MTH 133 Solutions to Exam 1 October 11, Without fully opening the exam, check that you have pages 1 through 11.

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1 MTH 33 Solutions to Exam October, 7 Name: Section: Recitation Instructor: INSTRUCTIONS Fill in your name, etc. on this first page. Without fully opening the exam, check that you have pages through. Show all your work on the stanar response questions. Write your answers clearly! Inclue enough steps for the graer to be able to follow your work. Don t skip limits or equal signs, etc. Inclue wors to clarify your reasoning. Do first all of the problems you know how to o immeiately. Do not spen too much time on any particular problem. Return to ifficult problems later. If you have any questions please raise your han. You will be given exactly 9 minutes for this exam. Remove an utilize the formula sheet provie to you at the en of this exam. ACADEMIC HONESTY Do not open the exam booklet until you are instructe to o so. Do not seek or obtain any kin of help from anyone to answer questions on this exam. If you have questions, consult only the proctor(s). Books, notes, calculators, phones, or any other electronic evices are not allowe on the exam. Stuents shoul store them in their backpacks. No scratch paper is permitte. If you nee more room use the back of a page. You must inicate if you esire work on the back of a page to be grae. Anyone who violates these instructions will have committe an act of acaemic ishonesty. Penalties for acaemic ishonesty can be very severe. All cases of acaemic ishonesty will be reporte immeiately to the Dean of Unergrauate Stuies an ae to the stuent s acaemic recor. I have rea an unerstan the above instructions an statements regaring acaemic honesty:. SIGNATURE Page of

2 MTH 33 Solutions to Exam October, 7 Stanar Response Questions. Show all work to receive creit. Please BOX your final answer.. Let R be the region in the first quarant boune by the curve y = ln x an y =. (a) (4 points) Sketch an label the region R. Solution: 4 y R y = y = ln x x (b) ( points) Fin the volume of the soli forme by revolving the region R aroun the y-axis. Solution: Using the expression V = b A(y) y, we fin a = an b =, with a A(y) = πx y = ln x = x = e y = x = e y = A(y) = πe y. So V = πe y y = π ey ] = π (e4 ). Page of

3 MTH 33 Solutions to Exam October, 7. (4 points) A tank fille with oil is in the shape of a ownwar-pointing cone with its vertical axis perpenicular to groun level. Assume that the height of the tank is 5 feet, the circular top of the tank has raius 5 feet an the oil insie the tank weighs 75 pouns per cubic foot. How much work woul it take to pump oil from the tank to a level 3 feet above the tank if the tank were completely full? Solution: Using the formula W = b ρa(y)d y. We have that a =, b = 5 are the bouns of a integration. The ensity we rea from the question to be ρ = 75. The istance to pump D = 8 y ft. The cross sectional area is A(y) = πr. Using that the raius is with y = an 5 when y = 5, for the circular cone we get that the raius r = y/3, so the area is A(y) = π(y/3). W = = π y (8 y) y 9 5πy 5 3 πy3 y = 5πy 3 5 ] 5 πy4 = π 5 54 π ft-lb. Page 3 of

4 MTH 33 Solutions to Exam October, 7 3. Evaluate the following integrals. 3x (a) (7 points) (x )(x + 4) x 3x Solution: Partial fraction ecomposition: (x )(x + 4) = A x + B x + 4. = 3x = A(x + 4) + B(x ) = (A + B)x + (4A B) { { A + B = 3 A = = 4A B = = B = So 3x (x )(x + 4) x = x + x + 4 x = ln x + ln x C. x (b) (7 points) x x Solution: Trig substitution x = sec θ. So x = sec θ tan θ θ, an we can simplify x = tan θ. x x x = tan θ sec θ sec θ tan θ θ = tan θ θ = sec θ θ = tan θ θ + C = x sec x + C Page 4 of

5 MTH 33 Solutions to Exam October, 7 4. (4 points) Evaluate the following limits. (If you use L Hopital s Rule, explicitly state your reasoning.) 6x (a) lim x ln (sec x) Solution: Directly evaluating the limits of the numerator an enominator gives 6 ln(sec()) = ln() = is ineterminate. Apply L Hopital [ 6x lim x ln(sec x) = lim x x = lim = ] x sec x tan x x tan x sec x another ineterminate form. So apply L Hopital again = lim x sec x = =. (b) lim x + xx Solution: Taking the limits of the base an the exponent gives ineterminate form. Since x x = e x ln x evaluate lim x ln x (which is an ineterminate form) by L Hopital: x + ln x lim x ln x = lim x + x + x So lim x + ex ln x = e lim x + x ln x = e =. = lim = lim x =. x x + x + /x Page 5 of

