Question

At a given moment, a plane passes directly above a radar station at an altitude of 6 km. Let θ be the angle that the line through the radar station and the plane makes with the horizontal. How fast is θ changing 12 min after the plane passes over the radar station? ________km/h.

Answer 1

- 187.237 km/hr fast is θ changing 12 min after the plane passes over the radar station

Explanation:

Let the distance x and angle θ be defined as in the figure below. Then

                  $\tan \theta=\frac{6}{x}$

Now, differentiate with respect to t, we get

                 $\sec ^{2} \theta \frac{d \theta}{d t}=-\left(\frac{6}{x^{2}}\right) \frac{d x}{d t}$

Now, calculate the travel distance from radar station to plane after 12min

Distance,  $x=800 \times \frac{12}{60}=160$

Substituting ‘x’ value, we get

                $\tan \theta=\frac{6}{160}=\frac{3}{80}$

Find the rate of change of theta after 12 min,

                $\frac{d \theta}{d t}=-\frac{1}{\sec ^{2} \theta} \times \frac{6}{x^{2}} \times \frac{d x}{d t}$

We know, the formula for,

                $\sec ^{2} \theta=1+\tan ^{2} \theta=1+\frac{3^{2}}{80^{2}}$

So, then, $\frac{d x}{d t}=800 \mathrm{km} / \mathrm{hr}(\text { let assume })$

               $\frac{d \theta}{d t}=-\frac{1}{\left(1+\frac{3^{2}}{80^{2}}\right)} \times \frac{6}{160^{2}} \times 800$

               $=-\frac{1}{\left(1+\frac{3^{2}}{80^{2}}\right)} \times \frac{6}{160^{2}} \times 800$

               $=-\frac{1}{1.0014} \times \frac{6}{25600} \times 800$

             $=-\frac{4800}{25635.84}=-0.187237 \mathrm{rad} / \mathrm{hr}$

When express the value in km/he, we get, the change in theta as

              $=-187.237 \mathrm{km} / \mathrm{hr}$