Question

Two cars start moving from the same point. One travels south at 60 mi/h and the other travels west at 25 mi/h. At what rate is the distance between the cars increasing two hours later?

Short answer

The rate at which the distance increases in 2 hours is 65 mi/h.

Step-by-step solution

Step 1:Find the rate of change of distance in two directions

Let x represents the distance traveled in the south and y is the distance traveled in the west.Therefore;

\(\frac{{{\rm{d}}x}}{{{\rm{d}}t}} = 60\) and \(\frac{{{\rm{d}}y}}{{{\rm{d}}t}} = 25\)

Find the rate of increase of distance between the curve

The figure below represents the distance traveled.

From the above triangle:

\({z^2} = {x^2} + {y^2}\)

Differentiate the equation with respect to t.

\(\begin{aligned}2z\frac{{{\rm{d}}z}}{{{\rm{d}}t}} &= 2x\frac{{{\rm{d}}x}}{{{\rm{d}}t}} + 2y\frac{{{\rm{d}}y}}{{{\rm{d}}t}}\\\frac{{{\rm{d}}z}}{{{\rm{d}}t}} &= \frac{1}{z}\left( {x\frac{{{\rm{d}}x}}{{{\rm{d}}t}} + y\frac{{{\rm{d}}y}}{{{\rm{d}}t}}} \right)\end{aligned}\)

Find the rate of increase two hours later

The distance traveled in south 120 mi whereas the distance traveled in the west is 50 mi, in two hours.

Using the equation \({z^2} = {x^2} + {y^2}\), find the value of z after two hours.

\(\begin{aligned}{z^2} &= {120^2} + {50^2}\\{z^2} &= 16900\\z &= 130\end{aligned}\)

Substitute 120 for x, 50 for y, 130 for z, 60 for \(\frac{{{\rm{d}}x}}{{{\rm{d}}t}}\), and 25 for \(\frac{{{\rm{d}}y}}{{{\rm{d}}t}}\) in the equation \(\frac{{{\rm{d}}z}}{{{\rm{d}}t}} = \frac{1}{z}\left( {x\frac{{{\rm{d}}x}}{{{\rm{d}}t}} + y\frac{{{\rm{d}}y}}{{{\rm{d}}t}}} \right)\).

\(\begin{aligned}\frac{{{\rm{d}}z}}{{{\rm{d}}t}} &= \frac{1}{{130}}\left( {\left( {120} \right)\left( {60} \right) + \left( {50} \right)\left( {25} \right)} \right)\\ &= \frac{{7200 + 1250}}{{130}}\\ &= 65\;{\rm{mi/h}}\end{aligned}\)

So, the rate of increase of distance in 2 hours is 65 mi/h.