COLUMBIA UNIVERSITY
Math S1101D
Calculus I
Summer 2013
2nd Midterm Exam
06.20.2013
Instructor: S. Ali Altug
Name:
Question:
1
2
3
4
5
6
7
Total
Points:
5
8
6
6
8
9
0
42
Score:
Instructions:
• There are 7 questions on this exam.
• Please write your name and UNI on top of every page.
• In order to get full credit you need to answer the first 6 questions correctly.
• The last question is a bonus question, and you do not have to answer it.
• Unless otherwise is explicitly stated SHOW YOUR WORK in every question.
• Write neatly, and put your final answer in a box.
• No calculators, cell phones, books, notebooks, notes or cheat sheets are allowed.
1. Determine whether the following statements are true (T) or false (F). You do not need to justify
your answer for this question.
(a) (1 point) If a function is differentiable at a point then it is continuous at that point.
√
(b) (1 point) The linear approximation to 4.002 is 2.
(c) (1 point) If c is a point in the domain of f (x) such that either f 0 (c) = 0 or f 0 (c) does not exist,
then f (c) is a local maximum or a minimum.
(d) (1 point) If limx→5
f (x)
(x−5)2
= 2 then limx→5 f (x) = 0.
(e) (1 point) If f (x) is one-to-one then f 0 (x) is one-to-one too.
Solution:
(a) True. (We have proved this in class)
√
(b) False. The linear approximation to 4.002 is 2.0005.
(c) False. For example f (x) = x3 satisfies f 0 (0) = 0 but 0 is neither a local max nor a min.
(d) True.
(x − 5)2
x→5
(x − 5)2
f (x)
= lim
lim (x − 5)2
x→5 (x − 5)2 x→5
=0
lim f (x) = lim f (x)
x→5
(e) False. Once again f (x) = x3 is a counter example.
Page 2
2. In each part, find f 0 (x) for the given function f (x).
(a) (2 points) f (x) = cos(2x) tan(x + 4)
(b) (2 points) f (x) = ln(ln(x))
(c) (2 points) f (x) =
1
sin(ln(x))
(d) (2 points) f (x) =
sin(x) cos(x)ex
x(x+1)(x2 +1)(x3 +1)2
Solution:
(a) By product rule and the chain rule
f 0 (x) = −2 sin(2x) tan(x + 4) + cos(2x) sec2 (x + 4)
(b) By the chain rule,
(ln(x))0
ln(x)
1
=
x ln(x)
f 0 (x) =
(c) By the quotient rule and the chain rule
−(sin(ln(x)))0
(sin(ln(x)))2
− cos(ln(x))
=
x(sin(ln(x)))2
f 0 (x) =
(d) Taking ln of both sides we get
sin(x) cos(x)ex
ln(f (x)) = ln
x(x + 1)(x2 + 1)(x3 + 1)2
= ln(sin(x)) + ln(cos(x)) + ln(ex ) − ln(x) − ln(x + 1) − ln(x2 + 1) − 2 ln(x3 + 1)
= ln(sin(x)) + ln(cos(x)) + x − ln(x) − ln(x + 1) − ln(x2 + 1) − 2 ln(x3 + 1)
Differentiating both sides then gives
f 0 (x)
6x2
cos(x)
sin(x)
1
1
2x
− 3
=
−
+1− −
− 2
f (x)
sin(x)
cos(x)
x x+1 x +1 x +1
Hence
0
f (x) = f (x)
=
cos(x)
sin(x)
1
1
2x
6x2
−
+1− −
− 2
− 3
sin(x)
cos(x)
x x+1 x +1 x +1
sin(x) cos(x)ex
x(x + 1)(x2 + 1)(x3 + 1)2
sin(x)
1
1
2x
6x2
cos(x)
−
+1− −
− 2
− 3
sin(x)
cos(x)
x x+1 x +1 x +1
Page 3
3. Let C be the curve implicitly given by
xy = ln(ex + y)
(a) (1 point) Show that the point (0, 0) is on the curve C.
(b) (3 points) Find y 0 .
(c) (2 points) Find the equation of the tangent line to C at the point (0, 0).
Solution:
(a)
0 · 0 = ln(1 + 0) = 0
(b) By implicit differentiation and the product rule
(ex + y)0
ex + y
x
e + y0
y + xy 0 = x
e +y
0
x
(y + xy )(e + y) = ex + y 0
y + xy 0 =
y(ex + y) − ex = y 0 − xy 0 (ex + y)
y(ex + y) − ex
= y0
1 − x(ex + y)
(c) The slope, m, of the tangent line to C at (0, 0) is the value of the derivative, y 0 , at the
point (0, 0). Using part (b) we find the slope to be
0−1
1−0
= −1
m=
Therefore the equation of the tangent line is
y − 21
= −1
x+1
1
y = −x −
2
Page 4
4. (6 points) A paper cup has the shape of a cone with height 10 cm and radius 3 cm (at the top). If water
is poured into the cup at a rate of 2 cm3 /s, how fast is the water level rising when the water is 5 cm
deep? (The volume formula for a cone of radius r and height h is 31 πr2 h.)
