π J. 5. Consider a thin rectangular lamina of water of thickness dx that is x feet from the bottom of the tank. Note that such a lamina has to move a distance of 3 x feet. Let dw be the work done to move this lamina of water. Then dw = h(x) df where h(x) = 3 x and df is the weight of the lamina. Then df = ρ dv where ρ is the density of water and dv is the volume of the lamina. As usual, dv = A(x) dx, where A(x) is the cross-sectional area. Observe that the cross-sections are rectangles so that A(x) = l(x)w(x), where l(x) and w(x) are the length and width, respectively, of the rectangular cross-section at a height of x feet. Observe that l(x) is a constant 5 ft. We can compute w(x) from similar triangles as follows: w(x) x = 4 3 w(x) = 4 3 x. ( ) 4 (Draw a picture to see why.) Therefore A(x) = 5 3 x = x. Hence W = h(x) df = (3 x)ρ dv = (6.4)(x)(3 x) dx = 456 ft-lb. (3 x)(6.4)a(x) dx =. Consider a thin rectangular lamina of oil of thickness dx that is x feet from the bottom of the tank. Note that such a lamina has to move a distance of + 6 x = 7 x meters. Let dw be the work done to move this lamina of oil. Then dw = h(x) df where h(x) = 7 x and df is the weight of the lamina. Then df = ρ dv where ρ is the density of oil and dv is the volume of the lamina. As usual, dv = A(x) dx, where A(x) is the cross-sectional area. Observe that the cross-sections are rectangles so that A(x) = l(x)w(x), where l(x) and w(x) are the length

13 and width, respectively, of the rectangular cross-section at a height of x feet. Observe that l(x) is a constant 8 m. We can compute w(x) using circular geometry. (I know that this will be difficult to follow, so definitely draw a picture to guide you through the following argument.) Indeed, the long sides of the lamina are chords of the circular sides of the tank. Let r(x) be the distance from the center of a circular side of the tank to either end of such a chord on the same side of the tank. (This length is the same for both ends since the segment connecting them is a radius of the circle.) Since r(x) is merely the radius of the circle, we see that r(x) is a constant 3 m. The perpendicular distance from the center of the circle to the chord is 3 x meters and, by the property of isosceles triangles, bisects the chord. By the Pythagorean Theorem, [ ] w(x) 3 = (3 x) + = [w(x)] 4 = 9 (3 x) w(x) = 4(9 (3 x) ). Therefore A(x) = 8 9 (3 x). Hence W = h(x) df = (7 3 3 x)ρ dv = (7 x)(96)a(x) dx = (96)(7 x) (6 ) 9 (3 x) dx = 44,56 3 (7 x) 6x x dx J

14 3. Using tabular integration u dv SIGN x e 4x + 4 e4x 6 e4x + the answer is 4 xe4x 6 e4x + C. 4. Using tabular integration u dv SIGN x cos x + x sin x cos x + sin x the answer is x sin x + x cos x sin x + C. 5. Note that the integral is the same as (ln x)(x ) dx. Since ln x is logarithmic and x is algebraic, LIATE suggests that we let u = ln x and dv = x dx. Then du = (/x)dx and v = (x 3 /3)dx. Hence 4

15 x ln x dx = u dv = uv 6. Note that the integral is the same as ( ) x 3 x 3 v du = (ln x) 3 3 x dx = 3 x3 ln x 9 x3 +C. (ln x)(x 4 ) dx. Since ln x is logarithmic and x 4 is algebraic, LIATE suggests that we let u = ln x and dv = (x 4 ) dx. Then du = (/x)dx and v = (x 3 /3) = /(3x 3 ). Hence ln x x dx = 4 u dv = uv ( v du = (ln x) ) ( ) ( ) dx. 3x 3 3x 3 x Note that the remaining integral above is positive. Now ( ) ( ) dx = 3x 3 x 3x dx = 4 9x + C, 3 and so the answer is ln x 3x 3 9x 3 + C. 7. Use the standard trick for integrating isolated inverse trigonometric functions (that is, multiply by ). Let u = sin x and dv = dx. Then du = dx and v = x. So x sin x dx = u dv = uv v du = (sin x)(x) ( ) (x) dx. x 5

16 To find x dx. Hence x x dx, let w = x so that dw = x dx dw = x dx = x dw w = w + C = x + C. The final answer is thus x sin x + x + C. 8. Using tabular integration u dv SIGN x sec x + tan x ln sec x + the answer is x tan x ln sec x + C = x tan x + ln cos x + C. 6

