FINAL EXAM 2, DECEMBER 2011 SOLUTIONS Directions and rules. The exam will last 2 hours and 15 minutes. No electronic devices of any kind will be allowed, with one exception: a music player that nobody else can hear, and whose controls you do not use during the exam (just put it on shuffle). Anything (else) with an off switch must be off. In particular, turn your cell phone off. There are 16 questions on the exam, counting equally. 1. Below is the graph of a function f (x). From the graph, read off the value (if any) of the following limits: lim f (x) = 0

x→−1−

lim f (x) = −1

x→−1+

lim f (x) = −∞

x→0−

lim f (x) = 0

x→0+

lim f (x) = 1

x→1−

1

2

FINAL EXAM 2, DECEMBER 2011 SOLUTIONS

2. James Bond drives his Aston-Martin car off a cliff on a twisting mountain road. He falls under the influence of gravity for fifteen seconds, and then, his car sprouts wings and becomes an airplane. The wings enable him to decrease his speed linearly while he descends to the sea below. Just in time, his car sprouts skis with which it can land on water, and he makes a smooth landing (which means there is no bump as he hits the water). “No bump” means no sudden change in velocity. In this problem, ignore his horizontal movement, and graph his vertical acceleration, vertical velocity, and height, as functions of time, on three separate graphs. Align the graphs vertically so that events that take place at the same time line up vertically. Solution: We start with the acceleration, which is a negative constant for fifteen seconds. So his (vertical) velocity during this period is a linear function, sloping downward from its initial value of zero. When the wings come out, his velocity is linear, but this time sloping upwards, because the velocity has to return to zero at the moment he lands. The speed decreases, but the velocity, which is negative, increases to zero. After he lands, of course, the velocity and acceleration are both zero. So the height is for the first 15 seconds a downwards-opening parabola, and then it is an upwards-opening parabola, with its lowest point at height zero (when he lands on the sea). Here are the graphs, with height y at the top followed by velocity y 0 and acceleration y 00 . The acceleration has to be a positive constant while he’s using the wings, to make the speed decrease linearly. Ignore the little circles in the graph.

FINAL EXAM 2, DECEMBER 2011 SOLUTIONS

3

3. Evaluate the following limits. lim (x2 + 5x + 1) = 32 + 5 · 3 + 1 = 25

x→3

since polynomials are continuous

(x + 2)(x + 1) x2 + 3x + 2 = lim = lim x + 2 = 1 x→−1 x→−1 x→−1 x+1 x+1 limx→π/2 sin x sin x 1 2 lim = = = x lim x π/2 π x→π/2 x→π/2 lim

4. (a) Fill in the blanks to make the following correct: p To prove (from the definition of limit) that 4 |x| approaches 0 as x approaches zero: Let  > 0 be given. Then choose δ = 4 p Then we verify that if |x| < δ then |4 |x|| < . p (b) Draw a picture showing part of the graph of y = 4 |x| and labeling  and δ. Here’s a similar-looking graph (with cube root instead of fourth root, but it wouldn’t look much different when sketched by hand). I don’t have a way to draw  and δ on this digital picture. It would just be a rectangle with center at the origin and two vertices on the curve, and the width is 2δ and the height is 2.

4

FINAL EXAM 2, DECEMBER 2011 SOLUTIONS

5. For each of the following functions, if the function is not continuous at every x, at which x is it not continuous? If it is continuous everywhere then just write “continuous everywhere.” x3 − 1 x2 − 1 The denominator is zero when x = ±1 so the function is not continuous there. p | sin x| Continuous everywhere. 1 tan x Discontinuous where tan x is zero, namely at odd multiples of π/2. ecos x Continuous everywhere 1 x Discontinuous at x = 0. (There are two reasons: the function is undefined, and the limit x sin

doesn’t exist). 6. Evaluate the following limits. √ √ lim (3 x − 4 x) =

x→∞

= =

lim (x1/3 − x1/4 )

x→∞

lim x1/4 (x1/12 − 1)

x→∞

lim x1/4 lim x1/12 − 1

x→∞

x→∞ 1/12

= ∞( lim x x→∞

− lim 1) x→∞

= ∞(∞ − 1) = ∞ 2t18 − 1 2t18 = lim = lim = 2 t→∞ t18 − t t→∞ t18 2→∞ since the leading terms determine the limit at ∞ of a rational function lim

