Math Practice exam 2 - solutions

C Roettger, Fall 13

Math 267 - Practice exam 2 - solutions

Problem 1 A solution of 10% perchlorate in water flows at a rate of 8 L/min into a tank holding 200L pure water. The solution is kept well stirred and flows out of the tank at a rate of 6 L/min.

a) Determine the volume of perchlorate in the tank after t minutes.

b) When will the percentage of perchlorate in the tank reach 6%?

Solution. a) Let V (t) and x(t) the volume of water and perchlorate in the tank at time t, respectively. So

V (t) = 200 + 2t

(time in minutes, volume in liters). Then the amount x(t) of perchlorate satisfies

x ⁰ (t) = 0.10 × 8 − 6x(t) V (t) . This is a linear equation, written in standard form as

x ⁰ (t) + 3

100 + t x = 0.8, so an integrating factor is

µ(t) = e R 3/(100+t) dt

= e 3 ln(100+t) = (100 + t) ³ . We multiply the entire equation by µ and integrate:

µ(t)x(t) = 0.8 Z

(100 + t) ³ dt = 0.2(100 + t) ⁴ + C.

so after solving for x, we get the general solution x(t) = 0.2(100 + t) + C

(100 + t) ³ . Since x(0) = 0, we can solve for C to get

C = −0.2 · 10 ⁸ = −20, 000, 000.

The amount of perchlorate after t minutes is

x(t) = 0.2(100 + t) − 20, 000, 000 (100 + t) ³ . b) We put

0.06 = x(t)

V (t) = 0.1 − 10, 000, 000 (100 + t) ⁴ and solve for t, getting first

(100 + t) ⁴ = 10 ⁷ 0.04 and then

t = p

10 ⁷ /0.04 − 100 ≈ 25.74 (in minutes).

Problem 2 A garage with no heating or cooling has a time constant of 2 hours. The outside temperature varies as a sine wave,

M (t) = 65 + 15 sin(π(t − 8)/12).

Determine the time when the building reaches its lowest and its highest

temperature, assuming that the exponential term has died off.

Note how the building temperature T (blue) lags behind the ambient tem- perature M !

Solution. The temperature T (t) of the building satisfies the differential equation

T ⁰ (t) = k(M (t) − T (t)).

Rewritten in standard form for a linear equation, this reads T ⁰ + kT = kM

so we have the integrating factor µ(t) = e ^kt , and the usual recipe gives e ^kt T (t) =

Z

ke ^kt (65 + 15 sin(π(t − 8)/12)) dt.

Integrating by parts twice gives for the hard part on the right-hand side Z

e ^kt sin(αt + β) dt = e ^kt [k sin(αt + β) − α cos(αt + β)]

α ² + k ² + C.

Integrating 65ke ^kt gives another 65e ^kt . We substitute k = 1/2 (1 / time constant), α = π/12, β = −2π/3. Altogether,

T (t) = 65 +

15 4 sin 

π(t−8) 12

 − ^5π ₈ cos 

π(t−8) 12



π ² /144 + 1/4 + Ce ^−t/2 .

For t large enough so that the exponential term ’has died off’ (meaning we can neglect it), we can find the maximum / minimum temperature by setting the derivative of the first term equal to 0, like in Calculus I. It helps to introduce a new variable u = π(t − 8)/12. So we differentiate

d du

 15

4 sin u − 5π 8 cos u



= 15

4 cos u + 5π

8 sin u = 0 which gives

u = − arctan  6 π



+ nπ ≈ −1.088 + nπ for any integer n. Going back to t,

π(t − 8)

12 = −1.088 + nπ

gives the times

t = 12.158 + 12n

for any integer n. A second-derivative test (no, we don’t want that) or first- derivative test (still, lots of work) or simply a graph (best! see above) shows that even integers n give times with minimal temperature, odd integers n give maximal temperature.

Problem 3 A parachutist whose mass is 75 kg drops from a helicopter hovering 2000 meters above the ground. She falls to the ground under the force of gravity and air resistance, which is proportional to her velocity, with proportionality constant b ₁ = 30 N sec/m when the chute is closed, and b 2 = 90 N sec/m when the chute is open. The chute does not open until the velocity reaches 20 m/sec. At what time will she reach the ground? Use g = 10N/kg.

Solution. Let y(t) be the parachutist’s height above ground. So y(0) = 2000, y ⁰ (0) = 0, and

my ⁰⁰ + by ⁰ = −mg

We write v = y ⁰ , so we get a first-order equation for the velocity, v ⁰ + bv

m = −g which has integrating factor µ = e ^bt/m ,

vµ = −g Z

e ^bt/m dt = − gme ^bt/m

b + C

and the solution

v = − gm

b + Ce ^−bt/m .

