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Mathematics 105 — Calculus I
Exam 1
February 13, 2009
Your Name: Solution Guide
There are 6 total problems in this exam. On each problem, you must show all your work, or otherwise thoroughly explain your conclusions. There is always at least one step preceding a final answer. Units may be requested for your final answer; a point deduction will apply if they are omitted.
On the portion of the exam marked N C, you will be allowed 30 minutes during which your calculator must be closed and put away. If you finish this section early, you may hand in your work early. However, only after you hand in the ”no calculators” section will you be permitted to use a calculator. You may not return to the ”no calculator” portion after handing it in.
Before beginning, ensure your calculator is set to Radians mode.
You will have 80 minutes to complete this exam.
Question Point Value Your Score
No Calc. 50
Total 150
N C P
Math 105-A (Salomone) Exam 1 Show all your work!
Name:
Score (50 possible):
Problem 1-NC. (25 points) Use the limit definition of derivative to compute f ′(x) for the function
f (x) = x ex.
Hint: Simplify using properties of exponentials. You will also need to know that lim h→ 0 e
h (^) − 1 h =^ 1.
The definition of f ′(x), if it exists, is that it takes the value given by the limit
lim h→ 0
f (x + h) − f (x) h
= lim h→ 0
(x + h)ex+h^ − xex h
= lim h→ 0
xex+h^ + hex+h^ − xex h
= lim h→ 0
xexeh^ + hexeh^ − xex h
= lim h→ 0
xex
eh^ − 1
= lim h→ 0
xex^
eh^ − 1 h
Now we just need to determine the behavior of the highlighted terms above. We are given in the problem that the first term approaches 1. The second term does as well, since eh^ is a continuous function of h and this allows us to substitute h = 0. Thus
f ′(x) = lim h→ 0
f (x + h) − f (x) h
= lim h→ 0
xex^
eh^ − 1
︸︷︷︸^ h → 1
+ex^ ︸︷︷︸eh → 1
= xex^ + ex.
Math 105 Exam 1 Solutions Problem 1. (25 points) This problem concerns the function
g(t) =
(t − 3)(t^2 − t + 1) t^2 − 4 t + 3
(a) (8 points) Determine the domain of this function.
This function will be defined anywhere, so long as its denominator is not zero: t^2 − 4 t + 3 , 0 (t − 3)(t − 1) , 0 t − 3 , 0 and t − 1 , 0 t , 3 and t , 1 The domain consists of all real numbers except 1 and 3: (−∞, 1) ∪ (1, 3) ∪ (3, ∞) or
{ t ∈ R : t , 1 and t , 3
} .
(b) (10 points) Using algebra, compute lim t→a
g(t) for each value of a not in the domain of g. Explain what each result
means about the continuity of g.
According to the result of part (a), we must compute lim t→ 1
g(t) and lim t→ 3
g(t). We begin by factoring the denominator: g(t) =
(t − 3)(t^2 − t + 1) (t − 3)(t − 1)
t^2 − t + 1 t − 1
(t , 3)
We can use this simpler form to compute the limits we want; after all, limits only see the behavior of g(t) the t values in question.
lim t→ 1
g(t) = lim t→ 1
t^2 − t + 1 t − 1
∼
The graph of g(t) has a vertical asymptote at t = 1.
lim t→ 3
g(t) = lim t→ 3
t^2 − t + 1 t − 1
=
The graph of g(t) has a hole at t = 3.
2
1 2
1
3 4
4
3
y = g(t)
t
(c) (7 points) At left is a partial graph of g(t). Fill in the gap, clearly indicating the nature of any discontinuities.
Math 105 Exam 1 Solutions
Problem 2. (25 points) A weight is attached to a spring and suspended in a container of motor oil. If it is allowed to oscillate, its vertical position (measured in cm above equilibrium) as a function of time t in seconds might be given by the function
p(t) = 3 e−t^ cos t.
(a) (10 points) Complete the data table below, and use your results to estimate the values of p′(0.9), p′(1), and p′(1.1). Include units in your answers.
t (sec) 0.85 0.9 0.95 1 1.05 1.1 1.
p (cm) 0.8462 0.7582 0.6749 0.5963 0.5224 0.4530 0.