FINAL EXAM 2, DECEMBER 2011 SOLUTIONS

5

7. Below is the graph of a function. Sketch the graph of its derivative on the second, blank graph given.

Every student knows how to sketch this graph, I think, so I did not spend the time required to make a digital image. 8. Evaluate the following derivatives d 87 (x + 2x + 1) = 87x86 + 2 dx d d d ((x187 + 3x8 )(sin x + 1)) = (x187 + 3x8 ) (sin x + 1) + (sin x + 1) (x187 + 3x8 ) dx dx dx = (x187 + 3x8 ) cos x + (sin x + 1)(187x186 + 24x7 ) d ex dx sin x

= =

d 1 4 dx x + 1

=

d x d sin x dx e − ex dx sin x 2 sin x ex sin x − ex cos x sin2 x d − dx (x4 + 1) 4x3 = (x4 + 1)2 (x4 + 1)2

d 2 d d (x tan x) = x2 tan x + tan x x2 = x2 sec2 x + 2x tan x dx dx dx

6

FINAL EXAM 2, DECEMBER 2011 SOLUTIONS

9. Evaluate the following derivatives d d sin x e = esin x sin x = esin x cos x dx dx √ d √ d sin √x = esin x (sin x) e dx dx √ √ d√ sin x cos( x) = e x dx   √ √ 1 sin x cos( x) √ = e 2 x √ √ 1 √ esin x cos x = 2 x

d d sin(ln x) cos(ln x) = sin(ln x) ln x = dx dx x √ 10. Suppose x2 + sin y = 1. Calculate dy/dx using implicit differentiation, getting an answer that involves both x and y. Solution : √ x2 + sin y = 1 d√ y = 0 2x + 2 cos y dx 2 cos y dy 2x + √ = 0 2 y dx dy dx

=

√ 2x x 2x3/2 = cos y cos y

11. At a place called Roopkund in the Indian Himalayas, more than five hundred human skeletons were found. Medical experts say they died by being caught without shelter in a hailstorm with large hailstones. Although these bones were known for decades, they were only radiocarbon dated in 2004. Assume the half-life of carbon 14 is 5730 years. The researchers found that 87% of the original carbon 14 was still present. What was the date of the fatal hailstorm? Here are some useful numbers: log2 (0.87) = −0.2009; ln(2) = 0.693; ln(0.87) = −0.1393. Please do the arithmetic required to get an actual date. The only acceptable answer is an integer for the year of the hailstorm. You only need to be accurate within ten years– radiocarbon dating is not more accurate than that anyway.

FINAL EXAM 2, DECEMBER 2011 SOLUTIONS

7

 t/5730 1 = 0.87 2 2−t/5730 t − 5730 t 2004 − 1146

= 0.87 = log2 0.87 = −0.2009 = 5730 · 0.2009 = 1146 years from death to dating = 858 the date of the hailstorm

12. At noon, ship A is 150 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 35 km/h. How fast is the distance between the ships changing at 4:00 p.m.? (a) Draw a diagram of this situation. Define three variables (other than t for time) and indicate their meaning on the diagram. B 

z 





y

  

A

x x is the distance of A from the point where B was at noon; y is the distance of B from where B was at noon; and z is the distance between the two planes. (b) What, in symbols, is the unknown quantity? dz/dt at 4 pm. (c) Write equations connecting the variables that is good all the time from noon to four pm. x2 + y 2 = z 2 x = 150 − 35t y = 35t (d) Finish solving the problem. dx dy + 2y dt dt dx dy x +y dt dt dz x(−35) + y · 35 = z dt 2x

dz dt dz = z dt = 2z

8

FINAL EXAM 2, DECEMBER 2011 SOLUTIONS

Now at 4 pm, x = 150 − 35 · 4 = 10 and y = 35 · 4 = 140, so z =

√

102

+

1402

q = 140 1 +

1 14

1 and the square root of 1 + 14 is very close to 1 + 1/28, or a little less that 1 + 1/25 = 1.04; so at 4 pm, z is about 1.04 times 140, or 145.6.

