Question

water is dripping slowly onto a metal countertop, creating a circular wet spot whose area is increasing at the constant rate of pie over two square inches per sec. how fast is its radius increasing when the radius is four inches

Answer 1

Answer-

The radius is increasing at a rate of $\dfrac{1}{16}$

Solution-

The rate of change of the area of the crater is $\dfrac{\pi}{2}$, i.e

$\dfrac{dA}{dt}=\dfrac{\pi}{2}$

We know that,  

$Area =\pi r^2$

Putting this,

$\Rightarrow \dfrac{d(\pi r^2)}{dt}=\dfrac{\pi}{2}$

Applying chain rule,

$\Rightarrow \dfrac{d(\pi r^2)}{dr}\times \dfrac{dr}{dt}=\dfrac{\pi}{2}$

$\Rightarrow (2\pi r)\times \dfrac{dr}{dt}=\dfrac{\pi}{2}$

Putting the value of radius as 4 (given),

$\Rightarrow 2\pi \times 4\times \dfrac{dr}{dt}=\dfrac{\pi}{2}$

$\Rightarrow 8\pi \dfrac{dr}{dt}=\dfrac{\pi}{2}$

$\Rightarrow \dfrac{dr}{dt}=\dfrac{\pi}{2\times 8\pi}$

$\Rightarrow \dfrac{dr}{dt}=\dfrac{1}{16}$