Question 562747
Hi, let A=(0,0).  Then, the second plane starts at (0,240).  Let x(t) be the distance from A to plane 1 at time t, y(t) be the distance from A to plane 2 at time 2, and z(t) be the distance between the planes at time t (I say distance because I know that plane two doesn't reach the airport in 15 minutes). We know,
{{{x^2+y^2=z^2}}}
{{{(dx)/(dt)=480}}}
{{{(dy)/(dt)=-320}}}
So by implicit differentiation,
{{{2x((dx)/(dt))+2y((dy)/(dt))=2z((dz)/(dt))}}}
So at time t=1/4 hr,
{{{x=480(1/4)=120}}}
{{{y=240-320(1/4)=160}}}
{{{z=sqrt(120^2+160^2)=200}}}
So, we get
{{{(dz)/(dt)=(1/200)(120*480+160*(-320))=32}}}
This is positive, so the distance between them is increasing, so the planes are getting further away at a rate of 32mph.