Math 126
Name:
Summer 2014
Score:
/34
Show all your work
Dr. Lily Yen
No Calculator permitted in this part. Read the questions carefully. Show all your work
and clearly indicate your final answer. Use proper notation.
Problem 1: Integrate the following.
Z
a. (x − 3)e−12x dx
Test 3
Integrating by parts,
Z
Z
−12x
−1 −12x
(x − 3)e
dx = (x − 3) 12 e
−
−1 −12x
e
12
dx
1
1 −12x
= − 12
(x − 3)e−12x − 144
e
+C
−12x
1
35 −12x
− 12 xe
+C
= 144 e
Score:
/3
Z
b.
6x + 4
dx
x2 − 1
A
B
Using partial fractions, if 6x+4
= x−1
+ x+1
, then 6x + 4 = (x + 1)A + (x − 1)B.
x2 −1
When x = 1 this yields that 10 = 2A, so A = 5. When x = −1, you get that
−2 = −2B, so B = 1. Therefore
Z
Z
5
1
6x + 4
dx =
+
dx
2
x −1
x−1 x+1
= 5 ln|x − 1| + ln|x + 1| + C
Score:
Z
c.
/3
ex cos(x) dx
Using
R x integration xby parts twice,
R x
R x
x
x
e cos(x) dx
=
e
cos(x)
+
e
sin(x)
dx
=
e
cos(x)
+
e
sin(x)
−
e cos(x) dx.
R x
x
x
Therefore 2 e cos(x) dx = e cos(x) + e sin(x), so
Z
ex cos(x) dx = 12 ex (cos(x) + sin(x)) + C.
Score:
/4
Problem 2: Does the following integral converge? If so, evaluate it. If not, show where
convergence fails.
Z π/2
cot(θ) dθ
0
First note that the integrand, cot(θ), is defined for all θ in the half-open interval (0, π2 ].
1
Then substitute u = sin(θ). Then du
= cos(θ), so dθ = cos(θ)
, and
dθ
Z
a
b
Z sin(b)
cos(θ)
1
cot(θ) dθ =
dθ =
du
a sin(θ)
sin(a) u
sin(b)
= ln|u|
= ln|sin(b)| − ln|sin(b)|
Z
b
sin(a)
Now limb→π/2 ln|sin(b)| = ln(1) = 0, but lima→0+ ln|sin(b)| = ∞.
R π/2
R π/2
Therefore 0 cot(θ) dθ is divergent, 0 cot(θ) dθ = +∞.
Score:
/4
Problem 3: Evaluate analytically the arc length of the astroid
x2/3 + y 2/3 = a2/3 ,
a > 0.
If x = a cos3 (t) and y = a sin3 (t), then
x2/3 + y 2/3 = a2/3 cos2 (t) + a2/3 sin2 (t) = a2/3 (cos2 (t) + sin2 (t)) = a2/3 , so the curve can be
parameterized as x = a cos3 (t), y = a sin3 (t), 0 ≤ t ≤ 2π. Therefore the arc length is
Z
2π
q
dx 2
dt
Z
dy 2
dt
2π
q
2
dt =
(−3a cos2 (t) sin(t))2 + 3a sin2 (t) cos(t) dt
0
0
Z 2π q
=
9a2 cos4 (t) sin2 (t) + 9a2 sin4 (t) cos2 (t) dt
0
Z 2π q
=
3a cos2 (t) sin2 (t)(cos2 (t) + sin2 (t)) dt
0
Z 2π q
Z π/2 q
Z π/2
2
2
2
2
=
3a cos (t) sin (t) dt = 12a
cos (t) sin (t) dt = 12a
cos(t) sin(t) dt
0
0
0
Z π/2
Z π/2
π/2
= 6a
2 cos(t) sin(t) dt = 6a
sin(2t) dt = −3a cos(2t) = 6a.
