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Mathematics 105 — Calculus I
F E S
December 11, 2008
Your Name:
There are 9 problems in this exam. Please complete only eight , clearly crossing out the one you do not wish to be graded. If you do not clearly indicate, your highest-scoring problem will be omitted! (Follow directions and avoid the grader’s wrath.) Extra credit will not be awarded for attempting a ninth problem.
On each problem, you must show all your work, or otherwise thoroughly explain your conclusions. Except where noted, include at least one complete English sentence in every solution. Units may be requested for your final answer; a point deduction will apply if they are omitted or incorrect.
You may use a calculator on any portion of this exam. However, this does not excuse you from showing your work at any time. Your calculating skills are at issue on this exam, not your calculator’s.
You will have two hours to complete this exam.
Question Possible Your Score
Total 200
xkcd.com/499 or, why I don’t believe in Scantron exams
Problem 2. (25 points) Shown below is a graph of the derivative of a function T. Answer each of the following questions; on this page only, you do not need to provide explanations except where indicated.
(a) (3 points) T is decreasing on what interval(s)?
(b) (3 points) T is concave up on what interval(s)?
(c) (3 points) T has a local minimum at what point(s)?
t = − 1
(Here, T switches from decreasing to increasing — even though T′^ doesn’t cross through zero!)
(d) (3 points) T has an inflection at what point(s)?
t = 6
(e) (3 points) How can you tell at which point(s) T(x) = 0? Explain.
In short, we cannot without more information! If all we know is the derivative T′, this only tells us the shape of the curve T(x) and not its location. T is just an antiderivative of T′, so if we do not know which antiderivative it is, we cannot answer this question.
x
y
2 3 4
T’
1
(f) (10 points) On the axes provided, sketch a possible graph of the function T. Do not worry about the scale on the vertical axis, only the major features of the graph’s shape.
x
y
T
1 2 3 4
Problem 3. (25 points) For the function g(x) =
x:
(a) (10 points) Complete the table of data below, and use average rates of change to speculate on the value of the derivative g′(0). (Do not compute g′(x) directly!)
x 0 0.001 0.01 0.1 1 g(x) 0 0.0316 0.1 0.316 1
Let’s look at the average rates of change on, say, the intervals [0, 0 .1], [0, 0 .01], and [0, 0 .001]:
g(0.1) − g(0)
- 1 − 0
g(0.01) − g(0)
- 01 − 0
g(0.001) − g(0)
- 001 − 0
You could make several predictions here, but the most tempting is to say that these average rates of change appear to be getting bigger and bigger. We suspect that the closer we get to x = 0, the larger the average rate of change gets. We speculate that g′(0) becomes infinite.
(b) (15 points) Using the limit definition of derivative, determine g′(0). Explain how this connects with your answer
from part (a).
If g′(0) exists, it is equal to the limit
lim h→ 0
g(0 + h) − g(0) h
= lim h→ 0
0 + h −
h
= lim h→ 0
h h
= lim h→ 0
h
∼
0 +^
This confirms that the closer h gets to zero, the larger the average rate of change on the interval [0, h] gets, without bound. The derivative g′(0) blows up to infinity (so it does not exist as a number).
Problem 5. (25 points)
(a) (10 points) The curve graphed below has equation sin(xy) + x^2 + y^2 = 2. Find an equation for a line tangent to this curve at the point (x, y) = (0. 827 , 0 .827).
We only need two things to determine an equation for a line: its slope and any point it passes through. The point is easy: (x, y) = (0. 827 , 0 .827). The slope is given by the derivative dy/dx. Using implicit differentiation with respect to x,
cos(xy)
y + x
dy dx
dy dx
dy dx
x cos(xy) + 2 y
= −y cos(xy) − 2 x
dy dx
y cos(xy) + 2 x x cos(xy) + 2 y
Evaluating at x = 0 .827 and y = 0 .827 gives
dy dx
0 .827 cos(0. 8272 ) + 2(0.827) 0 .827 cos(0. 8272 ) + 2(0.827)
Thus the tangent line has slope −1 and passes through (0. 827 , 0 .827), so its equation is
y − 0. 827 = −1(x − 0 .827) or y = 1. 654 − x.
(b) (15 points) As x → 0, both of the functions f (x) = ex
2 −1 and g(x) = cos(x^2 )−1 approach zero. Precisely determine which gets to zero more quickly, by computing an appropriate limit.
To judge this footrace, we compute the limit lim x→ 0
f (x) g(x)
. If this limit is zero, f has won the race. If it is infinite, g has won the race. If it is neither, the functions have ”tied.” Recall that l’H ˆopital’s rule assists us whenever a limit has the indeterminate form 00. Each use of l’H ˆopital’s rule is marked with a 00 below.
lim x→ 0
ex
2 − 1 cos(x^2 ) − 1
(^00) = lim x→ 0
2 xex
2
− 2 x sin(x^2 )
= lim x→ 0
ex
2
− sin(x^2 )
∼
0 −^
Since the limit is infinite, g(x) = cos(x^2 ) − 1 wins the race: it approaches zero strictly faster than f (x) does.
