(A) The area $A$ of a circle with radius $r$ is given by $A = \pi r^2$. Taking the derivative with respect to time $t$,we get: $\frac{dA}{dt} = \frac{

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(A) The area $A$ of a circle with radius $r$ is given by $A = \pi r^2$. Taking the derivative with respect to time $t$,we get: $\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2 \pi r \frac{dr}{dt}$. Given that the rate of change of the radius is $\frac{dr}{dt} = 4 \text{ cm/s}$. At the instant when $r = 10 \text{ cm}$,we substitute the values into the derivative formula: $\frac{dA}{dt} = 2 \pi (10 \text{ cm}) (4 \text{ cm/s}) = 80 \pi \text{ cm}^2/\text{s}$. Thus,the enclosed area is increasing at the rate of $80 \pi \text{ cm}^2/\text{s}$.

Class 12 Mathematics Application of Derivatives Rate of Change of Quantities

Medium

$A$ stone is dropped into a quiet lake and waves move in circles at a speed of $4 \text{ cm/s}$. At the instant when the radius of the circular wave is $10 \text{ cm}$,how fast is the enclosed area increasing?

A
$80 \pi \text{ cm}^2/\text{s}$
B
$40 \pi \text{ cm}^2/\text{s}$
C
$100 \pi \text{ cm}^2/\text{s}$
D
$160 \pi \text{ cm}^2/\text{s}$

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Application of Derivatives

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Rate of Change of Quantities

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$A$ stone is dropped into a quiet lake and waves move in circles at a speed of $4 \text{ cm/s}$. At the instant when the radius of the circular wave is $10 \text{ cm}$,how fast is the enclosed area increasing?

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$A$ spherical balloon is expanding. If at any instant the rate of increase of its volume is $16$ times the rate of increase of its radius,then its radius at that instant is:

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