If the volume of a sphere increases at the ra | vedclass

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

Question

(B) Given that the rate of change of volume is $\frac{dV}{dt} = 2 \pi 	ext{ cm}^3/	ext{s}$.We know that the volume of a sphere is $V = \frac{4}{3} \

Vedclass · Accepted Answer

$\frac{1}{72}$

Vedclass · Answer

(B) Given that the rate of change of volume is $\frac{dV}{dt} = 2 \pi 	ext{ cm}^3/	ext{s}$.We know that the volume of a sphere is $V = \frac{4}{3} \pi r^3$.Differentiating both sides with respect to $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.Given $V = 288 \pi$,we find the radius $r$ as:$288 \pi = \frac{4}{3} \pi r^3 \Rightarrow 216 = r^3 \Rightarrow r = 6 	ext{ cm}$.Substituting the values into the derivative equation:$2 \pi = 4 \pi (6)^2 \frac{dr}{dt}$$2 \pi = 4 \pi (36) \frac{dr}{dt}$$2 \pi = 144 \pi \frac{dr}{dt}$$\frac{dr}{dt} = \frac{2 \pi}{144 \pi} = \frac{1}{72} 	ext{ cm}/	ext{s}$.

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

Similar Questions

$A$ particle moves along the curve $y=x^2+2x$. Then, the point on the curve such that the $x$ and $y$ coordinates of the particle change with the same rate is

What will be the rate of change of the volume of a sphere with radius $r$,with respect to its diameter?

The sides of an equilateral triangle are increasing at the rate of $2 \, cm/sec$. The rate at which the area increases,when the side is $10 \, cm$ is:

$A$ spherical iron ball of $10 \ cm$ radius is coated with a layer of ice of uniform thickness that melts at a rate of $50 \ cm^3/min$. When the thickness of ice is $5 \ cm$,then the rate (in $cm/min$) at which the thickness of ice decreases is:

Find the rate of change of the area of a circle with respect to its radius $r$ when $r=5 \text{ cm}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module