The radius of a sphere is changing. At an ins | vedclass

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

Question

(B) Given that, at any instant of time, the rate of change in volume with respect to time is equal to the rate of change in surface area with respect

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given that, at any instant of time, the rate of change in volume with respect to time is equal to the rate of change in surface area with respect to time, i.e.,$\frac{dV}{dt} = \frac{dS}{dt}$ $\ldots(i)$ The volume of a sphere of radius $r$ is $V = \frac{4}{3} \pi r^3$. Differentiating with respect to $t$, we get $\frac{dV}{dt} = \frac{d}{dt}(\frac{4}{3} \pi r^3) = \frac{4}{3} \pi (3r^2) \frac{dr}{dt} = 4 \pi r^2 \frac{dr}{dt}$ $\ldots(ii)$ The surface area of a sphere of radius $r$ is $S = 4 \pi r^2$. Differentiating with respect to $t$, we get $\frac{dS}{dt} = \frac{d}{dt}(4 \pi r^2) = 4 \pi (2r) \frac{dr}{dt} = 8 \pi r \frac{dr}{dt}$ $\ldots(iii)$ Substituting the values from equations $(ii)$ and $(iii)$ into equation $(i)$, we get $4 \pi r^2 \frac{dr}{dt} = 8 \pi r \frac{dr}{dt}$. Assuming $\frac{dr}{dt} 
eq 0$, we divide both sides by $4 \pi r \frac{dr}{dt}$ to obtain $r = 2$.

The radius of a sphere is changing. At an instant of time, the rate of change in its volume and its surface area are equal. Then the value of the radius at that instant is?

Similar Questions

The radius of a cylinder is increasing at the rate of $5 \ cm/min$,so that its volume remains constant. When its radius is $5 \ cm$ and height is $3 \ cm$,the rate of decrease of its height is .......... $cm/min$.

For what values of $x$ does the rate of change of $x^3 - 5x^2 + 5x + 8$ become twice the rate of change of $x$?

If a particle is moving in a straight line so that after $t$ seconds its distance $S$ (in $cm$) from a fixed point on the line is given by $S = f(t) = t^3 - 5t^2 + 8t$, then the acceleration of the particle at $t = 5 \text{ sec}$ is (in $cm/sec^2$).

$A$ stone is dropped into a quiet lake and waves move in circles at the speed of $5 \text{ cm/s}$. At the instant when the radius of the circular wave is $8 \text{ cm}$,how fast is the enclosed area increasing?

$A$ particle moves along the curve $y = x^{3/2}$ in the first quadrant in such a way that its distance from the origin increases at the rate of $11$ units per second. The value of $\frac{dx}{dt}$ when $x = 3$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module