Worksheet 8 - Solutions
1. A population of insects in a region will grow at a rate that is proportional to their current population. In the absence of any outside
factors the population will triple in two weeks time. On any given day
there is a net migration into the area of 15 insects and 16 are eaten
by the local bird population and 7 die of natural causes. If there are
initially 100 insects in the area will the population survive? If not,
when do they die out?
Solution:
There’s a little bit to unpack here. The differential equation for our
population will be given by a rate in of kP + 15, and a rate out of 23.
This gives us the following differential equation:
dP
= kP − 8
dt
We don’t know the proportionality constant k, but can find it using
the rest of the information given, namely the fact that ”in the absence
of any outside factors, the population will triple in two weeks time.”
We read this to mean that k is such that solutions to the differential
equation dP
dt = kP have the property that if P (0) = P0 , then P (14) =
3P0 (this is exactly the property listed - if we just have proportional
growth, in 14 days, we have 3 times what we started with.) The
solution to this simpler differential equation is P = Aekt , and so we
use the information to find k by plugging in the two points and dividing
through, one equality by the other:
P (0) = Aek·0
P (14) = Aek·14
P0 = A
3P0 = Ae14k
3P0
Ae14k
=
P0
A
14k
3=e
ln 3
k=
14
Now we can use this information to find a solution to the original
1
equation, which is separable:
dP
= kP − 8
dt
dP
= dt
ˆ kP − 8 ˆ
dP
=
dt
kP − 8
1
ln |kP − 8| = t + C
k
ln |kP − 8| = kt + C
kP − 8 = Aekt
P = Aekt +
8
k
112
ln 3
In this solution, the constant 112/ ln 3 is just a hair over 100 (it’s about
102), and so when we solve P (0) = 100, we get that A is negative,
which means that the population will be decreasing. To find when it
will die out, we just solve the initial value problem (A ≈ 2) and set
equal to zero.
P = A3t/14 +
2. A 1000 gallon holding tank that catches runoff from some chemical
process initially has 800 gallons of water with 2 ounces of pollution
dissolved in it. Polluted water flows into the tank at a rate of 3 gal/hr
and contains 5 ounces/gal of pollution in it. A well mixed solution
leaves the tank at 3 gal/hr as well. Determine the amount of pollution
in the tank at any time t. How long does it take for the total amount
of pollution released into the environment from the holding tank to
reach 100 ounces?
Solution:
We write a differential equation for y(t), the quantity of pollution in
the tank at time t. The rate in is given by 3gal/hr · 5oz/gal = 15oz/hr.
The rate out is a proportion of the pollution in the tank at any time.
We drain 3 gal/hr, and there are a total of 800 gallons in the tank,
3
and so the rate out is given by 800
y. Combining these gives us:
dy
3y
= 15 −
dt
800
2
This is separable, and so we solve by:
dy
3y
= 15 −
dt
800
dy
= dt
15 − 3y/800
ˆ
ˆ
dy
=
dt
15 − 3y/800
−800
ln(15 − 3y/800) = t + C
3
−3t
+C
ln |15 − 3y/800| =
800
15 − 3y/800 = Ae−3t/800
3y/800 = 15 + Ae−3t/800
y = 4000 + Ae−3t/800
The solution, then, to our initial condition is
y = 4000 − 3998e−3t/800
The amount that has been spilled into the environment is just the
total amount released minus the amount still in the pool, that is
s(t) = 2 + 15t − (4000 − 3998e−3t/800 )
Feed this to a computer to find where s = 100, at about t = 62.
3. A 5 kg cannonball is shot straight up in the air at an initial velocity
of 100 meters/second at an initial height of 500 meters. Using the
fact that acceleration due to gravity is 9.8 m/s2 , and the force of air
resistance is Fair = −V :
(a) Write and solve a differential equation to find the velocity of our
cannonball at time t. (Hint: V 0 = A, F = M A)
(b) Write and solve a differential equation to find the height of our
cannonball at time t. (Hint: H 0 = V )
(c) When does the cannonball hit the ground?
3
Solution:
(a) We use the given hints to write a differential equation for V . There
are two different sources of acceleration, the first is gravity, which is
causing downward acceleration at a constant rate of 9.8 m/s2 . The
second source is wind resistance, which acts in the opposite direction
of velocity, in proportion to the velocity. Namely, the force is −V ,
and the acceleration due to this force is given by A = F/M = −V /5.
