Question

Two straight roads diverge from an intersection at an angle of \(60^\circ \). Two cars leave the intersection at the same time, the first traveling down one road at \(40\;{\rm{mi/h}}\) and the second traveling down the other road at \(60\;{\rm{mi/h}}\). How fast is the distance between the cars changing after half an hour? (Hint: Use the Law of Cosines (Formula 21 in Appendix D).)

Short answer

The distance between the cars is changing after half an hour is \(52.9\;{\rm{mi}}/{\rm{h}}\).

Step-by-step solution

Write the formula of Law of Cosines

Law of Cosines:\({c^2} = {a^2} + {b^2} - 2ab\cos \theta \)

Construct the diagram according to the given conditions

Step 3: Find how fast the distance between the cars is changing after half an hour

We need to find\(\frac{{dz}}{{dt}}\).

As the first car traveling down one road at the speed\(40\;{\rm{mi}}/{\rm{h}}\)that is\(\frac{{dx}}{{dt}} = 40\;{\rm{mi}}/{\rm{h}}\).

The second car traveling down the other road at the speed\(60\;{\rm{mi}}/{\rm{h}}\)that is\(\frac{{dy}}{{dt}} = 60\;{\rm{mi}}/{\rm{h}}\).

Apply the Law of Cosines.

\(\begin{array}{l}{z^2} = {x^2} + {y^2} - 2xy\cos 60^\circ \\{z^2} = {x^2} + {y^2} - 2xy\left( {\frac{1}{2}} \right)\\{z^2} = {x^2} + {y^2} - xy\end{array}\)

Differentiate the function w.r.t \(t\).

\(\begin{aligned}\frac{d}{{dt}}\left( {{z^2}} \right) &= \frac{d}{{dt}}\left( {{x^2}} \right) + \frac{d}{{dt}}\left( {{y^2}} \right) - \frac{d}{{dt}}\left( {xy} \right)\\2z\frac{{dz}}{{dt}} &= 2x\frac{{dx}}{{dt}} + 2y\frac{{dy}}{{dt}} - x\frac{{dy}}{{dt}} - y\frac{{dx}}{{dt}}\end{aligned}\)

As \(t = \frac{1}{2}{\rm{h}}\) then,

\(\begin{aligned}x &= 40\left( {\frac{1}{2}} \right)\\ &= 20\end{aligned}\)

And

\(\begin{aligned}y &= 60\left( {\frac{1}{2}} \right)\\ &= 30\end{aligned}\)

Now put \(x = 20\) and \(y = 30\) into \({z^2} = {x^2} + {y^2} - xy\).

\(\begin{aligned}{z^2} &= {\left( {20} \right)^2} + {\left( {30} \right)^2} - \left( {20} \right)\left( {30} \right)\\{z^2} &= 400 + 900 - 600\\z &= \sqrt {700} \end{aligned}\)

Therefore,

\(\begin{aligned}\frac{{dz}}{{dt}} &= \frac{{2x\frac{{dx}}{{dt}} + 2y\frac{{dy}}{{dt}} - x\frac{{dy}}{{dt}} - y\frac{{dx}}{{dt}}}}{{2z}}\\ &= \frac{{2\left( {20} \right)\left( {40} \right) + 2\left( {30} \right)\left( {60} \right) - 20\left( {60} \right) - 30\left( {40} \right)}}{{2\left( {\sqrt {700} } \right)}}\\ &= \frac{{2800}}{{2\sqrt {700} }}\\ \approx 52.9\;{\rm{mi}}/{\rm{h}}\end{aligned}\)

Thus, the distance between the cars is changing after half an hour is \(52.9\;{\rm{mi}}/{\rm{h}}\).