Final Exam Review #1

CP1 Calculus 2015-2016 Name______

1. Given the curve y = x 4 - 8x 2 +16. a. Find regions where y is increasing and where y is decreasing. b. Find (x, y) coordinates of local minima and maxima. c. Find regions of positive and negative concavity. d. Find all points of inflection. e. Sketch a graph from x=-4 to x = 4 identifying the features you found in (a.)-(d.).

2. A particle is moving along the x-axis with position function x(t)= 2t3 -14t2 + 22t - 5 t ³ 0 a. Find the velocity function. b. Find the acceleration function. c. Find when (on what intervals of t) the particle is moving left, moving right. d. Find when the particle changes direction. e. What does the acceleration function tell us about the graph of the position function?

3. Find the dimensions of the rectangle of largest area that fits atop the 400 y=256-x4 4 x-axis under the curve y = 256 - x . 200

–5 5

–200

–400 4. Two numbers x and y add to 600. Find the values of x and y so that xy2 is maximized. Also find the maximum value.

5. A box with a square base is needed to package 100 cm3 of powdered milk. Find the dimensions of the box that will use the minimum amount of material.

6. The radius of a spherical balloon is increasing by 2 cm/sec. At what rate is air being blown in at the 4 3 moment when the radius is 10 cm? Give units in your answer. Note that V = 3 pR .

7. A piece of ice cut in the shape of a cube melts uniformly so that its volume decreases at 3 cm3/sec. How fast is its surface area decreasing when the edge of the cube is 5 cm?

8. Ship A is traveling due west toward Lighthouse Rock at a speed of 15 kilometers per hour. Ship B is traveling due north away from Lighthouse Rock at a speed of 10 km/hr. Let x be the distance between Ship A and Lighthouse Rock at time t, and let y be the distance between Ship B and Lighthouse Rock at time t, as shown in the figure. a. Find the distance, in kilometers, between Ship A and Ship B when x = 4 km, and y = 3 km. b. Find the rate of change, in km/hr, of the distance between the two ships with x = 4 km, and y = 3 km.

1 9. Use the graph of f (x) to find the following: 4 a. f (x)dx 8 ò 2

5 b. f (x)dx 4 ò 0

2 10 c. f (x)dx 0 ò 7 0 1 2 3 4 5 6 7 8 9 10 -2 10 d. f (x)dx -4 ò 0

15 15 e. If f (x)dx = 14 then what is f (x)dx? ò 10 ò 0

10. The following table shows velocity data taken as a particle time (sec) velocity (m/sec)

accelerates from rest. Use the data to estimate the distance 0 0 traveled from time t = 0 to t = 10. Use a left-hand Riemann Sum with 3 rectangles. Include units in your answer. Hint: You may 4 13 want to sketch a graph. 6 17 10 29 11. You are trying to approximate the area under the graph of y = x3 + 2 on the interval [0,6]. You decide to approximate it using three rectangles. a. What is the area of the 2nd rectangle if you use left, right, or middle sums? b. For each left, right, and middle: will the approximation using three rectangles over-estimate the area, under-estimate it, or get it exactly correct? No calculations are necessary to answer this question.

12. The graph of the function f shown to the right consists of six line segments. Let g be the function given by x g(x) = f (t)dt ò0

a. Find g(4)

b. Find g'(4)

c. Find g''(4)

d. Does g have a relative minimum, a relative maximum, or neither at x = 1? Justify your answer.

e. How do you know that x = 0 is a point of inflection of g(x)?

2 Answers: x y y' y'' y = x 4 - 8x 2 +16 1. 0 16 0 local max 3 y'= 4x -16x = 0 -2 0 0 local min 2 0 0 local min 4x(x + 2)(x - 2) = 0 1.155 7.111 0 P.I. x = 0, 2, - 2 -1.155 7.111 0 P.I. y''=12x 2 -16 = 0 (-¥, -2)(-2, 0)(0, 2)(2, ¥) x = ± 4 /3 » ±1.155 y' - + - + dec inc dec inc

(-¥, -1.155)(-1.155, 1.155)(1.155, ¥) y'' + - + con. up con. down con. up

2. a. v(t) = 6t 2 - 28t + 22 b. a(t) =12t - 28 c. Moving to the right [0, 1) and (11/3, inf); Moving left (1, 11/3) d. Changes direction at 1 and 11/3 e. Deaccelerating (concave down) during (0, 7/3); Accelerating (7/3, inf). Not accelerating at t = 7/3 3. Width: 2 x 2.675 = 5.350, height: 204.8, area: 1095.664 4. x = 200, y = 400, M = 32,000,000 5. 4.64 cm x 4.64 cm x 4.64 cm 6. 800p » 2513 cm2 / sec 12 7. - cm2 /sec 5 8 a. 5 km b. dz/dt = -6 km/hr 9 a. 14 b. 29.5 c. -4 d. 28.5 e. 42.5

