Download Math 231.04, Problem Set 5 Solutions (Partial)

Math 231, Problem Set 5 Solutions
1
Math 231.04, Problem Set 5 Solutions (Partial)
Due Wednesday, Feb. 17, 2010
From Text Fundamentals of Differential Equations, by Nagle, Saff, and Snider
Section 3.2, # 9, 19
Section 3.3, # 1, 2
Section 3.4, # 1, 5
Additional Problems:
1.) Suppose that a bacteria colony in a petri dish is growing unrestrictedly. If it
takes 5 days for the population of bacteria to double, how long does it take for the
population to triple? Assume the Malthusian model.
Solution: Let 𝑃 (𝑡) denote the bacteria population at time 𝑡 days from the start.
The Malthusian model is that 𝑃 ′ (𝑡) = 𝑘𝑃 (𝑡). Solving this (which was done in class)
gives 𝑃 (𝑡) = 𝑃 (0)𝑒𝑘𝑡 . Because the doubling time (the time for the population to
double) is 5 days, we have 𝑃 (5) = 2𝑃 (0). Hence
2𝑃 (0) = 𝑃 (5) = 𝑃 (0)𝑒𝑘⋅5 .
Cancelling 𝑃 (0) gives
2 = 𝑒5𝑘 .
Taking the natural logarithm of both sides gives ln 2 = 5𝑘, or 𝑘 = 51 ln 2. Substituting
this value for 𝑘 in the equation 𝑃 (𝑡) = 𝑃 (0)𝑒𝑘𝑡 gives
1
𝑃 (𝑡) = 𝑃 (0)𝑒( 5 ln 2)𝑡 .
To find the tripling time (the time when the population is triple the original), we set
𝑃 (𝑡) = 3𝑃 (0) and solve for 𝑡:
1
3𝑃 (0) = 𝑃 (0)𝑒( 5 ln 2)𝑡 .
1
We divide by 𝑃 (0) to get 3 = 𝑒( 5 ln 2)𝑡 and then take the natural logarithm of both
sides to get
(
)
1
ln 3 =
ln 2 𝑡,
5
hence
5 ln 3
𝑡=
≈ 7.925 𝑑𝑎𝑦𝑠.
ln 2
2.) A spherical snowball melts so that its volume changes at a rate proportional to its
surface area. (This is reasonable since the heat comes in only through the surface.)
If the initial radius is 8 cm, and and in 1 hour the radius becomes 5 cm, when will
the snowball disappear? Hint: The volume of a sphere is 43 𝜋𝑟3 and the surface area
is 4𝜋𝑟2 , where 𝑟 is the radius. Find an equation expressing the rate of change of the
volume of the snowball in terms of the volume.
Math 231, Problem Set 5 Solutions
2
Solution: Let 𝑉 (𝑡) be the volume of the snowball at time 𝑡 hours, let 𝑆 be the
surface area, and 𝑟 the radius. The problem gives the information that
𝑉 ′ = 𝑘𝑆 = 𝑘4𝜋𝑟2 ,
(1)
4
𝜋𝑟3 ,
3
for some constant 𝑘. Since 𝑉 =
(
3
we have 𝑟 = 3𝑉 /(4𝜋), or 𝑟 =
3𝑉
4𝜋
)1/3
.
Substituting this value of 𝑟 in (1), we get
′
𝑉 = 𝑘4𝜋
where 𝐴 =
4⋅32/3 𝜋
𝑘
42/3 𝜋 2/3
[(
3𝑉
4𝜋
)1/3 ]2
(
= 4𝑘𝜋
3𝑉
4𝜋
)2/3
4 ⋅ 32/3 𝜋 2/3
𝑘𝑉
= 𝐴𝑉 2/3 ,
42/3 𝜋 2/3
=
is a constant. The equation 𝑉 ′ = 𝐴𝑉 2/3 is separable:
𝑑𝑉
= 𝐴𝑉 2/3 ,
𝑑𝑡
or
𝑑𝑉
= 𝐴 𝑑𝑡.
