Question

Car A is traveling west at 40 mi/h and car B is traveling north at 40 mi/h. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car A is 4 mi and car B is 3 mi from the intersection?

Answer 1

Explanation:

It is given that,

    $\frac{dx}{dt}$ = -40 mi/h,     $\frac{dx}{dt}$ = -40 mi/h

The negative sign indicates that x and y are decreasing.

We have to find $\frac{dz}{dt}$. Equation for the given variables according to the Pythagoras theorem is as follows.

              $z^{2} = x^{2} + y^{2}$

Now, we will differentiate each side w.r.t 't' as follows.

        $2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$

or,          $\frac{dz}{dt} = \frac{1}{z}(x\frac{dx}{dt} + y\frac{dy}{dt})$

So, when x = 4 mi, and y = 3 mi then z = 5 mi.

As,       $\frac{dz}{dt} = \frac{1}{z}(x\frac{dx}{dt} + y\frac{dy}{dt})$

                       = $\frac{1}{5}(4 \times (-40) + 3 \times (-40))$

                       = $\frac{-140 - 120}{5}$

                       = 52

Thus, we can conclude that the cars are approaching at a rate of 52 mi/h.