This analysis is valid both for open and closed parachute. Now use v(0) = 0 to get for the first, rapid fall

C ₁ = gm b 1

= 25, v = 25(e ^−b

^t/m − 1).

Let us find the time t ₁ when the parachute opens, so v = −20 (negative sign because the movement is down!!)

−20 = gm

b ₁ (e ^−b

^t/m − 1)

gives with g = 10N/kg

t ₁ = − m b ₁ ln



1 − 20b ₁ gm



≈ 4.02

(in seconds). After that time, we get v = − gm

b ₂ + C ₂ e ^−b

^t/m . and the initial condition v(t ₁ ) = −20 gives

C ₂ = −1458.33

Finally, find the position y(t) by integrating v. In both cases, y(t) = − gmt

b − Cm

b e ^−bt/m + D.

For the first case, y(0) = 2000 gives

D ₁ = 2000 + C ₁ m

b ₁ = 2062.5

so the position at time t ₁ is y(t ₁ ) ≈ 1949.41 meters. With the parachute open, we get

y(t) = − gmt

b ₂ − C ₂ m

b ₂ e ^−b

^t/m + D ₂

and from the initial condition y(t ₁ ) = 1949.41, we can determine D2 = 1973.22

Finally, the parachutist reaches the ground when 0 = y(t) ≈ − gmt

b 2

+ D ₂ so at time

t = D ₂ b ₂

gm ≈ 236.8

(in seconds). The exponential term at around this time is completely negli-

gible.

Note. This is 3.4 # 7, and the book’s answer is 241 seconds, because they used the more accurate value g = 9.81N/kg.

Problem 4 Find the general solution of y ⁰⁰ − 2y ⁰ + 5y = 0.

Solution. The auxiliary equation is

r ² − 2r + 5 = 0

which has roots 1 ± 2i. So we can write the general solution as y(t) = e ^t (c ₁ cos 2t + c ₂ sin 2t).

Problem 5 Consider the differential equation

x ⁰⁰ (t) − 2x ⁰ (t) + x(t) = cos t. (1) a) Find a particular solution of equation (1).

b) Find the complimentary function (general solution of the associated ho- mogeneous equation).

c) Find the unique solution satisfying both (1) and the initial conditions x(0) = 0, x ⁰ (0) = 1.

Solution. a) We try x _p (t) = A cos t + B sin t. This leads to A = 0, B =

−1/2.

b) The characteristic equation is (r − 1) ² = 0, with double root r = 1, so the complimentary function is

x _c (t) = (c ₁ + c ₂ t)e ^t

c) We get c ₁ = 0, c ₂ = 3/2 after substituting the general solution x _p + x _c for x.

Problem 6 A mass of 2 kg is traveling horizontally on wheels without

friction. It is hooked up to a spring with spring constant k = 162 N/m. Let

x(t) be the position of the mass with x = 0 being the equilibrium position.

a) Find the second-order linear differential equation governing x.

b) What is the natural frequency ω of this system?

c) Find the position function x(t) if x(0) = 12 and x ⁰ (0) = −45.

d) Suppose an external force F = 2 cos(10t) starts acting on the mass, and x(0) = x ⁰ (0) = 0. Find the resulting position function x(t).

Solution. a)

2x ⁰⁰ + 162x = 0.

b) ω = p162/2 = 9 (in radians per second).

c) The general solution here is

x _c (t) = c ₁ cos 9t + c ₂ sin 9t.

The initial conditions give c ₁ = 12 and c ₂ = −5, so x(t) = 12 cos 9t − 5 sin 9t.

d) Now we have to find a particular solution x _p for

2x ⁰⁰ + 162x = 2 cos 10t. (2)

There is no overlap between cos 10t, sin 10t and cos 9t, sin 9t., So we try x _p (t) = A cos 10t + B sin 10t,

and substituting this into (2) yields equations B = 0, −200A + 162A = 2, so A = −1/19. The general solution is now

x(t) = − 1

19 cos 10t + c ₁ cos 9t + c ₂ sin 9t.

The initial conditions give x ⁰ (0) = 9c ₂ = 0, so c ₂ = 0 and c ₁ = 1/19.

Problem 7 Consider the Euler-Cauchy equation t ² y ⁰⁰ + 3ty ⁰ + y = 0.

a) Find the general solution.

b) Then find the solution satisfying y(1) = 4, y ⁰ (1) = 3.

Solution. The characteristic equation here is

r(r − 1) + 3r + 1 = 0

which has a double root r = −1, so solutions are y = 1/t and y = ln(t)/t.

The general solution is

y = c ₁ + c ₂ ln t

t .

b) So c ₁ = 4 and from

y ⁰ (t) = c ₂ /t − (c ₁ + c ₂ ln t)

t ² ,

we get c ₂ − c ₁ = 3, so c ₂ = 7.