Get Answers to all your Questions

header-bg qa

Please solve RD Sharma class 12 chapter Derivative as a Rate Measure exercise 12.1 question 1 maths textbook solution

Answers (1)


                    4\pi r+2\pi h


To find the rate of change of total surface area of cylinder, we need to differentiate it with respect to radius

Total surface area of cylinder, A=2\pi r(r+h)


radius\; (r)\; and \: height\; (h)\; o\! f\; cylinder.


Here we have to find rate of change of the total surface area of cylinder. here radius=r, height=h
 so, total surface area of cylinder.

                                    A=2\pi r(r+h)

                               \therefore A= 2\pi r^{2}+2\pi rh\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(i)

Let’s differentiate equation (i) with respect to radius

                            \therefore \frac{d A}{d r}=\frac{d}{d r}\left(2 \pi r^{2}+2 \pi r h\right)

                               \frac{d A}{d r}=\frac{d}{d r}\left(2 \pi r^{2}\right)+\frac{d}{dr}(2\pi rh)

                            \therefore \frac{d A}{d r}=4\pi r+2\pi h                        [\because \frac{d (x^{n})}{d x}=nx^{n-1}]

Which is rate of change of total surface area with respect to radius.


Here we have took rate with respect to radius because radius was varying.

Posted by

Gurleen Kaur

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support