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MAS113 CALCULUS II SPRING 2008, QUIZ 4 SOLUTIONS
Quiz 4a Solutions
(1) Find the area of the surface obtained by rotating the curve y = x3 /6 + 1/(2x), 1/2 ≤
x ≤ 1 about the x-axis.
We have y ′ = x2 /2 − 1/(2x2 ). Therefore,
sµ
s
¶2
µ ¶2 r 4
x2
1
1
1
1
dy
x
x2
=
+ + 4 =
+ 2
+ 2.
=
1+
dx
4
2 4x
2
2x
2
2x
Therefore, the surface area is
µ 3
¶µ 2
¶
µ 6
¶1
Z 1
x
x
1
1
x2
8
x
263
2π
+
+ 2 dx = 2π
+
− 2
π.
=
6
2x
2
2x
72
6
x 1/2 256
1/2
(2) Solve the differential equation y ′ = −y + 5 with the initial condition y(0) = y0 .
The integration factor here is
µ(t) = exp
Therefore, we have
µZ
¶
1dt = et .
d ¡ t ¢
e y = 5et =⇒ y = 5 + ce−t .
dt
y(0) = y0 gives that c = y0 − 5. Consequently, the solution is
y(t) = 5 + (y0 − 5)e−t .
(3) Determine the values of r for which the differential equation y ′ + 2y = 0 has solutions
of the form y = ert .
y = ert gives that y ′ = rert . We must have
rert + 2ert = 0, or r + 2 = 0.
Therefore, we have r = 2.
Quiz 4b Solutions
(1) Find the length of the curve y 2 = 4x, 0 ≤ y ≤ 2.
We have 1 + (dx/dy)2 = y/2. Therefore, the length of the curve is
Z 2r
³ y ´2
dy.
1+
2
0
Making the change of variables u = y/1, we get that the above integral is
µ
¶1 √
Z 1√
√
√
√
1
2 1
u
1 + u2 du = 2
1 + u2 + ln(u + 1 + u2 ) =
2
+ ln(1 + 2).
2
2
2
2
0
0
1
(2) Solve the differential equation y ′ = −2y + 5 with the initial condition y(0) = y0 .
The integration factor here is
µ(t) = exp
Therefore, we have
µZ
¶
2dt = e2t .
d ¡ 2t ¢
5
e y = 5e2t =⇒ y = + ce−2t .
dt
2
y(0) = y0 gives that c = y0 − 5/2. Consequently, the solution is
¶
µ
5
5 −2t
e .
y(t) = + y0 −
2
2
(3) Determine the values of r for which the differential equation y ′′ − y = 0 has solutions of
the form y = ert .
y = ert gives that y ′ = rert and y ′′ = r2 ert . We must have
r2 ert − ert = 0, or r2 − 1 = 0.
Therefore, we have r = ±1.
Quiz 4c Solutions
(1) Find the area of the surface obtained by rotating the curve x = a cosh(y/a), −a ≤ y ≤ a
about the y-axis.
We have
¶2
dx
y
y
1+
= 1 + sinh2 = cosh2 .
dy
a
a
Therefore, the surface area is
¶
Z a
Z a
Z aµ
y
y
2y
2 y
cosh dy = 2πa
1 + cosh
2π
a cosh cosh dy = 4πa
dy
a
a
a
a
−a
0
0
¶a
µ
¶
µ
2y
1
a
2
= 2πa 1 + sinh 2 .
= 2πa y + sinh
2
a 0
2
µ
(2) Solve the differential equation y ′ = −2y + 10 with the initial condition y(0) = y0 .
The integration factor here is
µ(t) = exp
Therefore, we have
µZ
¶
2dt = e2t .
d ¡ 2t ¢
e y = 10e2t =⇒ y = 5 + ce−2t .
dt
y(0) = y0 gives that c = y0 − 5. Consequently, the solution is
y(t) = 5 + (y0 − 5)e−2t .
(3) Determine the values of r for which the differential equation
t2 y ′′ + 4ty ′ + 2y = 0
has solutions of the form y = tr for t > 0.
y = tr gives that y ′ = rtr−1 and y ′′ = r(r − 1)tr−2 . Therefore, we have
t2 r(r − 1)tr−2 + 4trtr−1 + 2tr = 0 or 0 = r(r − 1) + 4r + 2 = r2 − 3r + 2 = (r − 1)(r − 2).
Therefore, we have r = 1 or r = 2.
Quiz 4d Solutions
(1) Find the length of the curve y = ln(cos x), 0 ≤ x ≤ π/3.
In the original print out of the quiz the formula of the curve was
y 2 = ln(cos x)
which is a mis-print. The intended problem is as given here. You will be
given full credit so long as you set up the integral correctly.
We have
s
1+
µ
dy
dx
¶2
=
s
µ
sin x
1+ −
cos x
¶2
= sec x.
Therefore, the arc length is
¶
µ
Z π/3
1 √
π/3
+ 3 .
sec xdx = ln |sec x + tan x|0 = ln
2
0
(2) Solve the differential equation y ′ = 0.5y − 450 with the initial condition y(0) = 850.
The integration factor here is
µ(t) = exp
Therefore, we have
µZ
¶
1
− dt = e−t/2 .
2
d ¡ −t/2 ¢
e
y = −450e−t/2 =⇒ y = −450 + cet/2 .
dt
y(0) = 850 gives that c = 1300. Consequently, the solution is
y(t) = −450 + 1300et/2 .
(3) Determine the values of r for which the differential equation y ′′ + 2y ′ − 3y = 0 has
solutions of the form y = ert .
y = ert gives that y ′ = rert and y ′′ = r2 ert . We must have
r2 ert + 2rert − 3ert = 0, or r2 + 2r − 3 = (r + 3)(r − 1) = 0.