17 9. We use tabular integration. u dv SIGN e x cos 3x + e x sin 3x 3 4e x cos 3x + 9 We stop at the last row since it is a constant multiple of the first. The table tells us that e x cos 3x dx = 3 ex sin 3x + 9 ex cos 3x 4 9 e x cos 3x dx. Adding 4 9 e x cos 3x dx to both sides yields 3 9 e x cos 3x dx = 3 ex sin 3x + 9 ex cos 3x. Multiply both sides by 9/3 so that e x cos 3x dx = 3 3 ex sin 3x + 3 ex cos 3x + C.. First let w = x so that dw = x dx dw = x dx. Hence x 3 cos(x ) dx = x cos(x ) x dx = w cos w dw. 7

18 This new integral can be solved with integration by parts. Using tabular integration u dv SIGN w cos w + sin w cos w + this new integral is w sin w + cos w. Plugging back in w = x and remembering to multiply by /, we find that x 3 cos(x ) dx = [ x sin(x ) + cos(x ) ] + C.. Using tabular integration u dv SIGN 3x sin x + 3 cos x sin x + 8

19 π 3x sin x dx = [ 3x cos x + 3 sin x] π = 3π cos π = 3π.. Using tabular integration u dv SIGN x e x + x e x e x + e x x e x dx = [ x e x xe x + e x] = (e e + e) = e. 3. u dv = uv v du = u()v() u()v() 3 = (7)( 4) (5)() 3 = First let w = x. Then dw = x dx x dw = dx w dw = dx 9

20 so that sin x dx = w sin w dw. This new integral can be solved with integration by parts. Using tabular integration u dv SIGN w sin w + cos w sin w + this new integral is w cos w + sin w. Plugging back in w = x and remembering to multiply by, we find that sin x dx = x cos x + sin x + C. 5. FIRST SOLUTION: Use the standard trick for integrating isolated inverse trigonometric functions (that is, multiply by ). Let u = (sin x) and dv = dx. Then du = ( sin x/ x ) dx and v = x. So (sin x) dx = u dv = uv v du = (sin x) (x) ( ) sin x (x) dx. x

21 Now observe that the new integral ( ) sin x (x) dx = x sin x x x dx has as its integrand the product of two functions, one which we straightforward to differentiate (sin x) and one which is straightforward to integrate ( x/ x ). In particular, if we let w = x, then dw = x dx, and so x dx = x w dw = w + C = x + C. Therefore, to integrate sin x ( x)/ x, we can let u = sin x and dv = x/ x dx. Then du = (/ x ) dx and v = x. Hence sin x x dx = u dv = u v v du x = (sin x)( ( x ) ) ( ) x dx x = x (sin x) dx = x (sin x) x + C. Putting everything together, we find that (sin x) dx = x(sin x) + x (sin x) x + C. SECOND SOLUTION: Let w = sin x. Then sin w = x cos w dw = dx so that

22 (sin x) dx = w cos w dw. This new integral can be solved with integration by parts. Using tabular integration u dv SIGN w cos w + w sin w cos w + sin w this new integral is now w sin w + w cos w sin w. Plugging back in w = sin x, we find that (sin x) dx = (sin x) sin(sin x) + sin x cos(sin x) sin(sin x) + C = x(sin x) + x (sin x) x + C. e 6. This is simply (ln x ) dx. (This is because ln x = when x = e e and the graph of y = ln x is increasing.) Now use the standard trick for integrating isolated logarithmic functions (that is, multiply by ). Let u = ln x and dv = dx. Then du = dx and v = x. So x

23 e e e (ln x ) dx = u dv e e e = uv e v du e e = (ln x )x e e e x x dx = (ln e )e (ln e )e e = ( )e ( )e (e e) = e e + e = e. e dx 7. Since R is not touching the axis of revolution, we use washer method. Also, since the axis of revolution is horizontal, we integrate with respect to x. So dv = A(x) dx where A(x) = π[r (x)] π[r (x)]. Here r (x) = ln x + and r (x) = so that A(x) = π[(ln x + ) ] = π[(ln x) + ln x+ ]. Hence V = e dv = e A(x) dx = π e [(ln x) + ln x] dx. We already established earlier that an antiderivative of ln x is x ln x x. To find one for (ln x), first let w = ln x so that e w = x e w dw = dx. We now have (ln x) dx = w e w dw. This new integral can be solved with integration by parts. Using tabular integration 3

24 u dv SIGN w e w + w e w e w + e w This new integral is w e w we w + e w. Plugging back in w = ln x we find that (ln x) dx = (ln x) e ln x ln xe ln x +e ln x +C = x(ln x) x ln x+x+c. Hence, the volume is π[x(ln x) x ln x + x + x ln x x] e = [x(ln x) ] e = πe. 8. Because f is differentiable, we can let u = f(x) and dv = dx. Then du = f (x) dx and v = x. Hence f(x) dx = u dv = uv v du = xf(x) xf (x) dx. (This is a generalization of the technique we use to integrate logarithmic and inverse trigonometric functions.) 4