Combining these, we get
dV
= A = −9.8 − V /5
dt
We can solve this by separation to find V (t):
dV
= −9.8 − V /5
dt
dV
= dt
−9.8 − V /5
ˆ
ˆ
dV
=
dt
−9.8 − V /5
−5 ln | − 9.8 − V /5| = t + C
ln | − 9.8 − V /5| = −t/5 + C
−9.8 − V /5 = Ae−t/5
−V /5 = Ae−t/5 + 9.8
V = Ae−t/5 − 49
We can then plug in our initial condition of V (0) = 100 to find A =
149, giving
V (t) = 149e−t/5 − 49
(b) Here, we simply use
dH
= V (t) = 149e−t/5 − 49
dt
This is trivially separable giving
H(t) = −745e−t/5 − 49t + C
With our initial condition of H(0) = 500, this gives
H(t) = −745e−t/5 − 49t + 1245
4
(c) To see when our cannonball hits the ground, we simply plug in
H = 0 and solve (use a computer) giving t ≈ 25.3.
4. A 1500 gallon tank initially contains 600 gallons of water with 5 lbs
of salt dissolved in it. Water enters the tank at a rate of 9 gal/hr and
the water entering the tank has a salt concentration of 51 (1 + cos t)
lbs/gal. If a well-mixed solution leaves the tank at a rate of 6 gal/hr,
how much salt is in the tank when it overflows?
Solution:
We will write a differential equation for y(t), the amount of salt in the
tank at time t. Here, as often, we have a rate in and rate out. Both
are a little more complicated than in many of the problems that we’ve
had.
The rate in is given by
9
9 gal/hr · 1/5(1 + cos t) lb salt/gal = (1 + cos t) lb salt/hr.
5
The rate out is given by
6 gal H2O/hr ·
y lb salt
·
600 + 3t gal H2O
Combining these gives us the following first-order linear equation:
9
6y
dy
= (1 + cos t) −
dt
5
600 + 3t
6
9
dy
+
y = (1 + cos t)
dt
600 + 3t
5
Solving for the integrating factor gives:
´
m(x) = e
6
600+3t
dt
= e2 ln(600+3t)
= (600 + 3t)2
Multiplying both sides gives us
9
d
(600 + 3t)2 y = (600 + 3t)2 (1 + cos t)
dt
5ˆ
9
2
(600 + 3t) y =
(600 + 3t)2 (1 + cos t) dt
5
5
This last is solvable by integrating by parts, giving y(t), and the question asks for y(300)
5. Compute the second-, fourth-, and sixth-order Taylor polynomials of
ln x centered at x = 1. Use these to estimate ln(2).
Solution:
T61 ln x = (x−1)−
(x − 1)2 (x − 1)3 (x − 1)4 (x − 1)5 (x − 1)6
+
−
+
−
2
3
4
5
6
6. Compute the fifth-order Taylor polynomial of
1
1−x
centered at x = 0.
Solution:
T50
1
= 1 + x + x2 + x3 + x4 + x5
1−x
7. Compute the second order Taylor polynomial of sin(x2 ) around 0 and
use this to approximate sin( 41 ).
Solution: We need to compute two derivatives of f (x) = sin(x2 ).
f (x) = sin(x2 )
f 0 (x) = cos(x2 )(2x)
f (2) (x) = 2 cos(x2 ) − 4x2 sin(x2 )
Now we evaluate the function and its derivatives at zero.
f (0) = sin(0) = 0
f 0 (0) = cos(0)(0) = 0
f (2) (0) = 2 cos(0) − 4(0) sin(0) = 2
6
Therefore the degree two Taylor polynomial of sin(x2 ) is 2!2 x2 = x2 .
2
2
Since sin 14 = sin 12
, our approximation is 12 = 14 .
8. Compute the degree two Taylor polynomial of the function f (x) =
etan(x) around 0. Use this to estimate etan(.1) .
