Name: Instructor: Exam 3 Solutions. Multiple Choice. 3x + 2 x ) 3x 3 + 2x 2 + 5x + 2 3x 3 3x 2x 2 + 2x + 2 2x 2 2 2x.

Transcription

1 . Exam 3 Solutions Multiple Choice.(6 pts.) Find the equation of the slant asymptote to the function We have so the slant asymptote is y = 3x +. f(x) = 3x3 + x + 5x + x + 3x + x + ) 3x 3 + x + 5x + 3x 3 3x x + x + x x.(6 pts.) Calculate the indefinite integral x + 5 x x x + 5 x dx = x = dx ( x 5 x x + x ) dx ( + x 4/5 )dx = x + 5x /5 + C = x x + C.

2 3.(6 pts.) Estimate the area under the graph of y = f(x) between x = and x = 6 using the Riemann sum which gives the right end point approximation with 6 approximating rectangles whose bases are of equal length (i.e. use R 6 ) y y=f(x) x First, we draw our six rectangles with right endpoints. Now, since x =, we have R 6 = 6 f(i) = ( ) =. i= 4.(6 pts.) In finding the approximate solution to x 3 4 using Newton s method with initial approximation x =, what is x 3? Solutions: Let f(x) = x 3 4, the formula to find x is 3

3 Then, x = x f(x ) f (x ) = x x3 4 3x = 3 3 = x 3 = x f(x ) f (x ) = x x3 4 3x = 4 = 3 = 5 3 4

4 5.(6 pts.) A ball is thrown upwards from a height of feet above the surface of the planet Minerva with an initial velocity of 6 feet per second (at time t = ). The ball has a constant acceleration of ft/sec. What is the maximum height (from the surface of the planet) reached by the ball? Let h(t) denote the height of the ball, in feet, t seconds after it is thrown. We are told that the accelleration due to gravity on Minerva is ft/s, so we know f (t) =. Antidifferentiating, we get that f (t) = t + c, for some c. The initial velocity is 6ft/s, so f () = + c = c = 6, so f (t) = t + 6. Antidifferentiating again, we obtain f(t) = t + 6t + d, for some d. The initial position is feet above the surface, so f() = d =, and f(t) = t + 6t +. We are asked the maximum height. The maximum height will be attained when the ball has zero velocity, i.e. when f (t) = t + 6 =. This occurs when t = 3. At t = 3, the height is f(3) = = = 9ft. 6.(6 pts.) The graph of the piecewise defined function g(x) is shown below. The graph consists of part of a circle and straight lines. Use the graph to calculate y 8 g(x) dx x y = g(x) We need to compute the area between the function and the x-axis. Areas below the x-axis are counted as negative, and areas above the x-axis are counted as positive. From x = to x = 4, we have a quarter of a circle of radius 4 that lies below the x-axis, so 5

5 4 g(x)dx = ( ) π(4) = 4π. 4 From x = 4 to x = 7, we have a triangle with base 3 and height 6 that lies above the x-axis, so 7 g(x)dx = (3)(6) = 9. 4 From x = 7 to x = 8, we have a triangle that lies below the x-axis with base and height 6, so Thus, 8 7 g(x)dx = ()(6) = 3. 8 g(x)dx = 4 g(x)dx + 7 g(x)dx g(x)dx = 4π = 6 4π. 6

6 7.(6 pts.) Let x h(x) = 4 + sin (t) dt. Find h (x). Let f(x) = x 4 + sin t dt, and let g(x) = x. Then h(x) = (f g)(x). Therefore, by the chain rule, h (x) = f (g(x))g (x). Since f (x) = 4 + sin by the Fundamental Theorem x of Calculus, h (x) = 4 + sin (x ) (x) = x 4 + sin (x ) 8.(6 pts.) Calculate the indefinite integral x + sin( x) ; dx. x x + sin( x) sin(x First, rewrite dx = x / / ) dx + x x / x / dx = 3 x3/ + C. dx. We have To evaluate the other integral, we need to use substitution. Let u = x /, then du = x / dx dx = x / du. Then, sin(x / ) sin(x / ) dx = (x / du) = sin u du = cos(u)+c = cos( x)+c. x / x / x + sin( x) Therefore, dx = x 3 x3/ cos( x) + C. 7

