MTH132 Exam 2 Solutions Covers: Page Total. Max

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1 Name: PID: A Section #: Instructor: Page Total Score Max Instructions 1. You will be given exactly 90 minutes for this exam.. No calculators, phones, or any electronic devices. Put them away out of sight. 3. Nothing on your desk or lap but a pen or pencil and this booklet. 4. No outside scratch paper: use the back of the exam pages if needed. 5. Make sure you try all the problems: do not linger too long on hard ones. 6. Show your reasoning and calculations. Unsupported answers will not receive full credit. 7. Cross out (or erase) any incorrect statements in your answers. 8. Please raise your hand if you have any questions and we will come to you. 9. Do not open this booklet until you are instructed to do so. I have read and understood all of the above instructions: Signature: 1 3/31/14

2 1. Compute the derivative of each of the following functions using the Rules of Differentiation. You need not simplify your answers. (a) (10 points) fpxq 3? cos x fpxq 3? cos x fpxq pcos xq 1{3 ùñ f 1 pxq 1 3 pcos xq {3 p sin xq (b) (10 points) gpxq secpx 3 1q gpxq secpx 3 1q ùñ g 1 pxq secpx 3 1q tanpx 3 1qp3x q (c) (10 points) hpxq sin 3 ptan xq hpxq sin 3 ptan xq ùñ h 1 pxq 3 sin ptan xq cosptan xq sec x 3/31/14

3 (d) (10 points) F pxq tan x sec x F pxq tan x sec x ùñ F 1 pxq px sec x qpsec xq ` ptan x qpsec x tan xq. (1 points) Find an equation of the line tangent to the graph of the equation xy 3 x 3 ` y 1 at the point p 1, 1q. Lets use the equation y y 0 dy x 0 q for the equation of the tangent line giving us: px0,y 0qpx y ` 1 dy dy px ` 1q. Now we need to evaluate: p 1, 1q p 1, 1q ˆ p1qpy 3 q ` pxq ˆ 3y dy xy 3 x 3 ` y 1 3x ` dy p1qp 1q ` p 1q 3 dy 3 ` dy 0 p 1, 1q 1 3 dy 3 ` dy 0 p 1, 1q p 1, 1q dy 4 p 1, 1q dy p 1, 1q p 1, 1q (use implicit differentiation) 0 (plug in y x 1) (solve for dy p 1, 1q ) Yielding our final result: y ` 1 px ` 1q 3 3/31/14

4 3. (1 points) An aircraft is ascending at an angle of 30 0 with level ground. When its altitude is 1000 ft. its speed is 150 mi/hr. At that time how fast is its shadow moving along the ground? At that time, the sun is directly overhead. hptq 30 yptq xptq Consider the picture: Suppose that t 0 is the special time at which ypt 0 q 1000ft. Then we know that h 1 pt 0 q 150 mi hr. We want to solve for x 1 pt 0 q (since the shadow moves along the ground). So we can use the equation: cosp30q xptq (multiply by hptq and simplify) hptq? 3hptq xptq (differentiate)? 3h 1 ptq x 1 ptq (substitute t t 0 )? 3h 1 pt 0 q x 1 pt 0 q 75? 3 x 1 pt 0 q 4. (10 points) Use linear approximation to estimate? 150. (Hint:? 144 1) Consider the function fpxq? x and x Then we have the linearization of f through p144, 1q given by: Lpxq 1 ` f 1 p144qpx 144q Now lets solve for f 1 p144q and plug in x 150 to estimate fp150q? 150. Lpxq 1 ` d p? xq px 144q x 144 Lpxq 1 ` 1? px 144q x x 144 (differentiate) (evaluate) Lpxq 1 ` 1? px 144q (simplify) 144 Lpxq 1 ` 1 px 144q (evaluate at x 150) 4 Lp150q 1 ` 1 p q 4 Lp150q 1 ` 1 4 p6q Lp150q 1 ` /31/14

5 5. Let fpxq 9x x ` 9. Use that f 1 pxq 9 9 x px ` 9q and f pxq 18 xpx 7q px ` 9q 3 about the graph of f. to answer each of the following questions (a) (6 points) Find the interval(s) on which f is increasing and those on which it is decreasing. f 1 pxq 9 9 x ` 3qpx 3q px 9px ` 9q px ` 9q f 1 : f is increasing on r 3, 3s and decreasing on both p 8, 3s and r3, 8q. ` 3 3 (b) (6 points) Find the interval(s) on which f is concave up and those on which it is concave down. f pxq 18 xpx ` 3? 3qpx 3? f ` ` : 3q 3? 3 0 3? 3 px ` 9q 3 f in concave up on r 3? 3, 0s and on r3? 3, 8q. f is concave down on p 8, 3? 3s and on r0, 3? 3s. (c) (6 points) Sketch the graph of f. (Hint: What is the horizontal asymptote of f?) 3? ? 3 5 3/31/14

6 6. (10 points) On Exam 1 we used the Intermediate Value Theorem to show that vpxq x 5 ` 6x ` 1 is 0 for some x in 0 ď x ď. Use Newton s method with x 0 1 to approximate at what time the particle is stationary. (You can stop at x 1. You do not need to simplify.) vpx 0 q vp1q 1 ` 6 ` 1 6. v 1 pxq 5x 4 ` 1x; v 1 px 0 q v 1 p1q 5 ` 1 7. Thus x 1 x 0 ` vpx 1q v 1 px 1 q 1 ` 6 7. Answer: x 1 1 ` (1 points) Find and classify any local extrema for the function fpxq x 5{3 ` 0x {3. f 1 pxq 5 3 xp{3q ` 0 3 x p1{3q 5x ` 40 3x 1{3. f 1 : ` ` 8 0 f is increasing on p 8, 8s and r0, 8q. f is decreasing on r 8, 0s. consequently f has a local maximum at x 8 and a local minimum at x /31/14

7 8. Find the following indefinite integrals. ż (a) (8 points) x 3 sec x Answer: 1 3 x3 3 tan x ` C ż ` 3x 4 (b) (8 points) x ż ` 3x 4 ż x x ` 3x x 1 ` x 3 ` C Answer: x 1 ` x 3 ` C ż (c) (8 points) 3 cos x Answer: 6 sin ` x ` C 7 3/31/14

8 9. (14 points) A rectangular box with a square base and top must hold 8 cubic feet. What dimensions will minimize the material used? You must verify that you have found the minimum. h h x x Let M denote the amount of material used to construct the box. Then M x ` 4xh. Because the box is to hold 8 cubit feet, x h 8. Consequently h 8 x. Thus Mpxq x ` 4x 8 x x ` 3x 1. Because x must be a dimension, x ą 0. Since for any choice of x ą 0 letting h 8 results in a box holding 8 cubic feet. Thus the domain of the function M is p0, 8q. To find the minimum value x of M, its derivative must be found. M 1 pxq 4x 3x 4 x3 8 4 px qpx ` x ` 4q. So M 1 ă 0 on p0, s and hence decreasing there. Also M 1 ą 0 on r, 8q and hence increasing there. Thus M has a minimum value at x The corresponding value for h is h 8. x x 10. Bonus Question (10 points) Let f be continuous on an interval I and differentiable on the interior of I. Do ONE of the following. (a) Use the Mean Value Theorem to prove that if f 1 pxq 0 for all x in the interior of I, then f is constant on I. Let a be any fixed point in I. It will be shown that for any x in I, fpxq fpaq; that is, f is constant. Let x be any point in I other than a. Apply the Mean Value Theorem to the closed interval with endpoints a and fpxq fpaq x. Then there is a c between a and x such that fpcq. By assumption f 1 pcq 0. Thus fpxq fpaq. x a (b) Use the Mean Value Theorem to prove that if f 1 pxq ą 0 for all x in the interior of I, then f is increasing on I. Let x 1 and x be in I with x 1 ă x. Apply the Mean Value Theorem to the interval rx 1, x s to get that there is a number c between x 1 and x such that fpx q fpx 1 q f 1 pcq. By assumption f 1 pcq ą 0. Because x x 1 x 1 ă x, the denominator of the fraction is positive. Thus the numerator must be positive; that is, fpx q ą fpx 1 q. 8 3/31/14