Answer to Question #156055 in Trigonometry for anne

Question

A plane flying with a constant speed of 300 km/h passes over a ground radar station at an altitude of 1 km and climbs at an angle of 300. At what rate is the distance from the plane to the radar station increasing a minute later?

Answer 1

Explanations & Calculations

• When it is needed to check the rate of change of some variable it is necessary to check for the derivative of that with respect to time.
• like, the rate of change of displacement is called the velocity & the rate of change of velocity is called the acceleration.
• To derive a convenient equation, consider the figure below.

• Neglect the height of the radar station & consider the 1km is measured from there.
• Consider a moment after passing the at a time of t, then the distance covered is $\small vt$ .
• Then $\small r$ can be written as,

$\qquad\qquad \begin{aligned} \small \cos(120)&= \small \frac{(vt)^2+(1km)^2-r^2}{2(vt)(1km)}\cdots(1)\\ \small 2vt\cos (120)&= \small v^2t^2-r^2-1\\ \small 2v\cos(120)\times 1&= \small 2v^2t-2r\frac{dr}{dt}-0\\ \small \frac{dr}{dt}&= \small \frac{2v^2t-2v\cos(120)}{2r}\\ \small \frac{dr}{dt}_{t=\frac{1}{60}h}&= \small \frac{2(300kmh^{-1})^2(\frac{1}{60}h)-2(300)(-0.5)}{2r}\\ \small &= \small \frac{1650}{r} \end{aligned}$

• To calculate $\small r$ at one minute time, the same (1) equation can be used.

$\qquad\qquad \begin{aligned} \small -0.5&= \small \frac{(300\times \frac{1}{60})^2+1-r^2}{2(300\times\frac{1}{60})(1)}\\ \small r&= \small 5.568km \end{aligned}$

• Then,

$\qquad\qquad \begin{aligned} \small \frac{dr}{dt}_{t=\frac{1}{60}}&= \small \frac{1650\,km^2h^{-1}}{5.568km}\\ &= \small \bold{296.336\,kmh^{-1}} \end{aligned}$

• Note that the time— one minute— was expressed in hours, due to the speed being in the units of $\small kmh^{-1}$