FINAL EXAM CALCULUS MATH 00 FALL 08 Name PRACTICE EXAM SOLUTIONS Please answer all of the questions, and show your work. You must explain your answers to get credit. You will be graded on the clarity of your exposition! Date: December, 08.
. Consider the region bounded by the graphs of f (x) = x + and g(x) = x. 0 points.(a). (5 points) Write the integral for the volume of the solid of revolution obtained by rotating this region about the x-axis. Do not evaluate the integral. SOLUTION: We can see the region in question below. y f (x) = x + g(x) = x x Using the washer method, the volume integral is π g(x) f (x) dx = π ( x ) (x + ) dx..(b). (5 points) Write the integral for the volume of the solid of revolution obtained by rotating this region about the line x =. Do not evaluate the integral. SOLUTION: Now using the shell method, the integral is equal to π( x)(g(x) f (x)) dx = π = π ( x)( x ) dx ( x)(( x ) (x + )) dx
. MULTIPLE CHOICE: Circle the best answer. 6 points.(a). ( point) Is the integral dx an improper integral? x Yes No.(b). (5 points) Evaluate the integral: x dx = SOLUTION: The function /x is undefined at x = 0, so we we must evaluate the improper integral as a limit. c dx = lim dx + lim x c 0 x c 0 + c x dx = lim c c 0 x + lim c 0 + x c ( = lim c 0 c ) ( + lim c 0 + ) c ( ) ( = lim c 0 c + + lim ). c 0 + c Now, since ( ) lim c 0 c + = lim c 0 c and ( lim ) = lim c 0 + c c 0 + c both diverge to, and so the integral does not converge. Thus, the integral diverges.
. Consider the curve parameterized by { x = t + t + y = t t for 0 t 5..(a). (6 points) Find an equation for the line tangent to the curve when t =. 4 points SOLUTION: We first find a general formula for the slope using the chain rule, and then evaluate at t =, giving dy dx = dy/dt t= dx/dt = t t t= t + 6t = t= 7. Since x() = 4 and y() = 0, we need the formula for a line with slope /7 that passes through (4, 0). This equation is y = 7 x 4 7.(b). ( points) Compute d y at t =. dx SOLUTION: Again employing the chain rule, d y dx = d dy t= dx dx = t= d dt dy dx dx dt = t= (6t )(t +6t) (t+6)(t t) (t +6t) t + 6t = 0 t= 7..(c). (5 points) Write an integral to compute the total arc length of the curve. Do not evaluate the integral. SOLUTION: Arc length is given by 5 ( ) dx ( ) dy + dt = dt dt 0 5 0 (t + 6t) + (t t) dt. 4
4 4. Consider the function f (x) = x arctan(x). 8 points 4.(a). (5 points) Find a power series representation for f (x). SOLUTION: The power series of arctan(x) is ( ) n x n+, with interval of conver- n + gence x [, ]. Thus, for x [, ]. f (x) = x ( ) n x n+ n=0 n + n=0 = ( ) n x n+ n=0 n + 4.(b). ( points) What is f (8) (0), the 8rd derivative of f (x) at x = 0? SOLUTION: For a power series f (x) = n=0 c n(x a) n with positive radius of convergence, we have f (n) (a) = n!c n. In our power series representation f (x) = x arctan(x) = ( ) n n + xn+, which has radius of convergence, the coefficient of x 8 = x 40+ is n=0 ( ) 40 40+ = 8, so that f (8) (x) = 8! 8. Alternatively, using the above power series representation, and formally differentiating, we have Thus, f (8) (x) = n=40 (n + )! ( ) n x n+ 8 (n + 8)! n + = n=40 (n + )! ((n 40))! f (8) ( ) (0) = 40 (8)! = 8 8 (80!). 40 + ( ) n x (n 40). n + 5
5. A tank contains 00 L of salt water with a concentration of 4 g/l. Salt water with a concentration of g/l is being pumped into the tank 5 at the rate of 8 L/min, and the tank is being emptied at the rate of 8 L/min. Assume the contents of the tank are being mixed thoroughly 0 points and continuously. Let S(t) be the amount of salt (measured in grams) in the tank at time t (measured in minutes). 5.(a). ( points) What is the amount of salt in the tank at time t = 0? SOLUTION: S(0)g = 00L 4g/L = 800 g. 5.(b). ( points) What is the rate at which salt enters the tank? SOLUTION: 8L/min g/l = 4 g/min 5.(c). ( points) What is the rate at which salt leaves the tank at time t? SOLUTION: As the volume of water is a constant 00 L, this is S(t)g 00L 8L min = S(t) 5 5.(d). ( points) What is ds, the net rate of change of salt in the tank at time t? dt SOLUTION: Net change is given by gain minus loss, so using parts (b) and (c), g min. ds g S(t) g = 4 dt min 5 min.5.(e). (4 points) Write an initial value problem relating S(t) and ds. Solve the initial value dt problem. SOLUTION: The initial value problem is ds = 4 S(t), with S(0) = 800. Since this dt 5 differential equation is separable, we can solve by separating and then integrating: 4 5 S ds = dt 5 ln 4 5 S = t + C, ( ) Note that 4 5 S 0, so we can write this as 5 ln 5 S 4 = t + C, so that 5 S 4 = Ae 5 t. From this we get S = Ae 5 t + 600. Setting t = 0, and using (a), we find the answer is S = 00e 5 t + 600 6
