Question

The diameter of a circle is increasing at the rate of $ 1cm/\sec $ . When its radius is $ \pi $ , the rate of increase of its area is
A. $ \text{ }\!\!\pi\!\!\text{ c}{{\text{m}}^{2}}/\text{sec} $
B. $ \text{2 }\!\!\pi\!\!\text{ c}{{\text{m}}^{2}}/\text{sec} $
C. $ {{\text{ }\!\!\pi\!\!\text{ }}^{2}}\text{c}{{\text{m}}^{2}}/\text{sec} $
D. $ \text{2}{{\text{ }\!\!\pi\!\!\text{ }}^{2}}\text{c}{{\text{m}}^{2}}/\text{sec} $

Answer 1

Hint: Type of question is based on the simple differentiation, as question says that diameter is increasing at the rate of $ 1cm/\sec $ which means $ \dfrac{dD}{dt}=1cm/\sec $ in which ‘D’ is the diameter of circle. And to find out the rate of increase of area we had to differentiate the area and simplify it according to the given condition in question.

Complete step by step answer:
So according to question we need to calculate the rate of increase of area, as we know that rate of change is differentiation, so we can say that we need to find out the $ \dfrac{dA}{dt} $ in which ‘A’ is the area of circle. So we will put the formula of area and differentiate it, from where we will get the answer.
So moving ahead with the question as we know that;
Diameter of the circle in terms of radius can be written as 2r where ‘r’ is the radius of the circle. So we can write it as;
 $ D=2r $
On differentiation, the rate of change of differentiation is given to us. So we can write it as;
 $ \begin{align}
  & D=2r \\
 & \dfrac{dD}{dt}=\dfrac{d2r}{dt} \\
 & \dfrac{dD}{dt}=2\dfrac{dr}{dt} \\
\end{align} $
As we know that $ \dfrac{dD}{dt}=1cm/\sec $ , then it would be
 $ \begin{align}
  & \dfrac{dD}{dt}=2\dfrac{dr}{dt} \\
 & 1=2\dfrac{dr}{dt} \\
 & \dfrac{dr}{dt}=\dfrac{1}{2} \\
\end{align} $
So we got rate of change of radius i.e. $ \dfrac{dr}{dt}=\dfrac{1}{2} $ equation (i)
As we know, that area of circle of radius is equal to $ \text{ }\!\!\pi\!\!\text{ }{{\text{r}}^{2}} $ i.e. area of circle $ \text{= }\!\!\pi\!\!\text{ }{{\text{r}}^{2}} $
So now to find the answer we had to find out the rate of change of area at radius equal to $ \text{ }\!\!\pi\!\!\text{ } $ .
So we can write it as
 $ \begin{align}
  & \text{A= }\!\!\pi\!\!\text{ }{{\text{r}}^{2}} \\
 & \dfrac{dA}{dt}=\dfrac{d\text{ }\!\!\pi\!\!\text{ }{{\text{r}}^{2}}}{dt} \\
 & \dfrac{dA}{dt}=2\text{ }\!\!\pi\!\!\text{ }r\dfrac{d\text{r}}{dt} \\
\end{align} $
As we know that $ \dfrac{dr}{dt}=\dfrac{1}{2} $ , from equation (i), put it in the above equation. So it will be
\[\begin{align}
  & \dfrac{dA}{dt}=2\text{ }\!\!\pi\!\!\text{ }r\dfrac{1}{2} \\
 & \dfrac{dA}{dt}=\text{ }\!\!\pi\!\!\text{ }r \\
\end{align}\]
Area at radius equal to $ \text{ }\!\!\pi\!\!\text{ } $ , so put a radius equal to $ \text{ }\!\!\pi\!\!\text{ } $ in above equation, then we will get;
\[\dfrac{dA}{dt}={{\text{ }\!\!\pi\!\!\text{ }}^{2}}\]
Hence the answer is \[{{\text{ }\!\!\pi\!\!\text{ }}^{2}}\]i.e. the rate of increase of its area is $ {{\text{ }\!\!\pi\!\!\text{ }}^{2}}\text{c}{{\text{m}}^{2}}/\text{sec} $

So, the correct answer is “Option C”.

Note: As we had changed the diameter to radius and found its relation, because in area we are required radius. But if you don’t want than you can go by diameter but then you have to write area of circle in terms of diameter i.e. $ \dfrac{\text{ }\!\!\pi\!\!\text{ }{{\text{D}}^{2}}}{4} $ rather than writing it as $ \text{ }\!\!\pi\!\!\text{ }{{\text{r}}^{2}} $ and also last instead of putting a value of radius equal to $ \text{ }\!\!\pi\!\!\text{ } $ you have to put a value of diameter which will be $ \text{2 }\!\!\pi\!\!\text{ } $