10. 94 meters

11 a. left=20; right=132; middle=58 b. left and middle are too low; right is too high

12 a. 3 b. 0 c. -2 = g’(x) d. g has a relative min because the derivative is zero at x = 1 (slope is zero) and the function f (or g’) goes from negative to positive making g(1) a minimum. e. Since at x = 0, g’’(x) = f ‘ (x) does not exist, and changes sign.

3 Final Exam Review #2 1. Find the following anti-derivatives: x 2 - 5x + 7 a. (x 2 - 3x + 7)dx b. (5x 3 / 2 - 6x -2 / 3 )dx c. dx ò ò ò 2x d. ò (2x -1)11 dx e. ò (x 2 - 6)5 xdx f. ò (x 2 - 3)2 dx - 2dx g. h. e3cos(x)-1 sin xdx i. 6x 5x 2 - 9dx ò 3 7x -15 ò ò

2x2 +7 3 j. ò3xe dx k. ò (sin(5x) + 5cos x)dx l. ò sin x × cos xdx

6x 2 m. dx n. sin(x )xdx ò x 2 - 7 ò

2. Evaluate the following: 4 6 3 5 4 p a. dx b. c. dx d. ò 1 2x +13 dx (x + sin x) dx x ò -2 ò 2 2x - 3 ò 0

2 x 1 4 æ 3x ö ln 6 e dx e. x(1 - x)dx f. ò ç 3 ÷ g. dx ò 0 2 è x -1ø ò 0 2

3. An object’s velocity function is given by the equation v(t) = -t + 5 for 0 ≤ t ≤ 10. t is in seconds, v(t) is in meters per second. The object’s initial position is s(0) = -4.

You must answer the following questions by setting up an evaluating definite integral(s), not by using triangle areas. a) What is the total distance the object travels between t = 0 and t = 7?

b) What is the object’s position at t = 7? c) When does the object change direction? Is the change from positive-to-negative or negative-to- positive?

4. An object’s velocity function is given by the graph below. t is time in seconds, and v(t) is velocity in meters per second. 0 £ t £12. The numbers in circles are areas between the graph and the x-axis. The object’s initial position is s(0) = 6. NOTE: the graph is not to scale. Ignore the grid lines and use the circled values as the areas. Include the proper units in your answers to the following

questions. d) What is the object’s position at time t = 8? e) What is the object’s displacement between t = 0 and t =12? f) When is the object accelerating in the positive direction? g) When does the object change direction? List each time and whether the change is positive-to-negative or negative-to-positive direction. h) When is the object’s position a global maximum? Give a reason for your answer. 4 Answers: x 3 3x 2 x 2 5x 7 ln x 1 a. - + 7x + C b. 2x 5 / 2 -18x1/ 3 + C c. - + + C 3 2 4 2 2 (2x -1)12 (x 2 - 6) 6 x 5 d. + C e. + C f. - 2x3 + 9x + C (FOIL) 24 12 5 - 3 -1 2 g. (7x -15) 2 / 3 + C h. e3cos( x)-1 + C i. (5x 2 - 9)3 / 2 + C 7 3 5 4 3 2 1 (sin x) j. e 2 x +7 + C k. - cos(5x) + 5sin x + C l. + C 4 5 4 -1 m. 3ln(x 2 - 7) + C n. cos(x 2 ) + C 2

2 1 3 / 2 p 4 5 2 a. 12 b. ×19 - 9 c. 2ln 7 d. + 2 e. f. ln 9 g. 3 2 15 2

3 a. 29/2 = 14.5 meters b. 13/2 = 6.5 meters c, t = 5, positive-to-negative

4 a. 17 meters b. -1meter c. 3 < t < 4 d. t = 8, positive-to-negative e. t = 8 is a local maximum of position since the velocity (the derivative of position) changes from positive to negative. The end points t = 0 and t = 12 are both local minima of position. The other critical point is t = 3, but the velocity does not change sign there, so it is neither a local maximum nor minimum for position. So t = 8 must be the global maximum. OR Position is increasing from t = 0 to t = 8 and decreasing from t = 8 to t = 12, so t = 8 must be a global maximum

5 Final Exam Review #3

1. Find a function f that satisfies the conditions. a. f ''(x) = 2, f '(2) = 5, f (2) =10 b. f ''(x) = x2 , f '(0) = 6, f (0) = 3

c. f ''(x) = 4x - 7, f '(1) =11, f (0) = 31

2. A helicopter is rising straight up into the air. Its velocity t seconds after it starts is given by the equation v(t) = 3t +1. (in meters per second). Its initial height is 2 meters. a. Write the function describing the helicopter’s height at any given time. b. When is its height 50 meters? c. What is its average velocity over the first 4 seconds? d. What is its average height over the first four seconds?