𝑉 2/3
Integrating both sides,
∫
(2)
Note that
∫
∫
𝑑𝑉
=
𝐴 𝑑𝑡.
𝑉 2/3
∫
𝑉 1/3
𝑑𝑉
−2/3
=
𝑉
𝑑𝑉
=
= 3𝑉 1/3 ,
2/3
𝑉
1/3
with the constant of integration omitted. Hence (2) becomes
3𝑉
1/3
=
∫
𝐴 𝑑𝑡 = 𝐴𝑡 + 𝐵.
Multiplying by 3 gives
𝑉 1/3 = 3𝐴𝑡 + 3𝐵 = 𝐶𝑡 + 𝐷,
for constants 𝐶 = 3𝐴 and 𝐷 = 3𝐵. Hence
𝑉 = (𝐶𝑡 + 𝐷)3 .
The initial radius is given as 8 cm, so the initial volume is 𝑉 (0) = 34 𝜋𝑟(0)3 = 43 𝜋 ⋅ 83 .
Hence
4
𝜋 ⋅ 83 = 𝑉 (0) = (𝐶 ⋅ 0 + 𝐷)3 = 𝐷3 .
3
Hence
)1/3
(
)
(
4 1/3
4
3
=
𝐷=
𝜋⋅8
𝜋
⋅ 8.
3
3
Therefore
(
)3
(
)
4 1/3
(3)
𝑉 = 𝐶𝑡 +
𝜋
⋅8 .
3
We are also given that the radius is 5 when 𝑡 = 1, or 𝑟(1) = 5. Hence 𝑉 (1) =
4
𝜋𝑟(1)3 = 34 𝜋 ⋅ 53 cm, so
3
(
4
4
𝜋 ⋅ 53 = 𝑉 (1) = 𝑉 = 𝐶 ⋅ 1 +
𝜋
3
3
(
)1/3
)3
⋅8
(
4
= 𝐶+
𝜋
3
(
)1/3
)3
⋅8
.
Math 231, Problem Set 5 Solutions
3
Taking the cube root of both sides,
4
𝐶+
𝜋
3
(
(
or 𝐶 = 43 𝜋
(3) gives
)1/3
⋅5−
(
4
𝜋
3
)1/3
)1/3
4
⋅8=
𝜋 ⋅ 53
3
(
⋅8 =
(
(
4
𝜋
3
4
𝑉 = − 𝜋
3
((
=
4
𝜋
3
(
)1/3 )3
)1/3
)1/3
)1/3
4
=
𝜋
3
(
(5 − 8) = −
4
⋅ 3𝑡 +
𝜋
3
(
(
)1/3
)1/3
4
𝜋
3
⋅ 5,
)1/3
⋅ 3. Substituting into
)3
⋅8
4
(−3𝑡 + 8) = 𝜋(−3𝑡 + 8)3
3
The snowball disappears when 𝑉 = 0, or when (−3𝑡 + 8)3 = 0, hence
𝑡=
8
ℎ𝑟𝑠.
3
Although the approach above is straightforward, it is very messy computationally.
There is a faster solution based on related rates, as follows. Since
4
𝑉 = 𝜋𝑟3 ,
3
we have
𝑑𝑉 𝑑𝑟
𝑑𝑉
𝑑𝑟
=
= 4𝜋𝑟2 .
𝑑𝑡
𝑑𝑟 𝑑𝑡
𝑑𝑡
On the other hand, we are given that the rate of change of the volume is proportional
to the surface area 𝑆 = 4𝜋𝑟2 , so
𝑑𝑉
= 𝑘𝑆 = 4𝜋𝑟2 𝑘,
𝑑𝑡
for some constant 𝑘. Setting the two expressions for
4𝜋𝑟2
𝑑𝑉
𝑑𝑡
equal,
𝑑𝑟
= 4𝜋𝑟2 𝑘.
