1. There are 8 questions spanning 9 pages total (including this cover page). Please make sure that you have all 9 pages before starting.

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1 Instructor: K. Rotz Name: Solution PUID: Instructions and tips: 1. There are 8 questions spanning 9 pages total (including this cover page). Please make sure that you have all 9 pages before starting. 2. Show all work for full credit. Correct answers with insufficient work will not receive full credit. 3. Enter your final answers in the boxes provided. No credit will be given for answers not written inside the boxes. 4. Only calculators with one line are allowed. 5. Please turn off your cell phone, music player, and/or laptop. 6. If you get stuck, move on to the next problem and come back later if you have time. 7. Don t get frustrated. Getting frustrated will just make the exam that much harder. Instead, take a deep breath and a 15 second break to clear your head. You can do it! Good luck! Question Points Score Total: 100 Page 1 of 9

2 1. [6 points] Solve for x. Show work for full credit. log x 4 = 1 3 Solution: Turning the equation into exponential form, the statement is that 4 = x 1/3. Cubing each side gives x = 4 3 = 64. x = 64 Page 2 of 9

3 2. (a) [4 points] Use properties of logarithms to write the function f(x) given below as a sum and/or difference of logarithms without exponents or radicals. ( ) sin 2 x f(x) = ln 4. x Solution: Start by writing 4 x = x 1/4. The logarithm of a quotient is the difference of the logarithms, so f(x) = ln ( sin 2 x ) ln(x 1/4 ). Now bring down the exponents as powers. f(x) = 2 ln sin x 1 ln x. 4 f(x) = 2 ln sin x 1 4 ln x (b) [6 points] Compute f (x). Simplify your answer. Solution: It s easier to differentiate the answer from part (a) since it gets around using the quotient rule. The ln sin x part needs to chain rule: f (x) = 2 sin x (sin x) 1 4x = 2 cos x sin x 1 4x = 2 cot x 1 4x. f (x) = 2 cot x 1 4x Page 3 of 9

4 3. Compute the derivatives of the functions below. Factor and simplify your answer if possible. (a) [9 points] g(x) = ecos x x 3 Solution: From the quotient and chain rules, g (x) = x3 (e cos x ) e cos x (3x 2 ) x 6 = x3 e cos x ( sin x) e cos x (3x 2 ) x 6. The common term in the numerator is x 2 e cos x, so factor it out and cancel some x s: g (x) = x2 e cos x (x sin x + 3) x 6 = ecos x (x sin x + 3) x 4. g (x) = ecos x (x sin x + 3) x 4 (b) [9 points] f(x) = sin 3 ( 2x 3) Solution: Apply the chain rule in the following order: power rule (for the outer cube part), sine, then 2x 3. The result is f (x) = 3 sin 2 ( 2x 3 ) d dx sin( 2x3 ) = 3 sin 2 ( 2x 3 ) cos( 2x 3 ) d dx ( 2x3 ) = 3 sin 2 ( 2x 3 ) cos( 2x 3 )( 6x 2 ) = 18x 2 sin 2 ( 2x 3 ) cos( 2x 3 ). f (x) = 18x 2 sin 2 ( 2x 3 ) cos( 2x 3 ) Page 4 of 9

5 4. [10 points] Let f(x) = x 2 e x, x > 0. Find the critical point(s) of f(x). Write your answer(s) as ordered pair(s). If there aren t any, write NONE. You do not need to classify the point(s). Solution: To find critical points, find f (x) and set it equal to zero. From the product rule, f (x) = x 2 (e x ) + e x (x 2 ) = x 2 e x + 2xe x = e x (x 2 + 2x). Setting equal to zero, one sees that either e x = 0 (which is never true) or x 2 + 2x = 0. In the latter case, the solutions (either from factoring or using the quadratic formula) are x = 2, 0. The respective y-values are y = 4e 2 and y = 0. (0, 0) and ( 2, 4e 2 ) Page 5 of 9

6 5. [12 points] Compute the value of the following definite integral. Leave your answer in terms of logarithms, exponentials, and/or square roots, if applicable. e 2 e ln t t dt. Solution: Since both ln t and d (ln t) = 1 appear in the integrand, u = ln t is a good dt t choice for a substitution. Then du = dt, so the transformed integral (including new t u-bounds instead of t-bounds) is u du = 3 u3/2 = (23/2 1). 2 3 (23/2 1) Page 6 of 9

7 6. Compute the following integrals. If you use a substitution, clearly indicate what it is. (a) [8 points] 2 0 4s 3 e s4 ds (Leave your answer in terms of e s if applicable.) Solution: Let u = s 4. Then du = 4s 3 ds, so the integral becomes (with new u-bounds) 16 0 e u du = e u 16 = e 16 ( 1) = 1 e e 16 (b) [6 points] sec(e x ) tan(e x )e x dx Solution: The substitution u = e x yields du = e x dx, so the integral becomes sec u tan u du = sec u + C = sec e x + C sec e x + C (c) [6 points] cot 4x dx cos 4x Solution: If we write cot 4x =, then the substiution u = sin 4x gives du = sin 4x 4 cos 4x dx, which is (a multiple of) the numerator. Transforming into u s, u du = 1 4 ln u + C = 1 ln sin 4x + C. 4 1 ln sin 4x + C 4 Page 7 of 9

8 7. A mass attached to a spring undergoes simple harmonic motion and its position as a function of time is given by x(t) = 2 cos 3t. (a) [3 points] Find the velocity of the mass, v(t). Solution: The velocity is the derivative of position, and x (t) = 6 sin 3t. v(t) = 6 sin 3t (b) [5 points] Find the first time, t max, at which the velocity is maximum. Solution: To maximize v, we need to find v (t) = x (t) and set it equal to zero. This is v (t) = 18 cos 3t = 0, so cos 3t = 0. This happens when either 3t = π 2 or 3t = 3π 2. In the former case, t = π 6, which gives v( π 6 ) = 6 and in the latter, t = π 2, giving v( π 2 ) = 6. The first maximum velocity therefore occurs at t max = π 2 (t = π 6 corresponds to a minimum velocity). t max = π 2 (c) [4 points] Find the position of the particle when the velocity is maximum. Solution: Plugging the time value from (b) back into (a) gives x(t max ) = x( 3π) = 4 2 cos( 3 2) = 0. π x(t max ) = 0 Page 8 of 9

9 8. [12 points] Compute the integral below. (Hint: multiply both the numerator and denominator by sec x + tan x.) sec x dx Solution: Following the hint, write the integral as sec x + tan x sec 2 sec x sec x + tan x dx = x + sec x tan x sec x + tan x dx. Recall that the derivative of tan x is sec 2 x, and the derivative of sec x is sec x tan x (these can be derived from writing tan x = sin x 1 and sec x = if you forget them). Notice cos x cos x that the derivative of the denominator is exactly the numerator, so pick u = sec x+tan x. Then du = (sec x tan x + sec 2 x) dx and the integral transforms into 1 du = ln u + C = ln sec x + tan x + C. u ln u + C = ln sec x + tan x + C Page 9 of 9

Chapter 5: Integrals

Chapter 5: Integrals Section 5.5 The Substitution Rule (u-substitution) Sec. 5.5: The Substitution Rule We know how to find the derivative of any combination of functions Sum rule Difference rule Constant