If the area of a circle increases at the rate | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $A$ be the area and $P$ be the perimeter of the circle. We have, $\frac{dA}{dt} = \frac{1}{\sqrt{\pi}}$.Since $A = \pi r^2$, differentiating w

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Let $A$ be the area and $P$ be the perimeter of the circle. We have, $\frac{dA}{dt} = \frac{1}{\sqrt{\pi}}$.Since $A = \pi r^2$, differentiating with respect to $t$ gives $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.Substituting the given rate: $2\pi r \frac{dr}{dt} = \frac{1}{\sqrt{\pi}} \Rightarrow \frac{dr}{dt} = \frac{1}{2\pi r \sqrt{\pi}}$.The perimeter is $P = 2\pi r$. Differentiating with respect to $t$ gives $\frac{dP}{dt} = 2\pi \frac{dr}{dt}$.Substituting $\frac{dr}{dt}$: $\frac{dP}{dt} = 2\pi 	imes \frac{1}{2\pi r \sqrt{\pi}} = \frac{1}{r \sqrt{\pi}}$.Given $P = \sqrt{\pi}$, we have $2\pi r = \sqrt{\pi} \Rightarrow r = \frac{\sqrt{\pi}}{2\pi} = \frac{1}{2\sqrt{\pi}}$.Substituting $r$ into the expression for $\frac{dP}{dt}$: $\frac{dP}{dt} = \frac{1}{(\frac{1}{2\sqrt{\pi}}) \sqrt{\pi}} = \frac{1}{1/2} = 2$ units/sec.

If the area of a circle increases at the rate of $\frac{1}{\sqrt{\pi}}$ sq. units/sec, then the rate (in units/sec) at which the perimeter of the circle changes, when perimeter is $\sqrt{\pi}$ units, is

Similar Questions

The rate of change of the area of a circle with respect to its radius at $r = 3 \text{ cm}$ is . . . . . . $\text{cm}^2/\text{cm}$. (in $\pi$)

$A$ particle is moving in a straight line according to the equation $s = 45t + 11t^2 - t^3$. The time at which it will come to rest is ......... $sec$.

If water is being poured at the rate of $36 \text{ } m^3/sec$ into a cylindrical vessel with a base radius of $3 \text{ } m$,then the rate at which the water level is rising is:

$A$ stone is dropped in a quiet lake and it is observed that waves move in circles. If the radius of a circular wave increases at the rate of $2 \text{ cm/sec}$, then the rate of increase in its area at the instant when its radius is $10 \text{ cm}$, is in $\text{cm}^2\text{/sec}$: (in $\pi$)

If a spherical balloon has a variable diameter $d = 3x + \frac{9}{2}$ units, then the rate of change of its volume $V$ with respect to $x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module