Therefore, we have r = 1 or r = −3.
Quiz 4e Solutions
(1) Find the area of the surface obtained byg rotating the curve x = 1 + 2y 2 , 1 ≤ y ≤ 2
about the x-axis.
We have
1+
µ
dx
dy
¶2
= 1 + 16y 2 .
Hence the surface area is
µ
¶
Z 2 p
√ ´
¢3/2 2
π 2¡ 2
π ³ √
2
y 1 + 16y dy =
2π
16y + 1
65 65 − 17 17 .
=
16 3
24
1
1
(2) Solve the differential equation y ′ = y − 5 with the initial condition y(0) = y0 .
he integration factor here is
µ(t) = exp
Therefore, we have
µZ
¶
−1dt = e−t .
d ¡ −t ¢
e y = −5e−t =⇒ y = −5 + cet .
dt
y(0) = y0 gives that c = y0 + 5. Consequently, the solution is
y(t) = −5 + (y0 + 5)et/2 .
(3) Determine the values of r for which the differential equation y ′′ + y ′ − 6y = 0 has solutions of the form y = ert .
y = ert gives that y ′ = rert and y ′′ = r2 ert . We must have
r2 ert + rert − 6ert = 0, or r2 + r − 6 = (r + 3)(r − 2) = 0.
Therefore, we have r = 2 or r = −3.
Quiz 4f Solutions
(1) Find the length of the curve y = x2 /2 − (ln x)/4, 2 ≤ x ≤ 4.
We have y ′ = x − 1/(4x) and we see that
µ ¶2
¶2
µ
dy
1
1
1
2
1+
=x + +
.
= x+
dx
2 16x2
4x
Therefore, the arc length is
¶
µ 2
¶4
Z 4µ
ln x
x
ln 2
1
+
.
dx =
=6+
L=
1+
4x
2
4 2
4
2
(2) Solve the differential equation y ′ = 2y − 5 with the initial condition y(0) = y0 .
he integration factor here is
µ(t) = exp
µZ
¶
−2dt = e−2t .
Therefore, we have
d ¡ −2t ¢
e y = −5e−2t =⇒ y = −5 + ce−2t .
dt
y(0) = y0 gives that c = y0 + 5. Consequently, the solution is
y(t) = −5 + (y0 + 5)et/2 .
(3) Determine the values of r for which the differential equation y ′′ − 13y ′ + 12y = 0 has
solutions of the form y = ert .
y = ert gives that y ′ = rert and y ′′ = r2 ert . We must have
r2 ert − 13rert + 12ert = 0, or r2 − 13r + 12 = (r − 1)(r − 12) = 0.
Therefore, we have r = 1 or r = 12.
Quiz 4g Solutions
(1) Find the area of the surface obtained by rotating the curve y = 1 − x2 , 0 ≤ x ≤ 1 about
the y-axis.
We have y ′ = −2x. Therefore, the surface area is
Z 1
´
³√
p
√
1
2πx 1 + 4x2 dx = π (1 + 4x2 )0 = π
5−1 .
0
(2) Solve the differential equation y ′ = 2y − 10 with the initial condition y(0) = y0 .
he integration factor here is
µ(t) = exp
µZ
¶
−2dt = e−2t .
Therefore, we have
d ¡ −2t ¢
e y = −10e−2t =⇒ y = −10 + ce−2t .
dt
y(0) = y0 gives that c = y0 + 10. Consequently, the solution is
y(t) = −10 + (y0 + 10)et/2 .
(3) Determine the values of r for which the differential equation
t2 y ′′ − 4ty ′ + 4y = 0
has solutions of the form y = tr for t > 0.
y = tr gives that y ′ = rtr−1 and y ′′ = r(r − 1)tr−2 . Therefore, we have
t2 r(r − 1)tr−2 − 4trtr−1 + 4tr = 0 or 0 = r(r − 1) − 4r + 4 = r2 − 5r + 4 = (r − 1)(r − 4).
Therefore, we have r = 1 or r = 4.
Quiz 4h Solutions
(1) Find the length of the curve y = ex , 0 ≤ x ≤ 1.
We have y ′ = ex . Therefore, the arc length is
Z √1+e2
Z 1p
1 + (ex )2 dx = √
u2
du.
u2 − 1
2
0
√
The above arrives from the change of variables u = 1 + e2x . Now we have
¯
¯
¶
Z µ
Z
1 ¯¯ u − 1 ¯¯
1
1
1
u2
du = x + ln ¯
2−
du =
+
+ c.
u2 − 1
2
u+1 u−1
2
u + 1¯
Therefore, the arc length in question is
√
√
√
√
1 ( 1 + e2 − 1)( 2 + 1)
2
√
1 + e − 2 + ln √
.
2 ( 1 + e2 + 1)( 2 − 1)
(2) Solve the differential equation y ′ = 9.8 − y/5 with the initial condition y(0) = 0.
he integration factor here is
µ(t) = exp
Therefore, we have
µZ
¶
1
dt = et/5 .
5
d ¡ t/5 ¢
e y = 9.8et/5 =⇒ y = 9.8 + ce−t/5 .
dt
y(0) = 0 gives that c = −9.8. Consequently, the solution is
y(t) = 9.8 − 9.8et/2 .
(3) Determine the values of r for which the differential equation y ′′′ − 3y ′′ + 2y ′ = 0 has
solutions of the form y = ert .
y = ert gives that y ′ = rert , y ′′ = r2 ert and y ′′′ = r2 ert . We must have
r3 ert − 3r2 ert + 2rert = 0, or r3 − 3r2 + 2r = r(r − 1)(r − 2) = 0.
Therefore, we have r = 0, r = 1 or r = 2.