The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
The area of a circle (A) with radius (r) is given by,
A = πr2
Now, the rate of change of area (A) with respect to time (t) is given by,
\begin{align} \frac{dA}{dt}=\frac{d}{dt}(\pi r^2).\frac{dr}{dt}=2\pi r\frac{dr}{dt}\;\;\;[By\; Chain \;Rule]\end{align}
It is given that,
\begin{align} \frac{dr}{dt}= 3\; cm/s\end{align}
\begin{align} \therefore \frac{dA}{dt}= 2\pi r(3)=6 \pi r \end{align}
Thus, when r = 10 cm,
\begin{align} \frac{dA}{dt}= 6\pi(10)=60 \pi\; cm^2/s \end{align}
Hence, the rate at which the area of the circle is increasing when the radius is 10 cm is 60π cm2/s.
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Welcome to the NCERT Solutions for Class 12 Mathematics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 3: The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the are....
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