1.  The acceleration due to gravity is -9.8 m/s2 .

Suppose a ball is thrown upward from a height of 2 m, with an initial velocity of 15 m/s.

a.  Find a formula for the position of the ball as a function of time since it was thrown.

b.  When does it reach its peak?  How high does it get?

It reaches its peak when h'(t)=0, ie when v(t)=0.

-9.8t+15=0 -> t=  15/9.8 ~1.53 seconds

h(15/9.8)~13.48 m

2.  When the brakes are applied to a car, it undergoes a uniform acceleration of  -22 ft/sec2 (this is a realistic braking force on a dry road – in slippery conditions, the acceleration will not be as great).  Suppose you are driving a car at 60 mi/hr (=88ft/sec), and have to come to a quick stop.

a. How long will it take you to come to a complete stop?

b.  How far do you drive in that amount of time?

So you need 176 feet to come to a stop.

4.  Use Newton’s Method to approximate the value of.  (What is a good function to use for which this number will be a root?  What would be a good initial guess?)  Apply 2 steps of this method to get the next 2 improved estimates.

5.  A computer monitor shows a picture of a “growing” rectangle.  At a given time, it is 3 cm long and 2 cm high; the length is growing at a rate of 1 cm/sec, and the height is growing at a rate of 0.7 cm/sec.  At what rate is the area growing?

A = l*h

dA/dt = l*dh/dt + h*dl/dt

= 3cm(0.7cm/s) + 2cm(1cm/sec) = 4.1 cm2/sec

6. Find the point on the curve  nearest to (5,0).

The distance from (x,y) to (5,0) is .

To simplify computations, I'll minimize d2 instead of d (since one is minimized when the other is, and this avoids the square root).

Given the constraint, y2=x, so d2 = (x-5)2+x = x2 -9x + 25

This is minimized when 2x-9=0, or x=4.5

So the closest point to this curve is at

7. Maximize and minimize  subject to the constraints x+y=5, x,y≥0

I'll replace x with 5-y (I could also solve for y), and use this to re-write the objective function as a funciton of just y:

So the only cricitial point is when y=10/8=1.25

At this point .  Checking the endpoints,

when y=0, f(y) = 25, and y=5, f(y)=75.

So the minimum is 18.75, and this occurs when y=1.2 and x=3.75.  The maximum is 75, and this occurs when x=0 and  y=5.

8.  Find all critical points of f(x) = x+25/x.  Then find the absolute maximum and minimum values of this function on the domain [1,10]

f'(x) = 1-25/x2 .

f'(x)=0 if 1=25/x2 , or x2 =25, so x=-5 or 5

In the given domain, the absolute maximum and minimum values can only occur at 1,5 or 10, so we check those values:

f(1) = 26, f(5) = 10, f(10) = 12.5

So the maximum is 26, at x=1, and the minimum is 10, at x=5.

9.  F(x) is the antiderivative of the function shown in this graph.

Find the critical numbers of F, and indicate whether each is a local maximum, a local minimum, or neither.

The critical numbers of F are when F'(x)=0.  But F'(x) is the function graphed, so the critical numbers of F are at (approximately) 2 and 5

At x=2, F'(x) goes from positive to zero to negative, so by the first derivative test, this must be a local maximum.

At x=5, F'(x) goes from negative to zero to negative again, so by the first derivative test, this is not a local extremum.

10. You want to cut a rectangular beam from a cylindrical log of diameter 12 inches.

The strength of a beam is proportional to the product of the width and the square of its depth, so to make the beam as strong as possible you want to maximize this product.

a)  Express your objective as a function of 2 (or more) variables.

If S = the strength, then S=k(w)(d2), for some constant k

b)  What is/are your contraint equation(s)

w2 + d2 = 122, and w,d>0

c) Solve for the exact dimensions that will give the beam the greatest strength.

S(w) = k(w)(d2)= k(w)(144-w2)=k(144w-w3)

S'(w)= k(144-3w2), so S'(w) = 0 if w=Ö48=4Ö3

(I'll leave it to you to verify that this is in fact the maximum).

At this point d = Ö(144-48) = Ö96 = 4Ö6, with all measurements in inches.