A runner sprints around a circular track of radius 100 m at a constant speed of 7 mys. The runner’s friend is standing at a distance 200 m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200 m?

Answer:[tex]\frac{7\sqrt{15}}{4}\text{ meter per sec}[/tex]Step-by-step explanation:Let B represents the position of runner, A represents the position of the friend and C represents the position of centre of the circular track. ( shown  below ),We need to find : [tex]\frac{dc}{dt}[/tex]By the cosine law,[tex]c^2 = a^2 + b^2 - 2ab \cos C[/tex]Differentiating with respect to t ( time ),[tex]2c\frac{dc}{dt} = 2ab \sin C \frac{dC}{dt}[/tex][tex]\implies \frac{dc}{dt}=\frac{ab \sin C\frac{dC}{dt}}{c}-----(1)[/tex]Now, by arc length formula,Radius( say r ) × angle = arc length ( say l )[tex]r\times \angle C = l[/tex]Differentiating w. r. t. t,[tex]r\times \frac{dC}{dt}+\angle C\times \frac{dr}{dt}= \frac{dl}{dt}[/tex]Here, [tex]\frac{dl}{dt}=7\text{ m per sec}, \frac{dr}{dt}=0, r = 100[/tex][tex]\implies \frac{dC}{dt}=\frac{7}{100}----(2)[/tex],Now, again [tex]c^2 = a^2 + b^2 - 2ab \cos C[/tex][tex]200^2 = 100^2 + 200^2 - 2(100)(200)\cos C[/tex][tex]40000 = 10000 + 40000 - 40000\cos C[/tex][tex]-10000 = -40000\cos C[/tex][tex]\implies \cos C =\frac{1}{4}[/tex]We know that,[tex]\sin C = \sqrt{1-\cos^2 C}=\sqrt{1-\frac{1}{16}}=\sqrt{\frac{16-1}{16}}=\frac{\sqrt{15}}{4}---(3)[/tex]From equation (1), (2) and (3),[tex]\frac{dc}{dt}=\frac{(100)(200)\frac{\sqrt{15}}{4}\frac{7}{100}}{200}=\frac{7\sqrt{15}}{4}\text{ meter per sec}[/tex]Hence, the distance between the friends changing with the rate of [tex]\frac{7\sqrt{15}}{4}[/tex] meter per sec.