STAT 201(01)
Calculus for Statistics
Spring 2017
Midterm — Solution
Instructions: There are 8 problems worth 100 points in total. In order to receive full credit, your answer
must be complete, legible and correct. Show all of your work and give adequate explanations.
1. [12 points] Determine the following limits:
tan 3x
.
2x2 + 5x
sin(x + π/4) − 1
(b) lim
.
x − π/4
x→π/4
(a) lim
x→0
(Hint: You may use sin(a + b) = sin(a) cos(b) + cos(a) sin(b))
(Sol)
(a) limx→0
1
3x
sin 3x
3x cos 3x 2x2 +5x
=1·1·
3
5
= 53 .
(b) Let y = x − π/4. Then,
limπ
x→ 4
sin(x + π4 ) − 1
sin(y + π2 ) − 1
cos y − 1
= lim
= lim
π
y→0
y→0
x− 4
y
y
sin2 y
y
(cos y − 1)(cos y + 1)
= − lim
=0
2
y→0 y
y→0
y(cos y + 1)
cos y + 1
= lim
You may use L’Hopital’s rule to obtain the same result.
2. [12 points] Compute the following integrals:
Z 1 √
√
4
5
(a)
( x5 + x4 )dx
0
Z
(b)
0
π/4
√
sin 2x
dx
1 + cos 2x
(Sol)
√
R1 √
R1 5
4
4
5
(a) 0 ( x5 + x4 )dx = 0 x 4 + x 5 dx = 1.
(b) Let y = cos 2x + 1 then sin 2xdx = − 12 dy with x = (0, π4 ) corresponding to y = (2, 1), respectively.
√
R π/4 sin 2x
R2
Then, 0 √1+cos
dx = 12 1 y −1/2 dy = 2 − 1.
2x
2
x sin x1 , if x 6= 0,
3. [12 points] Let f (x) =
0,
if x = 0.
(a) Find f 0 (0).
(b) When x 6= 0, determine f 0 (x).
(c) Does f 00 (0) exist? If so, please find. If not, give reason to support your argument.
(Sol)
(a) f 0 (0) = limx→0
as x → 0.
f (x)−f (0)
x−0
= limx→0 x sin x1 = 0. The last equality holds because |x sin x1 | ≤ |x| → 0
(b) When x 6= 0, x2 and sin x1 are differentiable. Thus, f (x) = x2 sin x1 is differentiable and f 0 (x) =
2x sin x1 − cos x1 .
1
STAT 201(01)
Calculus for Statistics
Spring 2017
0
0
(0)
= limx→0 2 sin x1 − x1 cos x1 , which does not converge. To see this, let
(c) f 00 (0) = limx→0 f (x)−f
x−0
1
. As n → ∞, xn → 0, but 2 sin x1n − x1n cos x1n = −2nπ diverges to −∞. So, the limit
xn = 2nπ
does not exist.
4. [12 points] Use implicit differentiation to find (a) dy/dx and (b) the tangent line at (2, 2) of the graph
of the function 2x3 − 3y 2 = 4. Also, (c) find the second derivative d2 y/dx2 at (2, 2).
(Sol) (a) Implicit differentiation of 2x3 − 3y 2 = 4 gives x2 dx − ydy = 0 and y 0 = dy/dx = x2 /y. (b) At
dy
d 0
d x2
point (2, 2), dx
|(x,y)=(2,2) = 2 and the tangent line at (2, 2) is y = 2(x − 2) + 2. (c) y 00 = dx
y = dx
(y)=
2xy−x2 y 0
y2
and y 00 |(x,y)=(2,2) = 0.
5. [12 points] A particle is moving along the curve y 3 +xy 2 +1 = 0. As it passes through the point (−2, 1),
its x-coordinate x = x(t) is increasing at the rate of 2 units in time t. How fast is the y-coordinate y = y(t)
changing at this point?
(Sol) Again, apply implicit differentiation to y 3 + xy 2 + 1 = 0 with respect to t, which gives 3y 2 dy
dt +
dy
y 2 dx
dt + 2xy dt = 0. At (x, y) = (−2, 1),
dx
dt
= 2 and thus
dy
dt
=
−y 2
dx
3y 2 +2xy dt
= 2.
R1
6. [12 points] Let 0 ≤ a ≤ 1. Find the value of a such that 0 |x2 − ax|dx achieves its maximum.
ax − x2 , 0 ≤ x ≤ a
(Sol) For 0 ≤ a ≤ 1, |x2 − ax| =
. Then
x2 − ax, a ≤ x ≤ 1
Z
1
Z
a
1
Z
1 def
1 3 1
a − a + = f (a).
3
2
3
0
0
a
√
It follows from f 0 (a) = a2 − 12 = 0 that local max/min occurs at a = 1/ 2. We need to compare
√
1
f (0), f (1), f (1/ 2), which are 13 , 16 , and 31 − 3√
, respectively. Thus, the maximum takes place at a = 0
2
1
with the maximum f (0) = 3 .
|x2 − ax|dx =
(ax − x2 )dx +
(x2 − ax)dx =
1
7. [14 points] Let f (x) = 1+x
2 , x ∈ R. Consider a line L = P Q by picking two points P : (a, f (a)) and
√
Q : (b, f (b)) in the graph of this function. Let m denote the slope of L. Show that |m| ≤ 3 3/8.
(Sol) Note f is differentiable on R with f 0 (x) = −2x(1 + x2 )−2 . Given P and Q, the slope of the line L
(a)
is m = f (b)−f
. By the mean value theorem, there is a c ∈ (a, b), such that m = f 0 (c) = −2c/(c2 + 1)2 .
b−a
2
)
0
As c → ±∞, m → 0. Since m0 = f 00 (c) = − 2(1−3c
(c2 +1)3 = 0, m = f (c) attains its maximum or minimum
at c = ± √13 , that is, f 0 (− √13 ) =
√
3 3
8 ,
√
and f 0 ( √13 ) = − 3 8 3 . Thus, |m| ≤
√
3 3
8 .
8. [14 points] Suppose that lim an = 3 and lim bn = 2 for sequences {an } and {bn } for n = 1, 2, ....
n→∞
n→∞
Using the definition of the limit of a sequence, prove that lim an bn = 6.
n→∞
(Sol) Suppose limn→∞ an = 3 and limn→∞ bn = 2. Then, for any > 0, we can find some integer
values Na and Nb such that (i) n ≥ Na implies |an − 3| < and (ii) n ≥ Nb implies |bn − 2| < . It
suffices to show that for n large enough, |an bn − 6| becomes arbitrarily small. This is true, because when
n ≥ max(Na , Nb )
|an bn − 6| = |(an − 3)(bn − 2) + 3(an − 3) + 2(bn − 2)|
≤ |an − 3||bn − 2| + 3|an − 3| + 2|bn − 2| (by triangle inequality)
< 2 + 3 + 2 < 6.
2