(A) Given that the rate of change of the radius is $\frac{dr}{dt} = 5 \ cm/sec$. At $t = 0$,the radius $r = 0$. Integrating $\frac{dr}{dt} = 5$,we get

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(A) Given that the rate of change of the radius is $\frac{dr}{dt} = 5 \ cm/sec$. At $t = 0$,the radius $r = 0$. Integrating $\frac{dr}{dt} = 5$,we get $r = 5t$. At $t = 2 \ seconds$,the radius $r = 5(2) = 10 \ cm$. The area of the circular ripple is $A = \pi r^2$. Differentiating with respect to time $t$,we get $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$. Substituting the values $r = 10 \ cm$ and $\frac{dr}{dt} = 5 \ cm/sec$: $\frac{dA}{dt} = 2 \pi (10)(5) = 100 \pi \ cm^2/sec$.

Class 12 Mathematics Applications of Derivatives Rate of Change of Quantities

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$A$ stone is dropped into a pond. Waves in the form of circles are generated and the radius of the outermost ripple increases at the rate of $5 \ cm/sec$. The rate at which the area increases after $2 \ seconds$ is:

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A
$100 \pi \ cm^2/sec$
B
$40 \pi \ cm^2/sec$
C
$50 \pi \ cm^2/sec$
D
$25 \pi \ cm^2/sec$

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$A$ stone is dropped into a pond. Waves in the form of circles are generated and the radius of the outermost ripple increases at the rate of $5 \ cm/sec$. The rate at which the area increases after $2 \ seconds$ is:

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