6 6. Compute the following integrals. 6.(a). (4 points) sin (x) cos (x) dx 8 points SOLUTION: First, using the pythagorean identity, sin (x) cos (x) dx = sin(x)( cos (x)) cos (x) dx = sin(x) cos (x) dx sin(x) cos 4 (x) dx. Now, let u = cos(x), so that du = sin(x) dx. Then the above equation is equal to u du + u 4 du = u + u5 5 + C. Finally, reversing our substitution, we find that sin (x) cos (x) dx = cos (x) + cos5 (x) + C. 5 6.(b). (4 points) x + x (x ) dx SOLUTION: We start by using partial fractions: which gives x + x (x ) = A x + B x + C x, x + = Ax(x ) + B(x ) + Cx = (A + C)x + (B A)x B, from which we deduce A + C = 0, B A =, and B =. Therefore, B =, A =, and C =. Thus, x + x (x ) dx = x + x + x dx = ln x + + ln x + C. x 7
( 7. A slope field for the differential equation y = y y ) is shown below. y 7 6 points 4 (a) ( y = y y ) (0, ) 0 x (0, ) (b) 0 4 5 6 7.(a). ( points) Sketch the graph of the solution that satisfies following initial condition. Label the solution as (a). y(0) = 7.(b). ( points) Sketch the graph of the solution that satisfies following initial condition. Label the solution as (b). y(0) = 7.(c). ( points) Show that for y(0) = c 0, we have lim x y(x) is finite. SOLUTION: That this should be true is evident from the picture above. To see that it is in fact true, we argue as follows. First, if P 0 = 0, then P(t) = P 0 for all t, and lim t P(t) = 0. If P 0 = 0, consider the general solution to the logistics equation: dp dt = kp ( P M ) P(t) = M + ( M P 0 )e kt The function P(t) is well-defined, so long as the denominator is non-zero. We focus here on the case k, M > 0. If 0 < P 0 M, so that ( P M 0 ) 0, then the denominator is clearly never zero, and we have lim t P(t) = M. If P 0 > M, then it is also easy to see that the denominator is never zero for t 0, and so again, one easily computes lim t P(t) = M. ( ) ) Note however, that if P 0 < 0, then we have + MP0 e kt = 0 t = k ( log M P0 In fact, it is not hard to check that for P 0 < 0, we have lim ( ) t P(t) = k log P M 0 8
8 8. Consider the series n= n 4. 6 points 8.(a). ( points) Use the Remainder Estimate for the Integral Test to find an upper bound for the error in using S 0 (the 0th partial sum) to approximate the sum of this series. SOLUTION: If R 0 denotes the error described above, the Remainder Estimate for the Integral Test tells us that c R 0 dx = lim x4 c x = lim c c 0 = 000. 0 0 8.(b). ( points) How many terms suffice to ensure that the sum is accurate to within 0 6? SOLUTION: In order to ensure that the error in estimate is less than 0 6 with the Remainder Estimate for the Integral Test (REIT), we must solve Following the solution in (a), N N dx = lim x4 c x 4 dx 0 6. c N = N, and so we must solve N 0 6, So clearly it suffice to take N = 0. In other words, if we use 00 terms, we are ensured by the REIT that the error is at most 0 6. 9
points 9. Determine whether the series is convergent or divergent and circle the corresponding answer. Then write the test allows one to determine convergence or divergence 9.(a). ( points) n= 4 n 9 convergent divergent Test: p-series test 9.(b). ( points) ( ) n (n + ) n n= convergent divergent Test: alternating series test 9.(c). ( points) cos n= ( ) 5 n convergent divergent Test: test for divergence 9.(d). ( points) n= n + 5 (n + )! convergent divergent Test: ratio test 0
0 0. MULTIPLE CHOICE: Circle the best answer below. 6 points 0.(a). ( points) The sequence a n = 0. n converges to 0. converges, but not to 0. diverges. 0.(b). ( points) The sequence a n = n 4 n converges to 0. converges, but not to 0. diverges. 0.(c). ( points) The sequence a n = n + n converges to 0. converges, but not to 0. diverges.
. 8 points.(a). (4 points) Sketch the curves r = and r = + sin θ on the axes below. 5 6 π 4 6 π 6 π 6 π 6 π + sin θ π 0 4 0π 7 6 π 6 π 8 6 π 9 6 π 0 6 π.(b). (4 points) Write an integral that represents the area contained outside the first curve (r = ) and inside the second curve (r = + sin(θ)). Do not evaluate the integral. SOLUTION: From the graph (or via algebraic solution), the bound of integration should be π/6 to 7π/6. Thus, the integral is 7π/6 π/6 ( + sin(θ)) dθ
. MULTIPLE CHOICE: Circle the best answer below. 8 points.(a). ( points) Is the following statement ALWAYS, SOMETIMES, or NEVER true? If a n converges, then a n converges. ALWAYS SOMETIMES NEVER.(b). ( points) Is the following statement ALWAYS, SOMETIMES, or NEVER true? If a n converges, then a n converges. ALWAYS SOMETIMES NEVER.(c). ( points) The graph of { x = t y = t for < t < is a line parabola circle ellipse.(d). ( points) The graph of { x = t y = t for < t < is a line parabola circle ellipse