+
0
0
0
Score:
Page 2
/4
Math 126
Math 126
Name:
Summer 2014
Show all your work
Dr. Lily Yen
Calculators permitted in this part.
Problem 4: Find the volume of the solid S obtained by rotating the region R bounded by
1
and the x-axis on the interval [2, 7] about the y-axis. Sketch
the function f (x) =
(x − 3)4/5
the region R and a cross-sectional volume element as part of your solution.
Test 3
y
4
3
2
1
−7
−6
−5
−4
−3
−2
−1
0
1
2
3
4
5
6
7
x
By
of cylindrical
shells, the volume is R
R 7 the method
R4
4
−4/5
2πx(x − 3)
dx = −1 2π(u + 3)u−4/5 du = 2π −1 u1/5 + 3u−4/5 du =
2
Ra
R4
2π lima→0− −1 u1/5 + 3u−4/5 du + 2π limb→0+ b u1/5 + 3u−4/5 du =
a
4
2π lima→0− 56 u6/5 + 15u1/5 + 2π limb→0+ 65 u6/5 + 15u1/5 =
−1
b
2π(0 − 56 + 15) + 2π( 56 · 46/5 + 15 · 41/5 − 0) =
85
π
3
+
110
π
3
· 41/5 ≈ 241.0
Score:
/6
Problem 5: Find the orthogonal trajectories of the families of curves where k is a parameter
and sketch at least three members of each family. Label each curve with its parameter value.
kx3 − y 2 = 0
Note that k =
y2
.
x3
Now, differentiating the given equation with respect to x yields that
2
3kx2
2y
3 y 3 x2
x
2y
3y
=
= 2x
.
3kx − 2yy = 0, so y =
dy
Therefore the orthogonal trajectories satisfy the separable differential equation dx
= − 2x
,
3y
R
R
2
2
y
3 2
x
so 3y dy = − 2x dx, so 2 y = −x2 + C, so C + 2C/3 = 1, which is a family of ellipses.
2
0
0
y
k=
1 1
10 , 2 , 1, 2, 4
C = 1,
√ √
2, 3, 2
x
Score:
Page 3
/4
Math 126
Problem 6: Capilano Lake is stocked with 2000 rainbow trout, and after 1 year, the population has grown to 4500. Assuming logistic growth with a carrying capacity of 20 000, find
the growth constant k (including units) and determine when the population will increase to
10 000.
= kP (M − P ), so
P , follows logistic growth with capacity M , then dP
dt
RIf a population,
R
1
1
A
B
dP = k dt. Using partial fractions, if P (M −P ) = P + M −P , then
P (M −P )
1 = A(M
P ) + BP , so if P R= 0, 1 = AM , so A = M1 ; and if P = M , 1 = BM , so B =
R −
Hence M1P + M (M1−P ) dP = k dt, so M1 ln|M P | − M1 ln|M (M − P )| = kt + C, so
ln MP−P = M (kt + C), so MP−P = ±eM (kt+C) = a1 ebt , where a = ±e−M C and b = M k are
suitable constants. Thus M1−1 = a1 ebt , so M
− 1 = ae−bt , so M
= 1 + ae−bt , so
P
P
1
.
M
P
P =
M
.
1 + ae−bt
000
, so
In the case of the trout, M = 20 000, P (0) = 2000, and P (1) = 4500. Thus 2000 = 201+a
20 000
40
81
20 000
−b
1 + a = 10, so a = 9; and 4500 = 1+ae−b , so 1 + 9e = 4500 = 9 , so b = ln( 31 ). You need
20 000
1
1
−bt
to find when P (t) = 10 000, so 10 000 = 1+9e
, so
−bt , so 2 = 1+9e−bt , so 2 = 1 + 9e
ln(9)
ln(9)
1
−bt
= e , so bt = ln(9), so t = b = ln(81/31) ≈ 2.29 years.
9
Score:
Page 4
/6
Math 126