Midterm Exam I
Math 1310 - Engineering Calculus I
October 3, 2014
Answer each question completely in the area below. Show all work and explain your reasoning. If the work
is at all ambiguous, it is considered incorrect. No phones, calculators, or notes are allowed. Anyone found
violating these rules will be asked to leave immediately. Point values are in the square to the left of the
question. If there are any other issues, please ask the instructor.
By signing below, you are acknowledging that you have read and agree to the above paragraph, as well as agree
to abide University Honor Code:
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Signature:
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Solutions
Question
Points
1
15
2
15
3
15
4
15
5
15
6
15
7
15
Total:
105
Score
Note: there are a total of 105 points available on the exam, but it will be graded out of 100. There is also a
difficult bonus question at the end of the exam that is worth trying if you have extra time.
Midterm Exam 1
Math 1310 - Engineering Calculus I
October 3, 2014
1. The voltage across a charging capacitor with capacitance C (a positive number) can be described by:
t
V (t) = VS (1 − e − RC ),
where VS is the (constant) applied voltage used to charge the capacitor and R (also a positive number) is a
series resistance.
10
5
(a) Find the inverse function of V (t). Use it to find the time it takes for the capacitor to reach half of its
applied voltage, that is, 21 Vs .
(b) Calculate the following limit:
lim V (t).
t→∞
Solution:
(a) It seems as though a lot of students got flustered by the physics words in this problem, but if you read
closely, they truly have no impact on the problem whatsoever. The problem is simply asking to invert
the function V (t). Recall our technique for solving for the inverse: call V (t) = Vx (x for input), which
is the left hand side, we attempt to solve for t in terms of Vx . Physically, it represents the time it takes
to reach a certain voltage. Algebraically:
t
Vx = VS (1 − e − RC )
t
Vx
= 1 − e − RC
VS
t
Vx
− 1 = −e − RC
VS
t
Vx
1−
= e − RC
VS
Vx
t
ln 1 −
=−
VS
RC
Vx
=⇒ t = −RC ln 1 −
.
VS
Thus, we can conclude V −1 (Vx ) = −RC ln 1 −
V
−1
(VS /2) = −RC ln 1 −
= −RC ln 1 −
VS
2VS
1
2
Vx
VS
. We can plug in V = VS /2 to yield:
= −RC ln (1/2) .
(b) Many of you answered ∞ to this question, but notice that R, C are positive numbers. Thus, as t → ∞,
−t/RC becomes a large, negative number, which means that e −t/RC → 0. Thus, the limit of V (t) is
VS .
2/9
/15 pts
Midterm Exam 1
Math 1310 - Engineering Calculus I
October 3, 2014
2. For each of the following functions below, state the following limits if they exist. If they do not, explain why.
Hint: you can base your argument off of the leading order terms.
5
√
√
43x + x −3
√
(a) lim √
.
x→∞ 9 x − 5 3 x + 4
5
(b)
5
x 3 − 3x 2 + 5x − 7
.
x→−∞
4x 2 − x + 1
x 2 − 3x 2 + 5x − 7
(c) lim
.
x→∞
4x 3 − x + 1
lim
Solution: In all three of these problems, you did not need to do any more work than identifying the leading
coefficients of each quotient.
(a) In this case, the leading order terms are not immediately obvious because the order of the terms is
shuffled, but reasoning through which term is the largest is not difficult. The two candidates are either
√
√
the x√or 3 x terms.
Which grows
√
√
√ faster? This is easy to check by plugging in values of x. Is
8 > 3 8? Yes. 8 ≈ 2.82 and 3 8 = 2. In general, x a grows faster than x b (as x → ∞) if a > b.
√
Here we have a = 1/2 and b = 1/3. Thus, the x terms determine our limit. As we have them in
both the numerator and denominator, the the ratio grows at a constant rate which is the ratio of the
√
x coefficients, yielding 1/9.
(b) In this case, we see that the leading order term in the numerator, x 3 is larger than that in the denominator, x 2 . Thus, we know that as x → −∞, the limit must blow up to ±∞, but is it +∞or −∞?
Note, as x → −∞, x 3 becomes a large, negative number and x 2 becomes a large (but smaller) positive
number. Thus, the ratio of a negative number and positive number is always negative, telling us the
limit is −∞.
(c) Here, notice the numerator grows slower than the denominator, thus the denominator dominates and
the limit becomes 0.
3/9
/15 pts
Midterm Exam 1
15
Math 1310 - Engineering Calculus I
October 3, 2014
3. For the following function, identify all horizontal and vertical asymptotes. Also, identify all discontinuities
and classify them as: removable, jump, or essential/infinite:
f (x) =
x2 − x
.
x 2 + 4x − 5
Hint: there are two asymptotes and two discontinuities.