6 MTH 33 Solutions to Exam October, 7 5. (8 points) Solve the initial value problem y x = x y x + Solution: This is a separable ifferential equation, so we can write with initial value y() =. y y = x x. On the x + left we have y y = y + C an on the right we use u-substitution with u = x + an u = x x to get x x + x = u u = u + C = x + + C. Combining we get y = x + + C. Plugging in y() = yiels ( ) = + + C so C =. This means y = 4 x +. Taking square roots we have y = ± 4 x + ; we must take the negative of the ± to satisfy y() =. This leas to the final answer y = 4 x + or, equivalently, 4 x (6 points) Let f(x) = 3 tan (x). Using that f() = 3 π/4, compute (f ) (3 π/4 ). Solution: Using the formula (f ) (a) = f (f (a)), an that f (3 π/4 ) = is given, we compute f (). Writing f(x) = e tan (x) ln(3) we have f (x) = ln(3)3 tan (x) (tan (x)) = ln(3)3 tan (x) + x f () = ln(3)3 tan () + = ln(3) 3π/4 (f ) (3 π/4 ) = f () = ln 3 3 π/4 Page 6 of

7 MTH 33 Solutions to Exam October, 7 Multiple Choice. Circle the best answer. No work neee. No partial creit available. 7. (4 points) Shown is the graph of a force function (in Newtons) acting on a particle as it travels from x = m to x = m. How much work is one by the force? A. J B. 3 J C. 3 J D. 4 J E. 56 J 8. (4 points) If f(x) = log 3 x, fin f (e): A. 3 B. e ln 3 C. D. e E. 3 log 3 e 9. (4 points) Evaluate the integral A. e + 3 B. 7 e C. e 3 5 D. 3 4 e + 4 E. 4 e + 4 e t ln t t. Page 7 of

8 MTH 33 Solutions to Exam October, 7. (4 points) If f(x) = ( x) x, fin f (4). A. B. ln + C. D. (ln + )/4 E. 4(ln + ). (4 points) Fin y() if y (x) = sinh x, y () =, an y() =. A. B. sinh C. sinh 3 6 D. sinh E. sinh +. (4 points) Evaluate A. /4 B. /6 C. D. /6 E. /4 π sin 5 x cos 3 x x. Page 8 of

9 MTH 33 Solutions to Exam October, 7 3. (4 points) Evaluate the integral A. B. sin C. cos D. E. The integral is ivergent. ( ) x sin x. x 4. (4 points) Evaluate the integral A. B. ln C. e e x x. D. e E. The integral is ivergent. 5. (4 points) A roast turkey is remove from the oven when its temperature reache 95 o F. It is left on the counter where the air temperature is 75 o F. After hour, the temperature of the turkey roppe to 35 o F. What is the temperature of the turkey after another hour? (Remember: Newton s Law of Cooling is T (t) = k(t (t) T s )). A. 75 o F B. 5 o F C. 35 /95 o F D. 75 /35 o F E /35 o F Page 9 of

10 MTH 33 Solutions to Exam October, 7 DO NOT WRITE BELOW THIS LINE. Page Points Score Total: 6 No more than points may be earne on the exam. Page of

11 MTH 33 Solutions to Exam October, 7 Integrals FORMULA SHEET Derivatives Volume: Suppose A(x) is the cross-sectional area of the soli S perpenicular to the x-axis, then the volume of S is given by (sinh x) = cosh x x Inverse Trigonometric Functions: (cosh x) = sinh x x V = b a A(x) x x (sin x) = x x (csc x) = x x Work: Suppose f(x) is a force function. The work in moving an object from a to b is given by: W = b x = ln x + C x tan x x = ln sec x + C a f(x) x sec x x = ln sec x + tan x + C a x x = ax ln a + C for a Integration by Parts: u v = uv v u x (cos x) = x x (tan x) = x (sec x) = + x x (cot x) = x x + x If f is a one-to-one ifferentiable function with inverse function f an f (f (a)), then the inverse function is ifferentiable at a an (f ) (a) = f (f (a)) Hyperbolic an Trig Ientities Hyperbolic Functions sinh(x) = ex e x cosh(x) = ex + e x csch(x) = sinh x sech(x) = cosh x tanh(x) = sinh x cosh x coth(x) = cosh x sinh x cosh x sinh x = cos x + sin x = sin x = ( cos x) cos x = ( + cos x) sin(x) = sin x cos x sin A cos B = [sin(a B) + sin(a + B)] sin A sin B = [cos(a B) cos(a + B)] cos A cos B = [cos(a B) + cos(a + B)] Page of