𝑑𝑡
Cancelling the 4𝜋𝑟2 terms gives
𝑑𝑟
= 𝑘,
𝑑𝑡
which shows that the radius chages at a constant rate. Integrating in 𝑡 gives
𝑟 = 𝑘𝑡 + 𝐶,
for some constant 𝐶. Since 𝑟(0) = 8, we have 8 = 𝑘 ⋅ 0 + 𝐶, or 𝐶 = 8. Hence
𝑟 = 𝑘𝑡 + 8.
Since 𝑟(1) = 5, we have
5 = 𝑘 ⋅ 1 + 8 = 𝑘 + 8,
Math 231, Problem Set 5 Solutions
4
so 𝑘 = −3. Therefore
𝑟 = −3𝑡 + 8.
The snowball disappears when the radius is 0, or when −3𝑡 + 8 = 0, hence when
𝑡 = 8/3.
3.) The temperature of a living human being is 98.6 ∘ , where temperature is measured
in degrees Farenheit. A dead body with temperature 80 ∘ is found in a room with
constant temperature 70 ∘ . If the temperature of the body one hour later is 75 ∘ , how
long before the body was found did it die? Assume Newton’s Law of Cooling.
Solution: Let 𝑇 (𝑡) denote the temperature of the body at time 𝑡 hours, and let 𝑀
denote the room temperature, which is 70, measured in degrees Farenheit. Newton’s
Law of Cooling states that
𝑑𝑇
= 𝑘(𝑀 − 𝑇 ) = 𝑘(70 − 𝑇 ),
𝑑𝑡
for some constant 𝑘. This equation is separable:
𝑑𝑇
= 𝑘 𝑑𝑡.
70 − 𝑇
Integrating both sides gives
∫
∫
𝑑𝑇
= 𝑘 𝑑𝑡,
70 − 𝑇
or
− ln ∣70 − 𝑇 ∣ = 𝑘𝑡 + 𝐶.
Multiplying through the last equation by −1 and noting that the temperature of the
body will be greater than 70 and hence ∣70 − 𝑇 ∣ = 𝑇 − 70, we get
ln(𝑇 − 70) = −𝑘𝑡 − 𝐶.
Taking the exponential of both sides,
𝑇 − 70 = 𝑒ln(𝑇 −70) = 𝑒−𝑘𝑡−𝐶 = 𝑒−𝑘𝑡 𝑒−𝐶 = 𝐴𝑒−𝑘𝑡 ,
where 𝐴 = 𝑒−𝐶 is a constant. Hence
(4)
𝑇 = 70 + 𝐴𝑒−𝑘𝑡 .
It is easiest to break this problem into two parts, with separate time scales. In
order to find the constant 𝑘, which is a constant depending only on the object (in
this case the body), let 𝑡 = 0 correspond to the moment the body was found. Then
𝑇 (0) = 80 and 𝑇 (1) = 75. From 𝑇 (0) = 80, we get
80 = 70 + 𝐴𝑒−𝑘⋅0 = 70 + 𝐴𝑒0 = 70 + 𝐴,
so 𝐴 = 10. Hence (4) becomes 𝑇 = 70 + 10𝑒−𝑘𝑡 . Then using 𝑇 (1) = 75, we get
75 = 𝑇 (1) = 70 + 10𝑒−𝑘⋅1 = 70 + 10𝑒−𝑘 .
Math 231, Problem Set 5 Solutions
5
To solve for 𝑘, we subtract 70 from both sides to get 5 = 10𝑒−𝑘 . Dividing both
sides by 10, we get 1/2 = 𝑒−𝑘 . Taking the natural logarithm of both sides gives
ln(1/2) = −𝑘, or 𝑘 = − ln(1/2) = ln 2, using the fact that ln(1/𝑥) = − ln 𝑥 for any
𝑥 > 0.
Now we know that 𝑘 = ln 2, so substituting in (4) gives
𝑇 = 70 + 𝐴𝑒−(ln 2)𝑡 .
