Karl's Calculus Tutor - Solution to Exercise 8.2-1

Solution to Exercise 8.2-1KCT logo

© 1999 by Karl Hahn

The problem was that a northbound taxicab starts from the intersection of Park Ave. and 59th St. at 40 miles per hour. At the same time a westbound taxicab starts from one quarter mile west of Park and 59th at 30 miles per hour. When the northbound taxicab has gone one half mile from its starting position, how fast is the distance between the two taxicabs increasing?

Recall that in the original taxicab problem you came up with a relationship between y, which was how far north of Park and 59th the northbound taxicab was, and x, which was how far west of Park and 59th the westbound taxicab was. This relationship is the only thing that has changed from the original problem. By giving the westbound taxicab a head start, you have added a fixed amount to x. So, subtracting that fixed amount from x to get how far the westbound taxicab has traveled since the start, you now would have 

      y       x - 0.25 miles
           =                
   40 mph        30 mph
or equivalently 
   3       1
     y  +     mile  =  x
   4       4
That is, if you subtract the quarter-mile head start from x, then the speeds (30 mph and 40 mph) must govern the relationship between how far the two taxicabs have gone.

Everything else is the same in this problem. You still have the Pythagorean distance formula dictating that: 

   s2  =  x2 + y2
where s is the distance separating the taxicabs. And you can still take the derivatives of the two relationships: 
  3 dy     dx
        =    
  4 dt     dt


     ds        dx        dy
  2s     =  2x     +  2y   
     dt        dt        dt
and you can still make the substitutions suggested by the relationships. We cancel the 2's in the above. Then using substitutions based upon s2 = x2 + y2 and dx/dt = (3/4) dy/dt we have: 
            ds     3   dy       dy
   Öx2 + y2     =    x     +  y   
            dt     4   dt       dt
Now replace x with (3/4)y + (1/4) mile, multiply things out, and gather like terms: 



   










25
  
16




 y2




     3        1
  +    y  +    
     8       16









 ds
     =
 dt




  




25        3
   y  +    
16       16









 dy
   
 dt








Remember that y was given as half a mile, and dy/dt was
given as 40 miles per hour.  Poking the calculator indicates that the
above becomes:


             ds
   0.8003905     =  38.75
             dt


   ds
       =  48.4138662 mph
   dt








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