Solution:
We are given that the rate of change of V with respect to time is 2 cm3 /sec. i.e.
dV
=2
dt
We are asked
dh
dt
when h = 5cm. By the chain we have
dV
dV dh
=
dt
dh dt
(?)
dV
d 1
2
We need to find dV
dh . By the volume formula we have dh = dh ( 3 πr h). Note that r and h depend
on each other and we need to find the relation between them in order to differentiate the right hand
side.
Since the triangles ABC and ADE are similar, we have
we get
h
r
=
AD
DE
1 2
πr h
3
2
1
3h
= π
h
3
10
3π 3
h
=
100
V =
Therefore
dV
dh
=
9π 2
100 h .
Substituting this back into (?) we get
2=
9π 2 dh
dV
=
h
dt
100 dt
Finally at h = 5 we have
2=
9π dh
4 dt
Therefore we get
8
dh
=
cm/sec
dt
9π
Page 5
=
AB
BC
=
10
3 .
Therefore r =
3h
10 ,
and
5. Find the following limits.
(a) (2 points) limx→∞
sin2 (x)
x
(b) (2 points) limx→0+ (sin(x))x
1
(c) (2 points) limx→0+ xe cos(x)
(d) (2 points) limx→∞
√2x−1
x2 +5
Solution:
(a) Note that sin2 (x) oscillates between −1 and 1 when x approaches ∞ (i.e. it actually oscillates
between 0 and 1 but all we need is that it is bounded) and the denominator is going to infinity.
We have
0 ≤ sin2 (x) ≤ 1
sin2 (x)
1
≤ 2
x
x
1
sin2 (x)
≤ lim 2
⇒ lim 0 ≤ lim
x→∞ x
x→∞
x→∞
x
sin2 (x)
⇒ 0 ≤ lim
≤0
x→∞
x
Then by the Squeeze Theorem we get
⇒ 0≤
sin2 (x)
=0
x→∞
x
lim
(b) This is an indeterminacy of the form 00 (just try substituting in), therefore we can not directly
substitute x = 0.
Since for x > 0, x = eln(x) we have (since sin(x) > 0 for x > 0 and close to 0)
(sin(x))x = ex ln(sin(x))
On the other hand note that ex is continuous everywhere, therefore the limit exists if and only if
the limit limx→0+ x ln(sin(x)) exists, and if limx→0+ x ln(sin(x)) = L then limx→0+ (sin(x))x =
eL .
We will now compute limx→0+ x ln(sin(x)). Direct substitution shows that this is an indeterminacy of the form 0 · (−∞). We will rewrite the product as a quotient to apply L’Hospital’s
rule.
ln(sin(x))
lim x ln(sin(x)) = lim+
1/x
x→0+
x→0
Now this last limit is indeterminate of the form
lim
x→0+
−∞
∞
so we can apply L”Hospital’s rule and get,
ln(sin(x))
cos(x)/ sin(x)
= lim+
1/x
−1/x2
x→0
2
−x cos(x)
= lim+
sin(x)
x→0
Finally this last limit is also indeterminate of the form
L’Hospital’s rule once again and get
lim+
x→0
0
0
(by direct substitution). We apply
−x2 cos(x)
−2x cos(x) + x2 sin(x)
= lim
sin(x)
cos(x)
x→0+
=0
Therefore we have
lim (sin(x))x = elimx→0+ x ln(sin(x)) = e0 = 1
x→0+
Page 6
(c) By direct substitution we get,
1
lim+ xe cos(x) = 0
x→0
(d) Taking into x parenthesis we get
lim
x→∞
√2x−1
x2 +5
= lim
x(2−1/x)
√
x→∞ |x|
=2
Page 7
1+5/x2
6. Let f (x) =
x3
3
−
x2
2
− 2x + 3.
(a) (2 points) Find the critical point(s) of f (x) (if there are any).
(b) (2 points) Determine the interval(s) where f (x) is increasing and/or decreasing.
(c) (2 points) Determine the interval(s) where f (x) is concave up and/or concave down.
(d) (1 point) Find the inflection point(s) of f (x) (if there are any).
(e) (2 points) Find the absolute minimum and/or absolute maximum value(s) of f (x) in the interval
[−3, 3].
Solution:
(a) The derivative is
f 0 (x) = x2 − x − 2 = (x − 2)(x + 1)
The critical points are when the derivative is 0 or does not exist. Since the derivative is a
polynomial its domain is all of R, the critical points are x = −1, 2.
(b) We need to check the sign of the derivative on


> 0
f 0 (x) = < 0


>0
the intervals (−∞, −1), (−1, 2), (2, ∞). Since
if x < −1
if −1 < x < 2
if x > 2
We get that
(
is increasing on (−∞, −1) ∪ (2, ∞)
f (x) is
decreasing on (−1, 2)
(c) The second derivative of f (x) is
f 00 (x) = 2x − 1
Which is positive if x >
1
2
and negative if x < 12 . Therefore
(
concave down on (−∞, 12 )
f (x) is
concave up on ( 12 , ∞)
(d) The inflection points of f (x) are when f (x) changes from concave up to down or from concave
down to up. By part (c) the only inflection point is at x = 12 .