(5)
To finish the problem, it is convenient to use a different starting point for the time
scale. Let 𝑡 = 0 now be the time that the body died. (Changing the starting point
for 𝑡 doesn’t change 𝑘, which is why we can use the value of 𝑘 determined above, but
it does change 𝐴, which is why we treat 𝐴 as unknown.) Then 𝑇 (0) = 98.6 and if 𝑡0
denotes the time when the body was found, we have 𝑇 (𝑡0 ) = 80. Our goal is to find
𝑡0 . Using 𝑇 (0) = 98.6 in (5) gives
98.6 = 𝑇 (0) = 70 + 𝐴𝑒−(ln 2)⋅0 = 70 + 𝐴𝑒0 = 70 + 𝐴,
hence 𝐴 = 28.6. Thus we have
𝑇 = 70 + 28.6𝑒−(ln 2)𝑡 .
Since 𝑇 (𝑡0 ) = 80, we have
80 = 70 + 28.6𝑒−(ln 2)𝑡0 .
Subtracting 70 from both sides gives 10 = 28.6𝑒−(ln 2)𝑡0 , or 𝑒−(ln 2)𝑡0 = 10/28.6 =
1/2.86. Taking the natural logarithm of both sides yields
−(ln 2)𝑡0 = ln(1/2.86) = − ln 2.86.
Hence
𝑡0 =
ln 2.86
≈ 1.516 ℎ𝑟𝑠.
ln 2
4.) A baseball of mass .5 𝑘𝑔 is thrown directly upward from height 0 with an initial
velocity of 50 𝑚/𝑠𝑒𝑐. Assume that the force in newtons due to air resistance is −.3𝑣,
where 𝑣 is the velocity of the object. Determine the equation of motion of the baseball.
What is the maximum height that the baseball reaches? How does your answer
compare to what the maximum height would be if there is no air resistance? Warning:
if you set up coordinates so that the positive axis is in the up direction, then the force
due to gravity is downward, hence negative.
Solution: We use a coordinate system with origin at the point on the ground
where the baseball is thrown, oriented with the positive axis directly upward. Let
𝑥(𝑡) denote the height of the baseball at time 𝑡, let 𝑣(𝑡) denote its velocity, and 𝑎(𝑡) its
acceleration. My Newton’s Law, the acceleration of the baseball is related to the net
force 𝐹 acting on it by 𝐹 = 𝑚𝑎. The force acting on the baseball has two components.
One is due to gravity, and equals −𝑚𝑔 (it is negative because gravity is pulling the
object down, which is negative in our coordinate system), where 𝑔 = 9.8 𝑚/𝑠𝑒𝑐2 . The
other force is due to air resistance, and is equal to −.3𝑣. Thus
𝑚𝑣 ′ = −𝑚𝑔 − .3𝑣.
Math 231, Problem Set 5 Solutions
6
Putting in the mass 𝑚 = .5 and 𝑔 = 9.8, we have
.5𝑣 ′ = −.5(9.8) − .3𝑣.
Multiplying this equation through by 2, we have 𝑣 ′ = −(9.8) − .6𝑣, or
𝑣 ′ + .6𝑣 = −9.8.
∫
This is a first order linear equation, with integrating factor 𝑒
through by the integrating factor gives
.6 𝑑𝑡
= 𝑒.6𝑡 . Multiplying
𝑣 ′ 𝑒.6𝑡 + .6𝑒.6𝑡 𝑣 = −9.8𝑒.6𝑡 .
𝑑 ( .6𝑡 )
𝑒 𝑣 . Hence integrating both sides gives
𝑑𝑡
The left side is
𝑒.6𝑡 𝑣 =
∫
49
𝑑 ( .6𝑡 )
9.8
𝑒 𝑣 𝑑𝑡 = −9.8𝑒.6𝑡 𝑑𝑡 = − 𝑒.6𝑡 + 𝐶 = − 𝑒.6𝑡 + 𝐶.
𝑑𝑡
.6
3
∫
Multiplying the equation through by 𝑒−.6𝑡 gives
𝑣=−
(6)
49
+ 𝐶𝑒−.6𝑡 .
