Callahan:
    ch 3.5: 12, 16, 20
    ch 3.6: 2 (d) (f) (g), 3, 6, 11, 14
    ch 4.2: 1-5,17-23

12)  What is the slope of the graph of at ?

16)  If the radius of a spherical balloon is inches, its volume is cubic inches.

    a) At what rate does the volume increase, in cubic inches per inch, when the radius is 4 inches?

    b) Write the microscope equation for the volume when inches.

    c) When the radius is 4 in., approximately how much does it increase if the volume is increased by 50 ?

    d) Suppose someone is inflating the balloon at the rate of 10 cubic inches of air per second.  If the radius is 4 inches, at what rate is it increasing, in inches per sec.?

20)  A steel ball is rolling along a 20-inch-long straight track so that its distance from the midpoint of the track is inches after have passed.

    a) Find a formula for the velocity of the ball after .  What is happening when the velocity is positive; negative; zero?  Describe the motion of the ball.

    When the velocity is positive the ball is moving to the right.

    When the velocity is negative the ball is moving to the left.

    When the velocity is zero the ball is pausing instantaneously as it changes direction at the extreme left or right..

    b) How far from the midpoint of the track does the ball get?

    Where the ball changes direction it also reaches its maximum distance from the center point.  This is where the velocity is zero:

    c) How fast is the ball moving at the midpoint of the track?

    The midpoint of the track is where :

    This is the maximum speed because reaches its maximum value at .

3.6: 2 (d) (f) (g))  Find the derivatives of the following:

    d)

    f)

    g)

    You can use the rule:

    Or this program allows you to determine the value for any base:    (you could write a better program but this works OK)

from visual.graph import*

def f(x) :
    return 2**(x)

xinitial = -1
xfinal = 1
x = xinitial

numberofsteps = 1000
deltax = (xfinal - xinitial)/float(numberofsteps)

graph = gdisplay()
fcurve = gcurve(color = (1, 0, 0))
fprimecurve = gcurve(color = (0, 1, 0))

    
for k in range(1,numberofsteps+1) :
    y = f(x)
    fcurve.plot(pos = (x, y))
    
    fprime = (f(x+deltax)-f(x-deltax))/(2*deltax)
    fprimecurve.plot(pos = (x, fprime))

    if x>=-0.001 and x<=0.001:
        print 'x=', x, '   ', 'y=', y, '   ', 'Slope=', fprime
    
    x = x + deltax

3)  If , where is some function satisfying and , what is ?

6)  If , what are the numerical values of   and ?

11)

    a)  Write the microscope equation for the function at :

    The microscope equation is:

    So we can see that we need :

    b)  Using the microscope equation, estimate the following values:

    The real value is:

    The real value is:

14)

4.2: 1)  Verify that the given formula is a solution to the initial value problem:

    a)  

    First check the initial value:

    Now see if the derivative of the function matches:

    So it is a solution!

    b)  

    First check the initial value:

    Now see if the derivative of the function matches:

    So it is a solution!

    c)  

    First check the initial value:

    Now see if the derivative of the function matches:

    So it is a solution!

    d)  Write a general formula for the solution of the initial value problem for any integer :

    e)  Write a general formula for the solution of the initial value problem for any integer and any constant :

2)

3)

4)

5)

17)  Verify that is a solution to the differential equation:

    First check the initial value:

    Now see if the derivative of the function matches:

    So it is a solution!

18 & 19)  Since , and since , the positive of the body satisfies the differential equation and initial condition:

    Find a formula for that solves this differential equation.  This function describes how a body moves under the force of gravity.

    We know that the formula for has to satisfy the initial condition and has to have the correct derivative.  Lets start by making a formula that has the given expression for as its derivative.  Because of the rules for taking derivatives of power functions, we start out by guessing:

    We can see that we are off by a factor of in the first term, so lets multiply the first term in the guess by :

    Now everything works, but we still need to check that the initial condition is satisfied:

    We are only off by , so it looks like we need to add to our equation to make it satisfy the initial condition.  Because the derivative of a constant is zero, this will have no impact on the value of .  With this addition we now have a general solution to our differential equation:

20)

    a)  Suppose a body is held motionless 200 m above the ground, and then released.  What values do and have?  And what is the formula for the motion of this body as it falls towards the ground?

    So we can see that the formula for the motion of this body as it falls with the given initial conditions is:

    b)  How far has the body fallen in ?  In ?

    So the difference in position at is:

    So the difference in position at is:

    c)  How long does it take for the body to reach the ground?

21)  Show that the following is a solution to the differential equation (including air resistance):

    First check the initial value:

    Now see if the derivative of the function matches:

    We know from problem 3.6: 2 (g) that , so

    And if we plug the equation for into our equation for we have:

    Because they are identical, the solution is valid.

22)  Show that the position of a body that falls against air resistance from an initial height of is given by the formula:

    We can show this by showing that it is a solution to the differential equation:

    First check the initial value:

    Now we see if the derivative of the function matches.  In order to do this derivative we differentiate each term separately:

    We know from problem 3.6: 2 (g) that , so

    Because , the solution is valid.

    b)  Suppose the coefficient of air resistance is .  If a body is held motionless 200 m above the ground, and then released, how far will it fall in ?  In ?  How does this compare to the values if their was no air resistance?

    So the difference in position at is:

    So the difference in position at is:

    c)  How long does it take for the body to reach the ground?  How does this compare to the value if their was no air resistance?

    So there is a difference is: