A sphere increases its volume at the rate of | vedclass

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

Question

(A) Given the rate of increase in the volume of the sphere is $\frac{dV}{dt} = \pi 	ext{ cm}^3/	ext{s}$.We know that the volume of a sphere is $V =

Vedclass · Accepted Answer

$2 \pi 	ext{ cm}^2/	ext{s}$

Vedclass · Answer

(A) Given the rate of increase in the volume of the sphere is $\frac{dV}{dt} = \pi 	ext{ cm}^3/	ext{s}$.We know that the volume of a sphere is $V = \frac{4}{3} \pi r^3$.Differentiating with respect to $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.Substituting $\frac{dV}{dt} = \pi$,we have $\pi = 4 \pi r^2 \frac{dr}{dt}$,which implies $\frac{dr}{dt} = \frac{1}{4r^2}$.The surface area of a sphere is $S = 4 \pi r^2$.Differentiating with respect to $t$,we get $\frac{dS}{dt} = 8 \pi r \frac{dr}{dt}$.Substituting $\frac{dr}{dt} = \frac{1}{4r^2}$,we get $\frac{dS}{dt} = 8 \pi r \left( \frac{1}{4r^2} ight) = \frac{2 \pi}{r}$.When $r = 1 	ext{ cm}$,the rate of increase in surface area is $\frac{dS}{dt} = \frac{2 \pi}{1} = 2 \pi 	ext{ cm}^2/	ext{s}$.

$A$ sphere increases its volume at the rate of $\pi \text{ cm}^3/\text{s}$. The rate at which its surface area increases when the radius is $1 \text{ cm}$ is

Similar Questions

If the radius $r$ of a sphere is increasing at a rate of $2 \text{ cm/s}$,then the rate of change of its surface area is proportional to which of the following?

$A$ balloon,which always remains spherical,has a variable radius. Find the rate at which its volume is increasing with respect to its radius when the radius is $10 \text{ cm}$.

If the volume of a sphere increases at the rate of $2 \pi \text{ cm}^3/\text{s}$,then the rate of increase of its radius (in $\text{cm}/\text{s}$),when the volume is $288 \pi \text{ cm}^3$,is

$A$ particle is moving along a line according to the law $S = t^3 - 3t^2 + 4t - 2$,where $S$ is measured in meters and $t$ is measured in seconds. Then the velocity (in $m/s$) of the particle when its acceleration is zero is:

The rate of change of the volume of a sphere of radius $r$ with respect to its surface area is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module