2008 Solutions  (average scores 4.9, 3.4, 2.5, 2.6, 3.7, 3.1)

1a) Y1(x) = sin(πx);  Y2(x) = x3 – 4x

      Y1(x) = Y2(x)  at x = 0 & x = 2;  

 b) Y2(x) = -2 at r = 0.5391889 and s = 1.6751309; 

 

 c)     

      note: this is a less common type of volume problem.  The cross sections are squares

      instead of disks or washers.  So the area of each cross section is (y1 – y2)2.

 

 d) 

       note: You can imagine dividing the lake up into rectangular cross sections.  One

       dimension is (3-x) & the other is (y1 – y2).

 

 Note:  parts a) & b) are an easy 5 points

 

 

2a)   

 

  b)   

 

note:  we are just estimating  using the trapezoidal rule.

 

c) L is differentiable on [0, 9] so the Mean Value Theorem implies L′(t) is 56/3 for some t in (1, 3); 24/3 for some t in (4, 7); -50/1  for some t in (3, 4); & -70/1 for some t in (7, 8).

 

L’ is continuous, so the Intermediate Value Theorem implies that L′(t) = 0 for at least three values of t in [0, 9].

 

d) , so there were approximately 973 tickets sold by 3 P.M.

 

Note:  parts a) & d) are an easy 4 points        

 

 

3a) V’ = πr2h’ + 2πrr’h ;                                  b)  R(t) = 2000 when t = 25 minutes.

       r = 100, h = .5, V’ = 2000, r’ = 2.5                The slick reaches its maximum volume

       2000 = π(100)2h’ + 2π(100)(2.5)(.5)              at t=25 because R(t) < 2000 for t < 25

       h’(t) = .038 cm/min                                           & R(t) > 2000 for t > 25

 

c) The volume in cm3 at t = 25 minutes is 60,000 +

Note: fairly routine related rates  questions 4 pts for a) 3 for b) & 2 for c)        

 

4a)   V(t)<0 for 0<t<3 & 5<t<6 so look at t = 3 & t = 6.  Let P(t) be the position of the particle.    

       ;

        The particle is farthest left at t = 3 when its position is -10.

 

 b)   P(0) = -2 & P(3) = -10 & the particle was moving continuously to the left;

        P(3) = -10 & P(5) = -7 & the particle moved continuously to the right;

        P(5) = -7 & P(6) = -9 & the particle moved continuously to the left;

        By the Intermediate Value Theorem there are three values of t for which P(t) = -8.

 

 c)   The speed is decreasing for 2 < t < 3 because v < 0 and v is increasing.

 

 d)    The acceleration is negative on 0 < t < 1 & 4 < t < 6 since the velocity is decreasing there.

 

a & d  are a  fairly routine 5 points (make sure to show work/explain)

 

5a) Just draw in tangent lines with roughly the right slope

 

  b) 1/(y-1) y’ = x-2   ln(|y-1|) = -x-1 + C ;  ln(1) = -.5 + C C = .5

             

   c)    

 

Easy 5  pts for a) & writing the following for b):   1/(y-1) y’ = x-2   ln(y-1) = -x-1 + C

 

 6a)  f(e2) = ln(e2)/e2 = 2/e2 ; f ’(e2) = [1-ln(e2)]/(e2)2 = -1/e4;

 

            y – 2/e2 = -1/e4(x – e2)  or    y – ln(e2)/e2= [1-ln(e2)]/(e2)2 (x – e2)

  

b) f ‘(x) = 0 when x = e.  The function f has a relative maximum at x = e because f ‘(x)

       changes from positive to negative at x = e.

     

  c) ; f’’(x) = 0  -3+2ln(x)=0  x = e1.5    

       The graph has an inflection point at x = e1.5 because f’’ changes sign.

 d)  The limit does not exist   Note: 2 pts for a) & 2 pts for a formula for f ’’ on c)