Solution: The key to solving this problem is factoring the original equation:
f (x) =
x2 − x
x(x − 1)
=
.
x 2 + 4x − 5
(x + 5)(x − 1)
Notice the (x − 1) term in both the denominator and numerator. This means that x = 1 is a removable
discontinuity, as the limit exists from both sides, but the function is undefined at this point. On the other
hand, x = −5 is just an essential/infinite discontinuity as the function blows up near it, meaning it is also
a vertical asymptote. Thus, the last asymptote we need to identify describes the behavior as x → ∞,
which note, by the same logic as the previous problem is limx→∞ f (x) = 1, thus y = 1 is our last vertical
asymptote. A graph of the function can be seen below.
4/9
/15 pts
Midterm Exam 1
8
Math 1310 - Engineering Calculus I
October 3, 2014
4. (a) Consider the following function:
( 2
x −9
if x < 3
f (x) = x−3
.
3 + cx if x ≥ 3
For which value of c does f (x) become continuous? For full credit, your argument should involve limits
and the definition of continuity.
7
(b) Prove that the equation f (x) = x 3 + x 2 − 4 has a root in the interval (1, 2). For full credit, you must
state the theorem you are using.
Solution:
(a) We know that f (x) is a rational function when x < 3 and is therefore continuous and a polynomial
when x > 3 and is therefore continuous, but what about at x = 3? To be continuous, by definition,
we need:
lim f (x) = lim f (x) = lim f (x) = f (3)
x→3−
x→3+
x→3
In this case, we know the limit from the right is equal to f (3) = 3 + 3c. We need this to be equal to
the limit from the right. What is the limit from the right?
lim
x→3−
x2 − 9
(x − 3)(x + 3)
= lim
x −3
x −3
x→3−
(x
−
3)(x + 3)
= lim
x−
3
x→3−
= lim x + 3
x→3−
= 6.
Thus, for the limit on the left to match the limit on the right, we need 3 + 3c = 6, which implies that
c = 1.
(b) The implicit hint in this problem was that part (a) was about continuity, so part (b) is as well. Specifically, this is an application of the intermediate value theorem. To apply the intermediate value theorem,
we must first require that f (x) is continuous, which it definitely is because we have a polynomial. Thus,
we want to prove that there exists a c ∈ (1, 2) such that f (c) = 0. To do so, we must find that
either f (1) is negative and f (2) is positive or vice versa. In this case, plugging in 1 and 2 reveals
f (1) = −2 < 0 and f (2) = 8 > 0. Thus, by the intermediate value theorem, we are guaranteed a root
in the interval (1, 2).
5/9
/15 pts
Midterm Exam 1
7
Math 1310 - Engineering Calculus I
October 3, 2014
5. (a) The graph of f and three other functions, a, b, c are shown below. Which of the three choices, a, b, c
describes f 0 (x) and why?
f
(
x
)
a
b
c
8
(b) Below is the graph of h(t). On which intervals is h(t) continuous? Differentiable? Why?
Solution:
(a) To establish which function describes f 0 we examine three features of f : when it is increasing, decreasing, or flat. Notice when f is flat, both a, c are both zero, meaning they are possible candidates and
we can eliminate b immediately. Next, observe that f is increasing to the left of this point, meaning
our derivative must be positive, therefore identifying a as f 0 (x), as c is negative when f is increasing,
meaning it cannot be our derivative.
(b) First, to establish differentiability, we have three main problems: at t = 1 we have a corner, at t = 2 we
have a jump discontinuity, and at t = 3 we have a vertical line. Thus, h(t) is differentiable everywhere
but these three points. Continuity is a looser condition, as h(t) is continuous at t = 1 and t = 3 but
not at t = 2, thus we have continuity everywhere except t = 2.
6/9
/15 pts
Midterm Exam 1
Math 1310 - Engineering Calculus I
October 3, 2014
6. Consider the following function:
1
f (x) = √
x
10
(a) Compute the derivative of the following function using the limit definition of the derivative:
5
(b) State the power rule for derivatives of the form x n and use it to verify your answer to part (a).
