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Read MoreWelcome to the NCERT Solutions for Class 12 Mathematics - Chapter Application of Derivatives. This page offers a step-by-step solution to the specific question from Exercise 1, Question 6: . With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.
The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?
The circumference of a circle (C) with radius (r) is given by
C = 2πr.
Therefore, the rate of change of circumference (C) with respect to time (t) is given by,
\begin{align} \frac{dC}{dt}=\frac{dC}{dr}.\frac{dr}{dt}\;\;\; [By\; Chain\; Rule]\end{align}
\begin{align} =\frac{d}{dr}(2\pi r).\frac{dr}{dt}\end{align}
\begin{align} =2\pi.\frac{dr}{dt}\end{align}
It is given that
\begin{align} \frac{dr}{dt}=0.7\; cm/s\end{align}
Hence, the rate of increase of the circumference 2π(0.7)=1.4π cm/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 6: The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its ci....
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