The radius of a circle is increasing at a rate of $0.1 \mathrm{~cm} \mathrm{~s}^{-1}$. Then the rate of…
The radius of a circle is increasing at a rate of $0.1 \mathrm{~cm} \mathrm{~s}^{-1}$. Then the rate of change of area, when its radius is $5 \mathrm{~cm}$, is .........
$\pi^2 \mathrm{~cm}^2 \mathrm{~s}^{-1}$
$\pi \mathrm{cm}^2 \mathrm{~s}^{-1}$
$2 \pi \mathrm{cm}^2 \mathrm{~s}^{-1}$
$\frac{\pi}{2} \mathrm{~cm}^2 \mathrm{~s}^{-1}$
Solution
Let the circle have radius $r \mathrm{~cm}$ and area $A \mathrm{~cm}^2$ and it is given that $\frac{d r}{d t}=0.1 \mathrm{~cm} . \mathrm{s}^{-1}$ and as we know area $A=\pi r^2 \Rightarrow \frac{d A}{d t}=2 \pi r \frac{d r}{d t}$
$\therefore$ Rate of change of area, when radius is $5 \mathrm{~cm}$ is $\left.\frac{d A}{d t}\right|_{r=5 \mathrm{~cm}}=2 \pi(5)(0.1)=\pi \mathrm{cm}^2 \cdot \mathrm{s}^{-1}$
Alternative Solution:
Sure, let's break this down.
The formula for the area of a circle is $A = \pi r^2$, where $r$ is the radius of the circle. We want to find how fast the area is changing when the radius is $5$ cm.
Since the radius is changing, we'll want to use differentiation to find the rate of change of the area with respect to time. This is often called the derivative of the area with respect to time. In mathematical terms, we want to find $\frac{dA}{dt}$, when $r = 5$ cm.
Taking the derivative of both sides of the area formula with respect to time, we get $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
We know that $\frac{dr}{dt} = 0.1$ cm/s (the rate at which the radius is increasing), and we're looking for the rate when $r = 5$ cm.
Substituting these values into our derivative equation, we get $\frac{dA}{dt} = 2\pi \cdot 5 \cdot 0.1 = \pi$ cm²/s.
So, the rate of change of the area of the circle when its radius is $5$ cm is $\pi$ cm²/s. Therefore, the correct answer is option B) $\pi$ cm²/s.