7 9.(6 pts.) An underground beer pipeline in the city of Bruges has sprung a leak which is gradually worsening. Your statistics suggest that beer is leaking from the pipeline at a rate of 3t 4t + 3 gallons per day, where t denotes the number of days after the leak started. How many gallons of beer will have leaked in the first days after the leak started? The amount of beer leaked in the first two days is given by gallons. Solving, 3t 4t + 3dt 3t 4t + 3dt = ( t 3 t + 3t ) = = = 6 gallons.(6 pts.) Evaluate the following definite integral π 4 sin x cos 3 x dx π 4 sin x cos 3 x dx = = π 4 π 4 sin x cos x cos x dx tan x sec xdx Now we can use u-substitution. Let u = tan x. Then du = sec xdx, so 8

8 π 4 sin x cos 3 x dx = = = π 4 tan x sec xdx tan ( π 4 tan() udu ) udu = u =. 9

9 Partial Credit You must show your work on the partial credit problems to receive credit!.(3 pts.) (a) Evaluate the definite integral and the limit definition of the definite integral. ( n(n + ) ). Hint: n 3 = By the limit definition of the integral, we have x 3 dx = lim n = lim n i= x 3 dx using the right endpoint approximation n ( ) 3 i n n n 6i 3 i= 6 = lim n n 4 6 = lim n n 4 n 4 n i= i 3 ( n(n + ) ) = lim n 4(n + n + ) 4 + = lim n = 4. n 8 n + 4 n (b) Verify your answer using the fundamental theorem of calculus. We have x 3 dx = 4 x4 = 4 (4 ) = 4.

10 .(3 pts.) A manufacturer needs to make a cylindrical can (top included ) that will hold cm 3 of liquid. Find the dimensions of the can (values of r and h) that will minimize the amount of material used to make the can. (Exact values such as, 3, π, π, etc... should not be converted to a decimal approximation.) Note that the surface area of a cylinder with no top or bottom is πrh cm. To minimize the amount of material used to make the can, we need to minimize its surface area, including the top and bottom. The surface area function of the cylinder with top and bottom is given by S = πr + πrh. Now, in order to take the derivative of S, we need to rewrite S in term of one and only one variable. To accomplish this, we are given the fact that the volume of this can needs to be cm 3. So, V = πr h = h = πr. Then, ( ) S = πr + πr = πr + 4 πr r. So, S = 4πr 4 = 4πr 3 = 4 r 3 = 4 r = r 4π 3 π Note that we don t consider the critical point r = in this case since we want r >. And S = 4π + 8 > for all r >. Thus, at the critical point r = r 3, the surface 3 π area function attains its minimal value. So then h = πr = π ( 3 ) π = π/3 π = 3. π

11 Answer: r = 3 π and h = 3 π.

12 3.( pts.) Find the area of the bounded region between the curves Let y = x x + and y = 7 x + x. f(x) = (7 x + x) (x x + ) = x + 4x + 6 = (x x 3) = (x + )(x 3) be the difference between these two curves. The area between the curves is the area under f(x) between that function s x-intercepts. We choose the difference that we did, rather than subtracting in the opposite order, because it makes the leading coefficient negative. Consequently, the region of the resulting f between the x-intercepts is above the x-axis. Since the above factorization yields x-intercepts of and 3, the area of interest is given by 3 ( x + 4x + 6dx = ) 3 3 x3 + x + 6x ( = ) ( ) 3 ( )3 + ( ) + 6( ) ( ) = ( ) = = + 3 =

13 4.( pts.) You will earn points if your instructor can read your name easily on the front page of the exam and you mark the answer boxes with an X (as opposed to a circle or any other mark). Rough Work 4