Using derivatives and related rates in calculus?

Question

"Two cars start moving from the same point. One travels south at 48 mi/h and the other travels west at 20 mi/h. At what rate is the distance between the cars increasing two hours later?"

 

This question is supposed to be solved using related rates and derivatives.

Answer 1

Greta,

 

BTW, I have relatives next door in Endicott, NY. so please give me a thumbs up.

 

Because the cars are heading South and West, you have a 90⁰ right angle between them.  So this boils down to a right triangle problem using the distance formula:  d(t) = st  = constant speed times time traveled.

 

The length of the triangle side heading South from the origin along the -y axis is d¹(t) = 48t

The length of the triangle side heading West from the origin along the -x axis is d²(t) = 20t

Then for any t, the distance d(t) between the two cars is just the hypotenuse of the growing right triangle.

 

So   [d(t)]² = [d1(t)]² + [d2(t)]²  =  [48t]² + [20t]²  =  2304t² + 400t² = 2704t²

Taking square roots of both sides yields  d(t) = 52t mi  since only the positive square is physically meaningfull

 

Then at t=2 hours, the distance between the two cars is d(2) = 52(2) = 104 miles

And its rate of increase is d[d(t)]/dt = 52 mph