Estimate the derivatives using average rates of change:
p′(0.9) ≈
p(0.95) − p(0.85)
- 95 − 0. 85 ≈ 0.^6749 0 −.1 sec^0 .8462 cm
= −^0 .1713 cm 0 .1 sec = − 1. 713 cmsec
p′(1) ≈
p(1.05) − p(0.95)
- 05 − 0. 95 ≈ 0.^6749 −^0 .8462 cm 0 .1 sec = −^00 .1525 cm.1 sec
= − 1. 525 cmsec
p′(1.1) ≈
p(1.15) − p(1.05)
- 15 − 1. 05 ≈ 0.^3880 −^0 .5224 cm 0 .1 sec = −^00 .1344 cm.1 sec
= − 1. 344 cmsec
(b) (10 points) Use your answers to part (a) to estimate p′′(1), with units. What does this answer mean in practical terms?
Again, we’ll use an average rate of change — but not the rate of change of p, rather the rate of change of p′.
p′′(1) ≈
p′(1.1) − p′(0.9)
- 1 − 0. 9 ≈
− 1. 344 −−^1 .713 cm/sec 0 .2 sec =^
0 .369 cm/sec 0 .2 sec =^1.^845
cm/sec sec
(c) (5 points) Is it reasonable, based on your answers, to expect that p(t) satisfies the differential equation
p′′^ + 2 p′^ = − 2 p?
Why or why not?
We cannot predict this for all values of t, since this would require us to compute p′^ and p′′^ — but we don’t know how to do that just yet. (This will require something called the ”product rule.”) Instead, let’s see whether our estimates from parts (a) and (b) indicate that this differential equation is satisfied at t = 1. The left-hand side gives
p′′(1) ≈ 1. 845
- 2 p′(1) ≈ 2(− 1 .525) p′′(1) + 2 p′(1) ≈ 1. 845 + 2(− 1 .525) ≈ − 1. 205
Meanwhile, the right-hand side is
− 2 p(1) ≈ −2(0.5963) = − 1. 1926
These numbers are reasonably close to one another; within roundoff and estimation error, the differential equation appears to be satisfied at t = 1. (Answers may vary.)
Math 105 Exam 1 Solutions
Problem 4. (25 points) The latest press booklet for the 2009 Lotus Exige S-240 sports car claims it can accelerate from 0—60 mph in 4.0 seconds flat. A recent test-track run showed that under full throttle, the velocity of the car is modeled by the function
v(t) = 40
t − 5 t,
where v is measured in mph and t in seconds.
(a) (15 points) According to this model, after how many seconds will the car reach its maximum velocity, and what is the maximum velocity?
Note: do this symbolically, showing your work. You may include a graph or data table if you wish, but your answer must be exact.
f f’ f’’^ We wish to find a local maximum of^ v(t) — a point at which^ v′(t)^ =^ 0 and^ v′′(t) is negative.
v′(t) = 40 · 12 t−^1 /^2 − 5 = 0 20 √ t
= 5 √ t = 4 t = 16
v′′(t) = 20 · − 21 t−^3 /^2
= − 10 t^3 /^2 v′′(16) = − (^16103) / 2
= − 1064 < 0
According to this work, the car’s velocity has a local maximum after t = 16 seconds. The car’s velocity at this point will be v(16) = 40
√ 16 − 5(16) = 80 mph.
(b) (10 points) Determine the car’s distance function d(t) — an antiderivative of its velocity — and use it to find the distance the car traveled during the first 10 seconds of this time trial.
Note: write out the units of the antiderivative in your answer. Convert them if you wish.
An antiderivative of our function v(t) will be
d(t) = 40 ·
t^3 /^2 3 / 2
t^2 2
+C
t^3 /^2 −
t^2 +C
Does the +C matter? Not if we only care about how far the car travels between t = 0 and t = 10, as measured by the difference d(10) − d(0). (The +C will cancel out.) We may as well drop it, or assume that d(0) = 0. We then interpret d(t) to be the distance the car has traveled since t = 0, and the total distance during the first 10 seconds of the test will be merely
d(10) =
(10)^3 /^2 −
(10)^2 ≈ 593 .274 mph · sec
The units of this answer will be the product of the units of input and output of v — namely, mph·sec. To convert, we remember that there are 3600 seconds in one hour:
mi hr
· sec = 593. 274
mi 3600 sec
· sec ≈ 0 .1645 mi.
