Question 1082937
The distance between them is the hypotenuse of a right triangle who's legs are measured by the rate of each boat.
{{{A^2+B^2=H^2}}}
Let's measure time starting at 11 AM,
{{{A=20t+20=20(t+1)}}}
{{{B=9t}}}
So then,
{{{400(t+1)^2+(9t)^2=H^2}}}
{{{400(t^2+2t+1)+81t^2=H^2}}}
{{{H^2=481t^2+800t+400}}}
So at noon, {{{t=2}}},
{{{H^2=481(4)+800(2)+400}}}
{{{H^2=1924+1600+400}}}
{{{H^2=3924}}}
{{{H=sqrt(3924)}}}
{{{H=6sqrt(109)}}}
So now to find the rate of separation, implicitly differentiate the equation above with respect to time.
{{{2A(dA/dt)+2B(dB/dt)=2H(dH/dt)}}}
{{{A=20(2+1)=60}}}
{{{B=9(2)=18}}}
{{{dA/dt=20}}}
{{{dB/dt=9}}}
Substituting,
{{{2(60)(20)+2(18)(9)=2(6sqrt(109))(dH/dt)}}}
{{{12sqrt(109)(dH/dt)=2724)}}}
{{{dH/dt=227/sqrt(109)}}}
{{{highlight(dH/dt=(227/109)sqrt(109))}}}{{{kph}}}