SOLUTION: A boat is pulled toward dock by means of a rope wound on a drum that is located 6ft above the bow of the boat. if the rope is being pulled in at the rate of 6ft/sec, how fast is th

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Question 225993: A boat is pulled toward dock by means of a rope wound on a drum that is located 6ft above the bow of the boat. if the rope is being pulled in at the rate of 6ft/sec, how fast is the boat approaching the dock when it is 24ft from the dock?
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
A boat is pulled toward dock by means of a rope wound on a drum that is located 6ft above the bow of the boat. if the rope is being pulled in at the rate of 6ft/sec, how fast is the boat approaching the dock when it is 24ft from the dock?



We can describe the triangle by the Pythagorean theorem:

               x%5E2%2B6%5E2=r%5E2
                x%5E2%2B36=r%5E2

We take the derivative implicitly:

               2x%28dx%2Fdt%29+%2B+0+=+2r%28dr%2Fdt%29 
                   2x%28dx%2Fdt%29+=+2r%28dr%2Fdt%29
                    x%28dx%2Fdt%29+=+r%28dr%2Fdt%29
 
We are told that r (the rope length) is shrinking at the 
constant rate of 6 ft/sec.  Therefore %28dr%29%2F%28dt%29=-6
(it is negative because r is shrinking).  So we
substitute -6 for dr%2Fdt

                    x%28dx%2Fdt%29+=+r%28-6%29
                    x%28dx%2Fdt%29+=+-6r
                      dx%2Fdt+=+-6r%2Fx%29

Now we want to freeze the motion at the instant when the boat
is 24 feet from the dock.  That is, when x = 24.

Since               x%5E2%2B36=r%5E2

we substitute x=24

                24%5E2%2B36=r%5E2
                576%2B36=r%5E2
                   612=r%5E2
                  sqrt%28612%29=r
                sqrt%2836%2A17%29=r   
                  6sqrt%2817%29=r

So we substitute 6sqrt%2817%29 for r and
24 for x in

               dx%2Fdt+=+-6r%2Fx%29
               dx%2Fdt+=+-6%286sqrt%2817%29%2F24%29
               dx%2Fdt+=+-%2836sqrt%2817%29%2F24%29
               dx%2Fdt+=+-%2836sqrt%2817%29%2F24%29
               dx%2Fdt+=+-%283sqrt%2817%29%2F2%29
               dx%2Fdt+=+-6.2, approximately.

So x is shrinking at the rate of 6.2 ft/sec. Therefore
that is how fast the boat is approaching the dock at
that instant.              

Edwin