10(−35) + 140 · 35 =

√

197

dz dt dz dt

=

35 · 130 145.6

This is looking good, because if ship A were standing still and due south of B then B would be sailing away at 35 mph, but there are two reasons the distance is decreasing slower than that, because A is still a little bit west of B’s path, and because A is still sailing towards B’s path. So it’s good that we’re getting and answer that is 35 times a fraction a bit less than one. Now 130 is 15.6 less than 145.6, which is about ten percent less than a tenth, so 130/145.6 is about 0.91, so the answer is 35 · 0.91 = 31.2. By calculator I obtain 31.25 mph. 13. One side of a right triangle is known to be 20 cm long and the opposite angle is measured as 30◦ , with a possible error up ±1◦ . (a) Estimate the error in computing the length of the hypotenuse. Let z be the length of the hypothenuse. We have 20 z z sin θ = 20 dz sin θ + z cos θdθ = 0 sin θ =

side opposite over hypotenuse

√ Now a triangle with a 30◦ angle has sides in the ratio of 1 to 3/2 and the hypotenuse √ is twice the side opposite the 30◦ angle, so z = 40, and sin θ = 1/2, and cos θ = 3/2. Putting those values in we have √ 1 3 dz + 40 dθ = 0 2 2 √ dz = −40 3dθ √ ∆z ∼ = −40 3∆θ Now ∆θ is one degree, which is π/180 radians. Hence √ π ∆z ∼ = −40 3 √ 180 2π 3 = − 9 = 1.21 cm

FINAL EXAM 2, DECEMBER 2011 SOLUTIONS

9

(b) What is the percentage error? ∆z z

1.21 40 = 0.0302 = 3.02%

=

3.11 Hyperbolic functions 14. Find the exact numerical value of sinh(ln 10). Show your work, and be sure that your answer is a number. 1 ln 10 sinh(ln 10) = (e − e− ln 10 ) 2 1 ln 10 = (e − eln(1/10) ) 2 1 1 (10 − ) = 2 10 = 9.9/2 = 4.95

10

FINAL EXAM 2, DECEMBER 2011 SOLUTIONS

15. Consider the function f (x) = x3 − 72 x2 + 2x + 1. You can show your work below in the space provided for it, but please put your answers right after the questions where they will be easy to find. (a) For which x is f (x) increasing? (−∞, 1/3) and (2, ∞). (b) For which x is f (x) decreasing? (1/3, 2) (c) Find the values of x and f (x) at the local minima of f . (2, −1). 95 (d) Find the values of x and f (x) at the local maxima of f . ( 13 , 54 )

(e) Find the points of inflection of f on the given interval. x = 7/6 (f) Where is f concave up? (7/6, ∞) (g) Where is f concave down ? (−∞, 7/6) (h) Sketch the graph of f showing all the minima, maxima, and points of inflection.

f 0 (x) = 3x2 − 7x + 2 = (3x − 1)(x − 2) and f 00 (x) = 6x − 7, so f 00 changes sign at x = 7/6, and is positive on (7/6, ∞) and negative on (−∞, 7/6). That lets us answer (e), (f) and (g). The critical points (where f 0 is zero) are x = 1/3 and x = 2; at x = 1/3, f 00 is negative, so that is a local maximum, and at x = 2, it is positive, so that is a local minimum. We have 7 f (1/3) = (1/3)3 − (1/3)2 + 2/3 + 1 = (2 + 21 + 18 + 54)/54 = 95/54 2 and f (2) = 8 − 14 + 4 + 1 = −1. Finally, f 0 is negative on (1/3, 2) and positive elsewhere, which lets us answer (a) and (b).

FINAL EXAM 2, DECEMBER 2011 SOLUTIONS

11

16. Evaluate the following limits:  lim

x→1

x x − x2 − 1 x − 1





x − x(x + 1) x→1 x2 − 1 2 −x = lim 2 x→1 x − 1 =



lim

The numerator approaches 1 and the denominator approaches 0. Since the denominator changes sign at 1, the limit is “undefined” rather than ∞ or −∞. Too bad we can’t describe this situation by saying the limit is ±∞, but that is not the custom, so “undefined” is the right answer. lim

x→∞

ln ln x ln x

Here both numerator and denominator go to ∞, so we use L’Hospital’s rule. ln ln x x→∞ ln x lim

=

d dx ln ln x d x→∞ dx ln x

lim

1/(x ln x) x→∞ 1/x 1 = lim x→∞ ln x = 1/ lim ln x

=

lim

x→∞

= 1/∞ = 0 sinh 5x x→0 x Both numerator and denominator go to zero, so we use L’Hospital’s rule: lim

sinh 5x x→0 x lim

= =

lim

x→0

d dx

sinh 5x dx dx

lim 5 cosh 5x = 5 cosh 0 = 5

x→0