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?
<pre><font size = 4 color = "indigo"><b>
{{{drawing(400,300,-1,7,-1,5,locate(4.3,2.4,Drum),
locate(2,1.4,r),
line(0,0,6,0), line (0,0,4,2), line(4,2,6,2),locate(4.1,1.2,6),
line(4,0,4,2), circle(4,2.2,.2), locate(-.4,0,Boat),locate(2,0,x) )}}}

We can describe the triangle by the Pythagorean theorem:

               {{{x^2+6^2=r^2}}}
                {{{x^2+36=r^2}}}

We take the derivative implicitly:

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

                    {{{x(dx/dt) = r(-6)}}}
                    {{{x(dx/dt) = -6r}}}
                      {{{dx/dt = -6r/x)}}}

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^2+36=r^2}}}

we substitute {{{x=24}}}

                {{{24^2+36=r^2}}}
                {{{576+36=r^2}}}
                   {{{612=r^2}}}
                  {{{sqrt(612)=r}}}
                {{{sqrt(36*17)=r}}}   
                  {{{6sqrt(17)=r}}}

So we substitute {{{6sqrt(17)}}} for {{{r}}} and
{{{24}}} for {{{x}}} in

               {{{dx/dt = -6r/x)}}}
               {{{dx/dt = -6(6sqrt(17)/24)}}}
               {{{dx/dt = -(36sqrt(17)/24)}}}
               {{{dx/dt = -(36sqrt(17)/24)}}}
               {{{dx/dt = -(3sqrt(17)/2)}}}
               {{{dx/dt = -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</pre>