Question

A plane flying with a constant speed of \(296\;{\rm{km/h}}\) passes over a ground radar station at an altitude of \(1\) km and climbs at an angle of \(30^\circ \). At what rate is the distance from the plane to the radar station increasing a minute later?

Short answer

The distance from the plane to the radar station increases a minute later at the rate of \(296\;{\rm{km}}/{\rm{h}}\).

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 the rate at which the distance from the plane to the radar station increases a minute later

Given that \(\frac{{dx}}{{dt}} = 300\;{\rm{km}}/{\rm{h}}\).

Apply the Law of Cosines.

\(\begin{aligned}{y^2} &= {x^2} + {\left( 1 \right)^2} - 2\left( 1 \right)\left( x \right)\cos 120^\circ \\{y^2} &= {x^2} + 1 - 2x\left( { - \frac{1}{2}} \right)\\{y^2} &= {x^2} + x + 1................\left( 1 \right)\end{aligned}\)

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

\(\begin{aligned}\frac{d}{{dt}}\left( {{y^2}} \right) &= \frac{d}{{dt}}\left( {{x^2} + x + 1} \right)\\2y\frac{{dy}}{{dt}} &= 2x\frac{{dx}}{{dt}} + \frac{{dx}}{{dt}}\\\frac{{dy}}{{dt}} &= \frac{{2x + 1}}{{2y}} \cdot \frac{{dx}}{{dt}}...........\left( 2 \right)\end{aligned}\)

Put \(x = 5\) in the equation \(\left( 1 \right)\) then,

\(\begin{aligned}{y^2} &= {\left( 5 \right)^2} + \left( 5 \right) + 1\\y &= \sqrt {25 + 6} \\ &= \sqrt {31} \;{\rm{km}}\end{aligned}\)

Now put \(x = 5\), \(y = \sqrt {31} \) and \(\frac{{dx}}{{dt}} = 300\)into theequation \(\left( 2 \right)\).

\(\begin{aligned}\frac{{dy}}{{dt}} &= \frac{{2\left( 5 \right) + 1}}{{2\left( {\sqrt {31} } \right)}} \cdot 300\\ &= \frac{{1650}}{{\sqrt {31} }}\\ \approx 296\;{\rm{km}}/{\rm{h}}\end{aligned}\)

Thus, the distance from the plane to the radar station increases a minute later at the rate of \(296\;{\rm{km}}/{\rm{h}}\).