Two cars start moving from the same point. One travels south at 64 mi/h and the other travels west at 48 mi/h. At what rate is the distance between the cars increasing two hours later?

Answer 1

Let x = distance travelled by the southbound car in t hours and let y = distance travelled by the westbound car in t hours, and let z = distance between the cars after t hours.

Draw a right triangle with vertical side of length x, horizontal side of length y and hypotenuse z.

By the Pythagorean Theorem, x² + y² = z².

When t = 2, x = (64)(2) = 128 and y = (48)(2) = 96. So, z = √[(128)² + (96)²] = 160

Given: dx/dt = 64 and dy/dt = 48

Find: dz/dt when t = 2

z² = x² + y²

2z(dz/dt) = 2x(dx/dt) + 2y(dy/dt)

dz/dt = [x(dx/dt) + y(dy/dt)] / z = [(128)(64) + (96)(48)] / 160 = 80 mi/hr

Answer 2

This is a shortcut Physics solution, not the implicit differentiation that your Calculus textbook expects.

The rate of distance increase between the cars equals their relative velocity, which you find by subtracting the two velocities of the cars - one pointing south and one pointing west.

This vector subtraction results in a right triangle with sides 64 mi/h and 48 mi/h. The resulting relative velocity is along the hypotenuse and the distance between the cars is increasing at a rate:

relative velocity = √(64² + 48²) = 80 mi/h

Since the velocities are constant, so is the relative velocity and the result doesn't depend on time at all.