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

c

Explanation:

Answer 2

`(7√(15))/(4)\text{ meter per sec}`

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 :
`(dc)/(dt)`

By the cosine law,

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

Differentiating with respect to t ( time ),

`2c(dc)/(dt) = 2ab \sin C (dC)/(dt)`

`\implies (dc)/(dt)=(ab \sin C(dC)/(dt))/(c)-----(1)`

Now, by arc length formula,

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

`r* \angle C = l`

Differentiating w. r. t. t,

`r* (dC)/(dt)+\angle C* (dr)/(dt)= (dl)/(dt)`

Here,
`(dl)/(dt)=7\text{ m per sec}, (dr)/(dt)=0, r = 100`

`\implies (dC)/(dt)=(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 =(1)/(4)`

We know that,

`\sin C = √(1-\cos^2 C)=\sqrt{1-(1)/(16)}=\sqrt{(16-1)/(16)}=(√(15))/(4)---(3)`

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

`(dc)/(dt)=((100)(200)(√(15))/(4)(7)/(100))/(200)=(7√(15))/(4)\text{ meter per sec}`

Hence, the distance between the friends changing with the rate of
`(7√(15))/(4)` meter per sec.