Download Exam 3 Solutions

Math 2143 - Brief Calculus with Applications
Test #3 - 2013.04.24
Solutions
1. Compute the following integral:
Z
∞
1
dx
x (ln(x))2
First, we do the substitution u = ln(x), with du = x1 dx to get
Z
Z
1
1
dx
=
du
2
x (ln(x))
u2
1
=−
u
1
=−
ln(x)
Thus,
a
Z ∞
1 1
dx = lim −
a→∞
x (ln(x))2
ln(x) e
e
e
1 = lim
a→∞ ln(x) a
1
= 1 − lim
a→∞ ln(a)
=1
e
2. Consider the the function f (x) =
1 2
21 x
for 1 ≤ x ≤ 4.
(a) Verify that f (x) is a probability density function?
First, f (x) ≥ 0 for 1 ≤ x ≤ 4, and secondly,
Z
1
4
4
1 2
1 3 x dx =
x 21
63 1
1
=
(64 − 1)
63
=1
(b) Find P (1 ≤ x < 3).
Z
3
1 2
x dx
1 21
3
1 3 =
x 63
P (1 ≤ x < 3) =
1
1
=
(27 − 1)
63
26
=
63
3. Find the particular solution to the initial value problem:
dy
= (y − 1)2 ex−1 , y(1) = 2
dx
This is a separable equation:
dy
= ex−1 dx
(y − 1)2
2
Integrating both sides gives
−
1
= ex−1 + c
y−1
Using y(1) = 2 gives:
−
1
=1+c
1
Solving for c yields c = −2. Thus we now have
−
1
= ex−1 − 2
y−1
Solving for y gives
y(x) = 1 +
1
2 − ex−1
Using common denominators, this is
y(x) =
3 − ex−1
2 − ex−1
4. Consider the function f (x, y) = ex/y .
(a) State the domain of f (x, y).
The domain is D = {(x, y) |y 6= 0 }
(b) Compute fx
fx =
∂ x/y
1
e
= ex/y
∂x
y
(c) Compute fy
x
∂ x/y
e
= − 2 ex/y
∂y
y
5. Consider the function g(x, y) = 6x2 + 6y 2 + 6xy + 36x − 5.
fy =
(a) Find the critical points for g(x, y).
First, we must find gx and gy :
gx = 12x + 6y + 36, gy = 12y + 6x
Setting each of these to zero gives the system of equations
(
12x + 6y + 36 = 0
12y + 6x
=0
The second equation gives x = −2y, which plugged into the first equation gives y = 2. Thus the only critical point
for g(x, y) is the point (−4, 2).
(b) Use the D-test to classify the critical point(s) found in part (a).
To use the D-test, we need gxx , gyy and gxy :
gxx = 12,
gyy = 12,
gxy = 6
Thus
D((−4, 2)) = 122 − 62 = 108 > 0
Since D > 0 and gxx (−4, 2) > 0, we have g(x, y) has a relative minimum at the point (−4, 2).
6. There are two possible max/mins for the function h(x, y) = x3 + 2xy + 4y 2 subject to the constraint x + 2y = 12,
find them.
3
First, we define the function F (x, y) = x3 + 2xy + 4y 2 − λ(x + 2y − 12). Next we compute Fx , Fy and Fλ :
Fx = 3x2 + 2y − λ,
Fy = 2x + 8y − 2λ,
Fλ = x + 2y − 12
Setting Fx = Fy = Fλ = 0 gives

2

=λ
3x + 2y
2x + 8y
= 2λ


x + 2y = 12
If we solve for λ in both equations 1 and 2 from above, and set them equal, we get 3x2 + 2y = x + 4y. Using the
third equation, we have 2y = 12 − x. Plugging this into the previous equation gives
3x2 + 12 − x = x + 2(12 − x)
Solving for x gives x = ±2, which gives y = 5 and y = 7 respectively. This this possible max/mins are at (2, 5) and
(−2, 7).
7. Given the region R = {(x, y) |2 ≤ x ≤ 3, 0 ≤ y ≤ 2 }, compute
ZZ
2
yex+y dR
R
We can do this either way:
Z
3
Z
2
2
2
Z
2
yex+y dy dx,
0
0
Z
3
2
yex+y dx dy
2
We do the first one here:
Z
2
3
Z
0
2
2
yex+y dy dx =
Z
2
3
2
1 x+y2 e
dx
2
0
Z
1 3 x+4
=
e
− ex dx
2 2
3
1
= ex+4 − ex 2
2
1 7
=
e − e3 − e6 − e2
2