Problem 6. (25 points) After this exam is over, you plan to jump in a chartered Learjet-31 to fly home for the holidays. It’s currently parked on the tarmac at Lewiston/Auburn’s airport and waiting to take you to Las Vegas, a flight of 2400 miles.
Unfortunately, both the jet and the fuel it burns are expensive. The Learjet costs $1500 an hour to charter.
Furthermore, if it flies at an airspeed of x mph, it consumes
100 + x
2 900
gallons of jet fuel per hour — and jet fuel costs $5 per gallon.
What is the smallest total cost for this trip? Hint: how does the total time of the trip relate to the airspeed x?
Step 1: State your objective. We wish to find the minimum of the total cost function. We will use the airspeed x as the independent variable. Step 2: Visualize success. Here, we’ll use a quick diagram to organize the two parts of the total cost function. If h is the number of hours the flight will take, the total cost is made up of...
cost of charter and cost of fuel $1500 per hour for h hours, a total of
$ 1500h
( 100 + 900 x^2
) gallons of fuel per hour, at $5 each, for h hours, a total of
$ 5
( 100 +
x^2 900
) h
Therefore the total cost function can be written as
C = 1500 h + 5
( 100 +
x^2 900
) h =
( 2000 +
x^2 180
) h.
Step 3: Eliminate the competition. We have one too many variables: we need to either eliminate h or eliminate x. Either choice will work: for fun, let’s eliminate h. What relationship exists between the total time of the flight h and the airspeed x? The total distance, 2400 miles, is the product of the two:
2400 = hx so h =
2400 x
.
Substituting this into the total cost function gives us the single-variable function
C(x) =
( 2000 +
x^2 180
) (^2400) x
=
4800000 x
40 3
x.
Step 4: Optimize. The stationary points of C will occur when C′(x) = 0. Where is this?
C′(x) = −
4800000 x^2
40 3
= 0 4800000 x^2
=
40 3 x^2 = 360000 x = ±
√ 360000 = ± 600.
Since x, our airspeed, ought to be positive, we choose x = 600 mph. Is this a local minimum? A quick check of the second derivative shows it is, since C is concave up for all positive values of x: C′′(x) =
9600000 x^3
so C′′(600) =
2 45
Thus the cost is minimized at an airspeed of x = 600 mph. At this airspeed, the cost of the flight is
C(600) =
4800000 600
40 3
(600) = 16000
The cheapest possible charter to Vegas, then, will cost $16,000.
Problem 8. (25 points) Compute the following antiderivatives and definite integrals.
(a) (9 points) All antiderivatives of
x^2
x
− sin x
x
+ 2 ln
∣x
∣ + cos x+C
(b) (8 points) The area function
∫ (^) x
0
t^2 + 2 t^ dt
x^3
2 x
ln 2
(c) (8 points) The definite integral
− 1
18 x^8 − 2 dx
2 x^9 − 2 x
] 4
− 1
Problem 9. (25 points) While you have been taking this exam, a tiny device implanted in the ceiling has been measuring your happiness every 10 minutes. The results of its measurements are shown in the table below.
minutes since exam began t 0 10 20 30 40 50 60 70 80 90 h(t) 9.5 6.4 4.4 3.5 2.9 2.7 2.6 3.1 4.6 6. happiness points
(a) (15 points) Use the trapezoidal rule with 9 subintervals to estimate the integral
0
h(t) dt.
The trapezoidal approximation is the average of left- and right-hand approximations. We’ll compute each. With 9 subintervals on the interval [0, 90], each must have a width of ∆t = 909 − 0 = 10. Luckily for us, the data in the table are already spaced apart by this step size. The left- and right-hand sums are then
L 9 =
∑^8
n= 0
height ︷ ︸︸ ︷ h(10n)
width ︷︸︸︷ (10)
= 9 .5(10) + 6 .4(10) + 4 .4(10)
- 3 .5(10) + 2 .9(10) + 2 .7(10)
- 2 .6(10) + 3 .1(10) + 4 .6(10) = 397
R 9 =
∑^9
n= 1
height ︷ ︸︸ ︷ h(10n)
width ︷︸︸︷ (10)
= 6 .4(10) + 4 .4(10) + 3 .5(10)
- 2 .9(10) + 2 .7(10) + 2 .6(10)
- 3 .1(10) + 4 .6(10) + 6 .9(10) = 371
So the trapezoidal approximation is the average
T 9 =
L 9 + R 9
(b) (5 points) What are the units of your answer to part (a), and how can this integral be interpreted?
The units of an integral — being an area on the graph of a function — are the product of the units of input and units of output. So here, the units are ”happiness·minutes”. We might interpret the answer ”384 happiness minutes” as being the total amount of accumulated happiness you experienced while taking this exam.
(c) (5 points) If you are convinced that h(t) is a concave up function, what can you say about the approximation in part (a)?
The trapezoidal approximation will always overestimate the integral of a concave up function, since concave up functions lie beneath their secant lines. We conclude that your actual total accumulated happiness during the exam is probably less than 384 happiness·minutes.