(e) We need to check the value of f (x) at the critical points inside the interval [−3, 3] and at the
end points of the interval. By part (a) the critical points are at x = −1, 2. We therefore need
to compute f (−3), f (−1), f (2), f (3).
f (−3) = −4.5
f (−1) = 25/6
f (2) = −1/3
f (3) = 1.5
Therefore the minimum value of f (x) in [−3, 3] is −4.5 and the maximum value in [−3, 3] is
25/6.
Page 8
7. (4 points (bonus)) This is the bonus question. You do not have to answer this question to get
full credit from the exam. Each part is worth 2 points.
(a) Compute the following limit without using L’Hospital’s rule. (You will not get any points if
you use L’Hospital’s rule at any step.)
p
lim
x→0
1 + tan(x) −
x3
p
1 + sin(x)
(b) Let f (x) be a function that is differentiable twice (i.e. The second derivative f 00 (x) exists.) such
that f (x) has three distinct roots (i.e. There are three numbers x1 , x2 , x3 with xi 6= xj for i 6= j
such that f (x1 ) = f (x2 ) = f (x3 ) = 0.). Show that f 00 (x) has at least one root.
Solution:
(a) Recall that in the class we showed that
lim
x→0
sin(x)
=1
x
(?)
p
We will use this to compute the limit. We first multiply and divide the fraction by 1 + tan(x)+
p
1 + sin(x). This gives
p
p
p
p
p
p
1 + tan(x) − 1 + sin(x)
1 + tan(x) − 1 + sin(x) 1 + tan(x) + 1 + sin(x)
p
p
lim
= lim
x→0
x→0
x3
x3
1 + tan(x) + 1 + sin(x)
= lim
x→0
p
x3 (
tan(x) − sin(x)
p
1 + tan(x) + 1 + sin(x))
1
Since tan(x) − sin(x) = sin(x)( cos(x)−1)
we see that
1
sin(x)( cos(x)
− 1)
tan(x) − sin(x)
p
p
p
p
= lim
3
3
x→0 x ( 1 + tan(x) +
1 + sin(x)) x→0 x ( 1 + tan(x) + 1 + sin(x))
lim
1
sin(x)
cos(x) − 1
p
p
x→0
x x2 ( 1 + tan(x) + 1 + sin(x))
= lim
1
sin(x)
cos(x) − 1
p
p
lim
x→0
x x→0 x2 ( 1 + tan(x) + 1 + sin(x))
= lim
1
cos(x)
−1
p
p
x→0 x2 ( 1 + tan(x) +
1 + sin(x))
= lim
Where we used (?) in in the last line. Now this is still indeterminate of the form 00 (direct
1
substitution). Multiply and divide the above by cos(x)
+ 1. Using sin2 (x) + cos2 (x) = 1, this
gives
1
cos(x)
1
−1
cos(x) − 1
p
p
p
p
lim
= lim
x→0 x2 ( 1 + tan(x) +
1 + sin(x)) x→0 x2 ( 1 + tan(x) + 1 + sin(x))
1
cos(x)
1
cos(x)
+1
+1
1
cos2 (x)
−1
p
p
1
x→0 x2 ( 1 + tan(x) +
1 + sin(x))( cos(x)
+ 1)
= lim
= lim
x→0
sin2 (x)
cos2 (x)
p
p
1
+ 1)
x2 ( 1 + tan(x) + 1 + sin(x))( cos(x)
sin2 (x)
1
p
p
1
x→0
x2 cos2 (x)( 1 + tan(x) + 1 + sin(x))( cos(x)
+ 1)
= lim
= lim
x→0
cos2 (x)(
Page 9
p
1 + tan(x) +
1
p
1
1 + sin(x))( cos(x)
+ 1)
Where we used (?) once again in the last line. Now this last limit can be computed to be
direct substitution. Therefore we have shown
p
p
1 + tan(x) − 1 + sin(x)
1
=
lim
3
x→0
x
4
1
4
by
(b) This is a consequence of Rolle’s theorem. Because the f (x1 ) = f (x2 ) = 0, f (x) is continuous
and differentiable everywhere, Rolle’s Theorem implies that there is a c1 ∈ (x1 , x2 ) such that
f 0 (c1 ) = 0. Applying the same argument to the interval [x2 , x3 ] we see that there is another
c2 ∈ (x2 , 23 ) such that f 0 (c2 ) = 0. Furthermore note that since (x1 , x2 ) ∩ (x2 , x3 ) = ∅, x6 =
x2 . Now we apply Rolle’s theorem to f 0 (x) on [c1 , c2 ] noting that since f 00 (x) exists, f 0 (x) is
differentiable (and hence continuous) everywhere. The result follows.
Page 10