3. The instantaneous rate at which fans enter Fenway Park for a Red Sox game is given by the function f (t) = -0.1t 2 + 6t + 210 on the interval [0,80]. Here, t is the number of minutes after 6 pm and the rate is people per minute. FnInt is OK, but write the integral first. a. Write an integral describing how many fans entered between 6:10 pm and 6:20 pm and evaluate it. b. What was the average rate that fans entered Fenway Park over the [0,80] interval, and what was the 1 b total number of fans entering the park during this 80-minute period? av( f ) = ò f (x)dx b - a a c. Assuming that there were already 2,500 fans in the ballpark at 6 pm, write a function F(t) that shows how many fans were in the ballpark t minutes after 6 pm (assume no fans left the ballpark). d. Using this function from part c above, exactly when did the 10,000th fan enter the ballpark? [You may use your calculator in any way that you like to answer this part.]

4. A region R in the 1st quadrant is bounded by the x-axis, the line x=4, the line y = x + 2 and y = x 2 . (It is touching the x-axis, if you are unclear which region it is.) What is the area of R? You may use fnInt but set up integrals clearly first.

5. Find the area of the region bounded by the graph of f (x) = x3 - 2x 2 - 7x + 3 and g(x) = x + 3.

6. Determine the area bounded by the graphs of y = -x2 + 4x and y = 0.

7. Determine the area bounded by the graphs f (x) = x2 - 4x + 3 and g(x) = -x2 + 2x + 3

8. a. Sketch a graph of the region in quadrant I enclosed by the graph of y = x +1, the y-axis, and the line y = 3. b. Find the area in quadrant I given the region enclosed by the graph of y = x +1, the y-axis, and the line y = 3. c. Find the volume of the solid formed by rotating about the y-axis the region enclosed by the graph of y = x +1, the y-axis, and the line y = 3.

6 9. On your calculator, sketch the graphs of f (x) = x and g(x) = x. Define R as the region in the first quadrant between the two curves. a. Write an integral expressing this area. You must find the end-points algebraically. b. Evaluate this integral (no using fnInt!) c. Write an integral showing the volume that results from this area being rotated around the x-axis. Evaluate it with fnInt. d. Write an integral showing the volume that results from this area being rotated around the line y = -2. Evaluate it with fnInt.

10. Find the general solution to the differential equation. Simplify your answer. dy dy a. = x2 y b. = xy2 dx dx

11. Find the specific solution of the differential equation. dy 3x 2 dy x + 3 a. = and goes through the point (1, 4) b. = and goes through the point (2, 1) dx 2y dx y4

7 Answers: 1 2 7 1 a. f (x)= x 2 + x + 4 b. f (x)= x 4 + 6x + 3 c. f (x) = x 3 - x 2 +16x + 31 12 3 2

1 4 1 2 a. h(t) = 1.5t 2 + t + 2 b. 5.33 seconds c. (3t +1) dt = 7 OR [h(4) - h(0)] = 7 4 ò 0 4 1 4 d. (1.5t 2 + t + 2) dt = 12 4 ò 0

20 80 3 a. f (t) dt = 2767 b. total = f (t) dt = 18933 avg is this over 80 = 236.67 ò 10 ò 0 - 0.1 c. F(t) = t 3 + 3t 2 + 210t + 2500 d. F(t) = 10000 when t = 28minutes 3

2 4 38 4. x 2dx + (x + 2)dx = ò 0 ò 2 3

0 4 5. ( f (x) - g(x))dx + (g(x) - f (x))dx = 49.33 ò -2 ò 0

32 6. 3

8 32p 8. a. b. c. 3 5 4

1 1 é 2 ù 9 a. x - x dx b. 1/6 c. V = p ò x - (x)2 dx » 0.524 ò 0 ( ) ëê( ) ûú 0 1 é 2 ù V = p 2 + x - (2 + x)2 dx » 2.618 d. ò ëê( ) ûú 0

1 1 x3 y = 10 a. 3 b. -1 y = Ce x2 + C 2

5 1 11 a. y = ± x3 +15 b. y = ( x2 +15x - 39) 5 2