3
The initial velocity is 𝑣(0) = 50 𝑚/𝑠𝑒𝑐, as given in the problem. Hence
50 = −
49
49
+ 𝐶𝑒−.6⋅0 = − + 𝐶,
3
3
so 𝐶 = 50 + 49/3 = 199/3. Substituting this value into (6) gives
𝑣=−
Since 𝑣 =
𝑑𝑥
, we have
𝑑𝑡
∫
∫
𝑥=
𝑣(𝑡) 𝑑𝑡 =
−
49 199 −.6𝑡
+
𝑒 .
3
3
49 199 −.6𝑡
49
199 −.6𝑡
49 199 −.6𝑡
+
𝑒
𝑑𝑡 = − 𝑡−
𝑒 +𝐶1 = − 𝑡−
𝑒 +𝐶1 .
3
3
3
3(.6)
3
1.8
At time 0, the baseball is at ground level, so 𝑥(0) = 0. Hence
0=−
49
199 −.6⋅0
199
⋅0−
𝑒
+ 𝐶1 = −
+ 𝐶1 .
3
1.8
1.8
Hence 𝐶1 = 199/1.8. Substituting for 𝐶, we get
(7)
𝑥(𝑡) = −
49
199 −.6𝑡 199
𝑡−
𝑒
+
.
3
1.8
1.8
This is the equation of motion of the baseball.
𝑑𝑥
To find the maximum height the baseball reaches, we take
and set it equal to
𝑑𝑡
𝑑𝑥
49 199 −.6𝑡
0, to find the time when the maximum height occurs. But
=𝑣=− +
𝑒 .
𝑑𝑡
3
3
So we solve
49 199 −.6𝑡
− +
𝑒
= 0,
3
3
Math 231, Problem Set 5 Solutions
or
7
199 −.6𝑡 49
49
𝑒
= , or 𝑒−.6𝑡 =
. Taking the natural logarithm of both sides, we get
3
3
199
−.6𝑡 = ln
49
199
= − ln
,
199
49
1 199
ln
. Substituting this 𝑡 into equation (7) (it is easier if we recall from
.6
49
49
), we get
solving for 𝑡 that 𝑒−.6𝑡 =
199
so 𝑡 =
49 1 199 199 49
199
49 199
49
199
𝑥=−
ln
−
+
=−
ln
−
+
3 .6
49
1.8 199
1.8
1.8
49
1.8
1.8
(
)
≈ −27.2222(1.40148) − 27.2222 + 110.5555 ≈ 45.17 𝑚.
If there is no air resistance, then the only force acting on the baseball is due to
gravity, so
𝑚𝑣 ′ = 𝑚𝑎 = 𝐹 = −𝑚𝑔.
Cancelling 𝑚, we get 𝑣 ′ = −𝑔 = −9.8. Integrating gives
𝑣 = −9.8𝑡 + 𝐶.
Since 𝑣(0) = 50, we get 50 = 0 + 𝐶, or 𝐶 = 50. Hence
𝑣 = −9.8𝑡 + 50.
(8)
Hence the maximum height is reached when 𝑣 =
𝑡 = 50/9.8. To find the equation of motion,
𝑥(𝑡) =
∫
𝑣(𝑡) 𝑑𝑡 =
∫
𝑑𝑥
= 0, which is when 9.8𝑡 = 50, or
𝑑𝑡
−9.8𝑡 + 50 𝑑𝑡 = −4.9𝑡2 + 50𝑡 + 𝐶1 .
Since 𝑥(0) = 0, we get 0 = 0 + 0 + 𝐶1 , so 𝐶1 = 0. Therefore
𝑥(𝑡) = −4.9𝑡2 + 50𝑡.
At the maximum height, where 𝑡 = 50/9.8, we get
50
𝑥(𝑡) = −4.9
9.8
(
)2
50
+ 50
≈ 127.55 𝑚.
9.8
(
)