Practice Midterm 1

Name

MATH 106 PRACTICE MIDTERM 1

Please work out each of the given problems. Credit will be based on the steps that you show towards the final answer. Show your work.

PROBLEM 1 Please answer the following true or false. If false, explain why or provide a counter example. If true, explain why or state the proper theorem.

A) (10 points) The differential equation is a homogeneous differential equation

Solution

True, dividing numerator by y² gives

which is a function of x/y.

B) (10 points) If f is a differentiable function such that both f and f ' are positive for all x, then g(x) = ln(f(x)) is increasing for all values of x.

Solution

True, since

                           f '(x)
        g '(x) =
                            f(x)

is positive if both f ' and f are positive

PROBLEM 2 Calculate the derivatives of the following functions.

A) (10 points) d
(2x)^1-x dx

Solution

(2x)^1-x = e^{(1 - x)ln(2x)}

Now use the chain and product rules. The derivative of

(1 - x)ln(2x)

(1 - x)(1/x) - ln(2x) = 1/x - 1 - ln(2x)

Hence the derivative of

e^{(1 - x)ln(2x)}

[1/x - 1 - ln(2x)](2x)^1-x

B)    (10 points)      d
                                 e^ln(sin
x)                     dx

Solution

First use the inverse property of e and ln to get

e^ln(sin
x) = sin x

Now the derivative is simply

cos x

C) (10 points) d 2^3t
dx t

Solution

We use the quotient rule to get

t(2^3t)' -2^3t t²

Now use the chain rule and the fact that

b^x ' = b^x ln b

to get

3t ln2 (2^3t)- 2^3t t²

PROBLEM 3 Find the following integrals

A) (10 points)

Solution

Let

u = 1 - x du = -x dx x = 1 - u

The substitution produces

B) (10 points)

Solution

Let

u = 2 - 3x du = -3 dx

The substitution produces

PROBLEM 4 You have been called as an expert witness in the case involving the recent murder that occurred in room E106. It is clear that at the time of death the victim was healthy with a temperature of 98.6 degrees. It is also know that a human body in this situation will cool down to 90 degrees in one hour. When the body was discovered at 10:00 PM the corpse had a body temperature of 85 degrees. During the entire day, the temperature of the room was a constant 65 degrees.

A) (10 points) Use the Newton’s Law of Cooling (the rate of change of the temperature of the body is proportional to the difference between the body’s temperature and the ambient temperature) to write down a differential equation for this situation.

Solution

        dT
                = k(T - 65),        T(0) = 98.6,    T(1) = 90
        dt

B) (10 points) Solve the differential equation from part A.

Solution

Separating variables gives

           dT
                       = kdt
        T - 65

Now integrate

ln|T - 65| = kt + C

Now use T(0) = 98.6

ln(98.6 - 65) = C

Now use T(1) = 90

ln|90 - 65| = k + ln(65)

Solving gives

k = ln(25/33.6)

Exponentiating the solution produces

T - 65 = 33.6 e^ln(25/33.6)t

C) (10 points) What was the time of death?

Solution

The body temperature is 85 degrees, hence

ln|85 - 65| = ln(25/33.6)t + ln(33.6)

Now solve for t

                   ln(20) - ln33.6
        t =                                  = 1.75476
                      ln(25/33.6)

Now convert 0.75476 to minutes by multiplying by 60 to get that the murder occurred one hour and 45 minutes ago or at 8:15 PM.

PROBLEM 5 Let f(x) = 2x³ + 4x + 5

(10 Points) Prove that f(x) has an inverse.

Solution

We calculate

f '(x) = 6x² + 4

which is always positive, hence f(x) has an inverse.
(10 Points) Find d
f^-1(11) dx

Solution

We use the inverse formula

d 1
f^-1(11) = dx f '(f^-1(11))

Since

f(1) = 11

We have

f^-1(11) = 1

and

f '(1) = 6(1)² + 4 = 10

Hence

d 1 1
f^-1(11) = = dx f '(f^-1(11)) 10

PROBLEM 6 Solve the following differential equations

A. (15 Points) y(1 + x²)y' - x(1 + y²) = 0, y(0) =

Solution

Solving for dy/dx gives

        dy               x(1 + y²)
                  =
        dx                y(1 + x²)

Separating variables gives

         y dy                  x dx
                       =
        1 + y²               1 + x²

Integrating both sides yields

ln(1 + y²) = ln(1 + x²) + C

The initial value y(0) = gives

ln 4 = C

Using the sum to product property of ln gives

ln(1 + y²) = ln(4 + 4x²)

Exponentiating both sides produces

1 + y² = 4 + 4x²

B. (15 Points) x³ + y³
y' = xy²

Solution

This is a homogeneous differential equation. We can write

1 + (y/x)³
y' = (y/x)²

Letting

v = y/x, y = xv, y' = v + xv'

gives

1 + v³
v + xv' = v²

x dv 1
= dx v²

Separating gives

dx
v² dv = x
Integrating both sides gives

1/3 v³ = ln|x| + C₁

Multiplying by 3, letting C = 3C₁, and taking the 1/3 power produces

v = (3ln|x| + C)¹³

Substituting v = y/x and multiplying by x gives

y = x(3ln|x| + C)^1/3