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?

Respuesta :

Explanation:

It is given that,

    [tex]\frac{dx}{dt}[/tex] = -40 mi/h,     [tex]\frac{dx}{dt}[/tex] = -40 mi/h

The negative sign indicates that x and y are decreasing.

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

              [tex]z^{2} = x^{2} + y^{2}[/tex]

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

        [tex]2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}[/tex]

or,          [tex]\frac{dz}{dt} = \frac{1}{z}(x\frac{dx}{dt} + y\frac{dy}{dt})[/tex]

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

As,       [tex]\frac{dz}{dt} = \frac{1}{z}(x\frac{dx}{dt} + y\frac{dy}{dt})[/tex]

                       = [tex]\frac{1}{5}(4 \times (-40) + 3 \times (-40))[/tex]

                       = [tex]\frac{-140 - 120}{5}[/tex]

                       = 52

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