Question 682777
Two ants are at a common point at time t=0, the first ant starts crawling along a straight line at the rate of 4 ft/min. Two minutes later, the second ant starts crawling in a direction perpendicular to that of the first, at a rate of 5 ft/min. How fast is the distance between them changing when the first insect has traveled 12 feet?

<br> Let r be the distance between them, x be the distance of the first and y be the distance of the second. Then according to the Pythagorean theorem:<br>
{{{r^2=x^2+y^2}}}
Taking the derivative of the equation with respect to time yields:
{{{2r(dr/dt)=2x(dx/dt)+2y(dy/dt)}}}
Everything is divisible by 2 so divide the whole thing by 2 to make things easier.
{{{highlight(r(dr/dt)=x(dx/dt)+y(dy/dt))}}}
Now use a little algebra to find the missing components.
Given:
{{{dx/dt=4}}}the rate of the first ant
{{{dy/dt=5}}}the rate of the second ant
{{{x=12}}} the distance the first ant traveled.
<br>distance = rate*time<br> 

We get t from the first ant.
{{{12=4t}}}
{{{12/4=4t/4}}}
{{{3=t}}}
Therefore the second ant traveled.
{{{y=5(t-2)}}}
{{{y=5(highlight(3)-2)}}}
{{{y=5}}}

<br> We use the Pythagorean Theorem to find r <br>
{{{r^2=x^2+y^2}}}
{{{r^2=12^2+5^2}}}
{{{r^2=144+25}}}
{{{r^2=169}}}
{{{r=sqrt(169)}}}
{{{r=13}}}

Take all the parts and put them into the equation you took the derivative of.
{{{highlight(r(dr/dt)=x(dx/dt)+y(dy/dt))}}}
{{{13(dr/dt)=12(4)+5(5)}}}
{{{13(dr/dt)=48+25}}}
{{{13(dr/dt)=73}}}
{{{dr/dt=73/13}}}
{{{dr/dt=5.6}}}ft/min

That should be it barring any careless mistakes. Happy Calculating!