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Midterm 3 Sample A Solutions
Loyola University Chicago
Math 162-003, Spring 2014
Problem 1.(10 points) Find the following integrals. Show work. Put a box around your final answer .
Z ∞
a.(5 points)
e−3x dx
2
Solution:
∞
Z
e−3x dx = lim
b→∞ 2
2
b.(5 points)
4
Z
1
Z
b
1
1
1
e−3x dx = lim − e−3b − − e−3·2 = e−6
b→∞ 3
3
3
1
dx
(x − 2)2/3
Solution: the function has a discontinuity/vertical asymptote at x = 2. Hence
Z
1
4
Z 4
1
1
dx
+
dx
2/3
(x
−
2)
(x
−
2)2/3
1
2
Z 4
Z b
1
1
dx
+
lim
dx
= lim
a→2+ a (x − 2)2/3
b→2− 1 (x − 2)2/3
1
dx =
(x − 2)2/3
=
Z
2
lim 3(x − 2)1/3|b1 + lim 3(x − 2)1/3|4a
a→2+
b→2−
| − 3(1 − 2)1/3 + lim 3(4 − 2)1/3 − 3(a − 2)1/3
a→2+
√
3
1/3
= −3(−1) + 3 2 = 3(1 + 2)
=
lim 3(b − 2)
1/3
b→2−
Problem 2.(10 points) Say whether the following integrals converge or not. Show work. Put a
box around your final answer .
a.(5 points)
Z
π/2
tan x dx
0
Solution: diverges!
Z
π/2
tan x dx =
0
b.(5 points)
Z
1
∞
2x
dx
3x + 1
lim − ln | cos x| |b0 = − lim ln(cos b) = ∞
b→π/2−
b→π/2−
Solution: converges! The integrand is positive and
x
Z ∞
Z ∞ x
Z ∞ x
2x
2
2
1
2
1
2
b
dx
<
dx
=
dx
=
lim
|
=
−
1
x+1
x
3
3
3
b→∞
ln(2/3)
3
ln(2/3)
3
1
1
1
Problem 3.(10 points total) A colony of bacteria is grown under ideal conditions in a laboratory,
so that the population increases exponentially in time. At the end of 3 hours there were 10,000
bacteria. At the end of 5 hours there are 40,000. How many bacteria were present initially?
Solution: Exponential growth: b(t) = b(0)ekt. Use data: 10000 = b(0)e3k, 40000 = b(0)e5k , divide
side by side, get 4 = e2k , ln 4 = 2k, k = .5 ln 4 = .5 ln 22 = ln 2. Hence b(t) = b(0)et ln 2 = b(0)2t. Use
data again, 10000 = b(0)23 so b(0) = 10000/8 = 1250.
Problem 4.(10 points) Find the equation of the line tangent to the implicitly given curve t =
ln(x − t), y = tet at the point corresponding to t = 0.
Solution: Set t = 0 to find x(0), y(0): 0 = ln(x − 0), so 0 = ln x, so x = 1; then y = 0e0 = 0, so
1
y = 0. Differentiate: 1 = x−t
(x0 − 1), y 0 = et + tet . Set t = 0 to get x0 (0), y 0 (0), and remember that
0
0
0
0
0
0
x(0) = 1, y(0) = 0: 1 = 1−0
( x − 1), so x = 2; then y = e + 0e = 1, so y = 1. Then
Use point slope formula: y − 0 = 21 (x − 1).
dy
dx
=
y0
x0
= 12 .
Problem 5.(10 points) Find parametric equations for the semicircle x2 + y 2 = 1, y > 0, using as
the parameter the slope t = dy/dx of the tangent to the curve at (x, y).
Solution: the slope changes from −∞ at the point (1, 0), through 0 at (0, 1), to ∞ at the point
(−1, 0). At each point of the circle, implicit differentiation gives 2xx0 + 2yy 0 = 0 and so t = dy/dx =
y 0 /x0 = −x/y. So x = −ty, plug to equation of circle to get (−ty)2 + y 2 = 1, solve for y, get
y 2 = t21+1 , and because y > 0, y = √t21+1 . Then x = √t−t
2 +1 . Answer:
x= √
−t
1
, y=√
,
t2 + 1
t2 + 1
t ∈ (−∞, ∞).