Solution:
(a) Here, we just apply the limit definition and manipulate using some algebra:
f (x + h) − f (x)
h
h→0
√1
− √1x
x+h
= lim
h
h→0
f 0 (x) = lim
= lim
√ √
√x− x+h
√
x+h x
(common denominator)
h
√
√
x − x +h
= lim √ √
h→0 h x x + h
√
√
√
√
x − x +h
x + x +h
√
= lim √ √
·√
(multiply by conjugate)
h→0 h x x + h
x + x +h
x − (x + h)
√
= lim √ √
√
h→0 h x x + h
x + x +h
−h
√
= lim √ √
√
h→0 h x x + h
x + x +h
h→0
−h
√
= lim √ √
√
h→0 h
x + x +h
x x +h
−1
√
= lim √ √
√
h→0 1 x x + h
x + x +h
−1
= √ √ √
√ 1 x x
x+ x
1
√
=−
x(2 x)
1
= − 3/2 .
2x
While this looks scary, I’ve shown every algebra step in excruciating detail, which you did not necessarily
need to do. If you showed that you were on the right path for the problem and messed up the algebra,
you received a very hefty chunk of the points for the problem.
√
d
(b) Notice that we can rewrite 1/ x = x −1/2 , and the power rule states dx
(x n ) = nx n−1 , suggesting our
0
−3/2
derivative should be f (x) = −1/2x
, which is exactly the answer above.
7/9
/15 pts
Midterm Exam 1
15
Math 1310 - Engineering Calculus I
October 3, 2014
7. For the following function, compute the equation of the tangent line at the point (0, 1):
f (x) = x 2 e x + 4x + 1
Note, you do not need to use the limit definition of the derivative.
Solution: Here, we can compute f 0 (x) using any means, but the hopefully obvious solution is to apply the
d
d
(4x) = 4 and dx
(1) = 0 by the power rule, meaning we need only consider the
product rule. We know dx
first term, by the product rule:
d 2 x
d
d 2 x
(x e ) =
(x )e + x 2 (e x ) = 2xe x + x 2 e x = (x 2 + 2x)e x .
dx
dx
dx
Thus, our derivative in total becomes:
f 0 (x) = (x 2 + 2x)e x + 4.
We know, in general, the equation of a tangent line through the point (x0 , y0 ) is:
y − y0 = f 0 (x0 )(x − x0 ),
and here we want to consider x0 = 0, y0 = 1. Plugging in x0 = 0, we find f 0 (0) = 4, thus our tangent line
equation becomes:
y − 1 = 4(x − 0) =⇒ y = 4x + 1.
8/9
/15 pts
Midterm Exam 1
10
Math 1310 - Engineering Calculus I
October 3, 2014
8. (Bonus) Consider the following limit:
√
1 + cx 2 − 1 + x 2
.
lim
x→0
x4
For which value of c is the limit finite (that is, a number, not ±∞)? What is the limit in this case?
Hint: multiply by the conjugate and try to choose c to cancel terms that would blow up to ±∞.
Solution: This problem is difficult for algebra reasons, but not necessarily conceptually difficult. Applying
the limit definition:
√
√
√
1 + cx 2 − 1 + x 2
1 + cx 2 − 1 + x 2 1 + cx 2 + 1 + x 2
√
lim
= lim
·
(multiplying by the conjugate)
x→0
x→0
x4
x4
1 + cx 2 + 1 + x 2
(1 + cx 2 )2 − (1 + x 2 )
√
= lim
x→0 x 4 (1 + cx 2 + 1 + x 2 )
(1 + 2cx 2 + c 2 x 4 − 1 − x 2 )
√
= lim
(expanding out the numerator)
x→0
x 4 (1 + cx 2 + 1 + x 2 )
(2c − 1)x 2 + c 2 x 4
√
= lim
x→0 x 4 (1 + cx 2 + 1 + x 2 )
2c − 1)x 2
c 2x 4
√
√
= lim
+ lim
x→0 x 4 (1 + cx 2 + 1 + x 2 )
x→0 x 4 (1 + cx 2 + 1 + x 2 )
2c − 1)x 2
c 2x 4
+ lim
√
√
x→0 x 4 (1 + cx 2 + 1 + x 2 )
x→0 x 4 (1 + cx 2 + 1 + x 2 )
2c − 1
c2
√
√
= lim
+ lim
.
x→0 x 2 (1 + cx 2 + 1 + x 2 )
x→0 1 + cx 2 + 1 + x 2
= lim
Let’s examine each term separately. First, the rightmost term:
c2
c2
√
√
=
= c 2,
2
x→0 1 + cx 2 + 1 + x 2
1 + c(0) + 1 + 0
lim
thus, this limit is finite as long as c is finite, meaning this term is fine.
What about the other term? Notice, if c 6= 1/2, the x 2 in the denominator of always causes this term to
blow up to ∞. Thus, we need the first term to cancel, meaning 2c − 1 = 0 =⇒ c = 1/2. In this case,
the limit of the first term is 0 and the limit of the second term is c 2 = 1/4.
9/9
/10 pts