If the radius of a circle be increasing at a uniform rate of `2cms^(-1)`. The rate of increasing of area of circle, at the instant when the radius is 20 cm is

Options

A. `70 pi cm^(2)s^(-1)`
B. `70 cm^(2) s^(-1)`
C. `80 pi cm^(2)s^(-1)`
D. `80 cm^(2)s^(-1)`

Answer

C

Explanation

Given `(dr)/(dt)=2 cm //s`
and `r=20 cm`
Let area of circle `A=pir^(2)`
`implies (dA)/(dt)=2pir(dr)/(dt)=(dA)/(dt)=2pi.20.2`
`implies(dA)/(dt)=80 pi cm^(2)//s`
Given `x=a(cos theta +theta sin theta)`
and `y=a (sin theta-theta cos theta)`
`implies(dx)/(d theta)=a(- sin theta + sin theta +theta cos theta)`
`implies(dx)/( d theta) =a theta cos theta`
and `(dy)/(d theta)=a (cos theta -cos theta +theta sin theta) `
`implies(dy)/( d theta) =a theta sin theta :. (dy)/(dx) = tan theta`
Then slope of normal `=(-1)/(tan theta) =-(cos theta)/(sin theta)`
So equation of normal is
`y-a sin theta + a theta cos theta=-(cos theta)/(sin theta )(x-a cos theta-a theta sin theta)`
`impliesy sin theta -a sin^(2) theta + a theta cos theta sin theta`
`=-x cos theta +a cos^(2) theta +a theta sin theta cos theta `
`implies x cos theta +y sin theta =a`
It is always at a constant distance a from origin.