2005 AP Solutions

1a) f & g intersect at x = 0.178218 & x=1;                                          

 

  b) 

 

 

c) Y1 = R(x) = f(x) + 1 V =                                                         

    Y2 = r(x) = g(x) + 1                                         

                                                              

 

2 a)       (reminder: integrate a rate to get a change in amount)

   

   b) Y(t) = 2500 + or

     (amount at at time 0 + amount gained – amount lost)

 

   c) S(4) – R(4) = -1.908 or -1.909 yd3/hr

 

 

   d)  t = 5.11787 is a relative min as Y’(t)  change from negative to positive.  Also check the

         endpoints: Y(0) = 2500; Y(5.11787) = 2492.364; Y(6) = 2493.2766.  The minimum

         amount of sand is 2492.364 yd3 at t = 5.11787. (AP folks - in this case no need to check

        endpoints, but easier to check than explain why you don’t need to)

 

        

3a)      T’(7) ≈  oC/cm

 

 

  b)       

 

               (notice that the intervals for the trapezoid method have different widths)

      

  c)  =T(8) – T(0) = -45 oC     (when you integrate a rate you get a change in amount)

       

        The temperature of the wire drops 45oC when you go 8 cm from the heated end.

 

 

 

4a) f has a relative max at x = 2 because f ‘ changes from + to -.

 

  b) Plot points (0,-1), (1,0),(2,2), (3,0); make a corner at x = 2; and

                for 0 < x < 1  below x-axis;increasing; concave down

                for 1 < x < 2  above x-axis;increasing; concave up

                etc.

 

c) g’(x) = f(x), so rel min at x =1 as g’ goes from – to +;  rel max at x = 3 as g’ goes from + to - .

 

d) g’’(x) = f ‘(x), so I.P. at x = 2 as g’’ changes sign.

 

 

 

 

5a)  = 360 meters.  The car traveled 360 meters from t = 0 to t =  24 sec. 

(the velocity is all positive so no need to worry about distance vs displacement)

 

  b) v’(4) does not exist because: ; v’(20) = (0-20)/(24-16) = -20/8 m/sec2

 

  c) and a(t) does not exist at t = 4 & t = 16

 

d) .  No guarantee because v’(16) does not exist.

 

 

 

6a) slope at the point (-1,1) is 2 so draw small line segment with slope of around 2 there.  Do the

       same thing for the other points on the grid.

 

  b) y’ =-2x/y, so at the point (1,-1), y’ = 2;the tangent line is y + 1 = 2(x – 1).

       plug x = 1.1 into the tangent line eq. to get tangent line estimate of y =  -0.8

 

  c)