Question

find the rate of change of circumference of a circle with respect to it's radius.
(a) π
(b) 2π
(c) 2πr
(d) 3π​

Answer 1

Question :-

Find the rate of change of circumference of a circle with respect to it's radius.

(a) π

(b) 2π

(c) 2πr

(d) 3π

$\large\underline{\sf{Solution-}}$

Let assume that radius of circle be 'r' units.

Let assume that C represents the circumference of a circle.

We know,

Circumference, C of a circle of radius r is given by

$\rm \: C \: = \: 2 \: \pi \: r$

On differentiating both sides w. r. t. r, we get

$\rm \: \dfrac{d}{dr}C \: = \dfrac{d}{dr}[\: 2 \: \pi \: r ]$

$\rm \: \dfrac{d}{dr}C \: = 2 \: \pi \: \dfrac{d}{dr}r \:$

$\rm \: \dfrac{d}{dr}C \: = 2 \: \pi \: \times 1 \:$

$\rm \: \dfrac{dC}{dr} \: = 2 \: \pi \: \:$

Hence,

$\rm\implies \:\boxed{\sf{ \: \: \rm \: \dfrac{dC}{dr} \: = 2 \: \pi \: \: }}$

So, option (b) is correct.

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Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$