Solution:
f (x) = etan(x)
f 0 (x) = sec2 (x)etan(x)
f (2) (x) = 2 sec2 (x) tan(x)etan(x) + sec4 (x)etan(x)
We evaluate these at zero.
f (0) = etan(0) = 1
f 0 (0) = sec2 (0)etan(0) = 1
f (2) (0) = 2 sec2 (0) tan(0)etan(0) + sec4 (0)etan(0) = 1
This gives the Taylor polynomial 1 + 1!1 x + 2!1 x2 = 1 + x + 21 x2 . We
can then approximate etan(.1) ≈ 1 + .1 + 12 (.1)2 = 1.105.
9. Compute the second order Taylor polynomial of sin(ex − 1) around 0
1
and use this to approximate sin(e 2 − 1).
Solution: We first need to compute two derivatives of sin(ex − 1)
f (x) = sin(ex − 1)
f 0 (x) = cos(ex − 1)ex
f (2) (x) = cos(ex − 1)ex − sin(ex − 1)e2x
We evaluate these at zero.
f (0) = sin(0) = 0
f 0 (0) = cos(0)e0 = 1
f (2) (0) = cos(0)e0 − sin(0)e0 = 1
7
Combining these, we find that the second order Taylor polynomial of
1
sin(ex − 1) is x + 12 x2 . This gives the approximation sin(e 2 − 1) ≈
1
1 3
= 58 .
2 + 2
10. Find the second and fourth order Taylor expansions around 1 for the
function f (x) = x3 + 5x + 1.
Solution: We can first observe that since this function is a third order
polynomial, the fourth order Taylor expansion of f (x) is f (x). To find
the second order Taylor expansion, we need to differentiate:
f (x) = x3 + 5x + 1
f 0 (x) = 3x2 + 5
f (2) (x) = 6x
Now we evaluate at zero.
f (1) = 7
f 0 (1) = 8
f (2) (1) = 6
So the second order Taylor polynomial around 1 is 7 + 8(x − 1) + 26 (x −
1)2 = 7 + 8(x − 1) + 3(x − 1)2 .
11. Find the second order Taylor polynomial around 0 for f (x) =
and use this to estimate f (.1).
Solution:
ˆ
f (x) =
0
f (x) = e
x
2
e−t dt
0
−x2
f (2) (x) = −2xe−x
8
2
´x
0
2
e−t dt
so that
ˆ
0
f (0) =
0
f (0) = e
f
(2)
2
e−t dt = 0
0
−02
=1
2
(0) = −2(0)e−0 = 0
so that the degree 2 Taylor polynomial for f (x) is x. Our estimate for
f (.1) is therefore .1.
12. Find the first order Taylor polynomial for the function f (x) =
and use this to find an approximation for f ( 12 ).
´ sin(x)
0
Solution:
ˆ
sin(x)
f (x) =
0
f (x) = e
3
e−t dt
0
− sin3 (x)
cos(x)
so that
ˆ
f (0) =
0
sin(0)
3
0
− sin3 (0)
f (0) = e
ˆ
e−t dt =
0
3
e−t dt = 0
0
cos(0) = 1
so the first order Taylor polynomial for f (x) is given by x and our
approximation for f ( 21 ) is 12 .
13. Find the second order Taylor polynomial of cos(x) around 0 then integrate this polynomial. Additionally, find´the third order Taylor polynomial of sin(x) around 0 . Recall that cos(x)dx = sin(x) + C and
compare your answer to the previously computed Taylor polynomial
for the integral of cos(x).
9
3
e−t dt
Solution: We begin by calling f (x) = cos(x) and g(x) = sin(x). We
need to compute two derivatives of f and three derivatives of g.
f (x) = cos(x)
f 0 (x) = − sin(x)
f (2) (x) = − cos(x)
g(x) = sin(x)
g 0 (x) = cos(x)
g (2) (x) = − sin(x)
g (3) (x) = − cos(x)
so that
f (0) = 1
f 0 (0) = 0
f (2) (0) = −1
g(0) = 0
g 0 (0) = 1
g (2) (0) = 0
g (3) (0) = −1
The degree two Taylor polynomial of cos(x) around 0 is then 1 − x2
3
and the integral of this is C +x− x3 . Similarly, we find that the degree
3
three Taylor polynomial of sin(x) is x − x3 . For C = 0, these agree–
for nice functions, we can exchange the operation of taking Taylor
polynomials and integration.
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