Question

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 1

Answer:

$\frac{7\sqrt{15}}{4}\text{ meter per sec}$

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 : $\frac{dc}{dt}$

By the cosine law,

$c^2 = a^2 + b^2 - 2ab \cos C$

Differentiating with respect to t ( time ),

$2c\frac{dc}{dt} = 2ab \sin C \frac{dC}{dt}$

$\implies \frac{dc}{dt}=\frac{ab \sin C\frac{dC}{dt}}{c}-----(1)$

Now, by arc length formula,

Radius( say r ) × angle = arc length ( say l )

$r\times \angle C = l$

Differentiating w. r. t. t,

$r\times \frac{dC}{dt}+\angle C\times \frac{dr}{dt}= \frac{dl}{dt}$

Here, $\frac{dl}{dt}=7\text{ m per sec}, \frac{dr}{dt}=0, r = 100$

$\implies \frac{dC}{dt}=\frac{7}{100}----(2)$,

Now, again $c^2 = a^2 + b^2 - 2ab \cos C$

$200^2 = 100^2 + 200^2 - 2(100)(200)\cos C$

$40000 = 10000 + 40000 - 40000\cos C$

$-10000 = -40000\cos C$

$\implies \cos C =\frac{1}{4}$

We know that,

$\sin C = \sqrt{1-\cos^2 C}=\sqrt{1-\frac{1}{16}}=\sqrt{\frac{16-1}{16}}=\frac{\sqrt{15}}{4}---(3)$

From equation (1), (2) and (3),

$\frac{dc}{dt}=\frac{(100)(200)\frac{\sqrt{15}}{4}\frac{7}{100}}{200}=\frac{7\sqrt{15}}{4}\text{ meter per sec}$

Hence, the distance between the friends changing with the rate of $\frac{7\sqrt{15}}{4}$ meter per sec.

Answer 2

Answer:

c

Step-by-step explanation: