1. The acceleration due to gravity is -9.8 m/s^{2} .

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/sec^{2} (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 cm ^{2}/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 d ^{2}
instead of d (since one is minimized when the other is, and this avoids the
square root).*

*Given the constraint, y ^{2}=x, so d^{2}
= (x-5)^{2}+x = x^{2} -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/x ^{2} .*

*f'(x)=0 if 1=25/x ^{2} , or x^{2} =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)(d ^{2}), for some constant k*

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

* w ^{2}
+ d^{2 }= 12^{2}, and w,d>0*

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

*S(w) = k(w)(d ^{2})= k(w)(144-w^{2})=k(144w-w^{3})*

* S'(w)= k(144-3w ^{2}), so S'(w) = 0 if
w=*

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

*At this
point d = **Ö**(144-48) = **Ö**96 = 4**Ö**6, with all measurements in inches.*