Solutions to Exam II - - Spring 1996

Solutions to Exam II - Math 151, Spring 1996

Part I - - Multiple Choice

1. c)    2. c)   3. e)   4. b)   5. a)   6. e)

7. d)   8. c)    9. a)   10. c)

Part II - - Work Out Part

11. The graph of the function is given below.

12. The volume of the slick is given by 

> v:=.005*Pi*r^2;

                                         2
                           v := .005 Pi r

Differentiating this with respect to t gives
dv/dt=.01*Pi*r*(dr/dt). Inserting dv/dt=400 and r=300 and solving for
dr/dt gives   dr/dt=400/3Pi  meters per hour.

13. The function 

> f:=x->x^3-6*x^2+9*x+1;

                               3      2
                    f := x -> x  - 6 x  + 9 x + 1

has derivative and second derivative equal to
> D(f); D(D(f));

                                 2
                         x -> 3 x  - 12 x + 9


                            x -> 6 x - 12

The inflection point is where the second derivative changes
sign, which occurs at x=2 and y=f(2)=3. When x>2, the graph
is concave up and when x<2 the graph is concave down.

14. The function is

> f:=x-> 3*(x-2)^(2/3)-2*x;

                                        2/3
                     f := x -> 3 (x - 2)    - 2 x

> 
The derivative of f is

> D(f);

                                  2
                         x -> ---------- - 2
                                     1/3
                              (x - 2)

When x=2 the derivative does not exist and when x=3, the derivative is zero.
Therefore, x=2 and x=3 are critical points. The derivative is negative for x<2
and for x>3. The derivative is positive for x between 2 and 3. 
A graph is given below.

15.  The derivative of f(x)= x-2sin(x) is 1-2 cos(x), which is zero
at x=Pi/3. So Pi/3 is the only critical point. The value of the function f
at x=Pi/3 is about -.685. The values at the endpoints x=0 and x=Pi
are 0 and Pi, respectively. So the absolute max occurs at x=Pi (with value Pi)
 and  the absolute min occurs at x=Pi/3 (with value -.685).

16.  The increase in a student's grade is given by f(x)=10*sqrt(x)-x,
where x represents the original grade. The derivative of f is

> D(f);

                                   5
                           x -> ------- - 1
                                sqrt(x)

which is zero at x=25. Since the value of f at the endpoints, x=0 and x=100
is zero and the value of f at x=25 is 25, clearly the maximum increase in 
score occurs at x=25.

17. The function describing the yo-yo is 
> y:=t->4*(t-1)^2;

                                            2
                         y := t -> 4 (t - 1)

The velocity is given by the  derivative 8(t-1)
The velocity is zero when t=1. The acceleration is 8 for all time t.