The rate of change of the volume of a sphere | vedclass

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

Question

(C) Let the radius of the sphere be $r$. Volume of the sphere $V = \frac{4}{3} \pi r^3$. Surface area of the sphere $A = 4 \pi r^2$. We need to find t

Vedclass · Accepted Answer

$1 	ext{ cm}^3 / 	ext{ cm}^2$

Vedclass · Answer

(C) Let the radius of the sphere be $r$. Volume of the sphere $V = \frac{4}{3} \pi r^3$. Surface area of the sphere $A = 4 \pi r^2$. We need to find the rate of change of volume with respect to surface area,which is $\frac{dV}{dA}$. Using the chain rule,$\frac{dV}{dA} = \frac{dV/dr}{dA/dr}$. First,differentiate $V$ with respect to $r$: $\frac{dV}{dr} = \frac{d}{dr}(\frac{4}{3} \pi r^3) = 4 \pi r^2$. Next,differentiate $A$ with respect to $r$: $\frac{dA}{dr} = \frac{d}{dr}(4 \pi r^2) = 8 \pi r$. Now,calculate $\frac{dV}{dA} = \frac{4 \pi r^2}{8 \pi r} = \frac{r}{2}$. Given the radius $r = 2 	ext{ cm}$,substitute this value into the expression: $\frac{dV}{dA} = \frac{2}{2} = 1 	ext{ cm}^3 / 	ext{ cm}^2$.

The rate of change of the volume of a sphere with respect to its surface area,when its radius is $2 \text{ cm}$,is

Similar Questions

$A$ tank in the shape of a rectangular parallelepiped has a volume of $27 \ m^3$. This tank is filled with water such that the rate of change of the water level is thrice the rate of change of the water quantity falling into the tank. The height of the tank (in meters) is:

The total revenue in Rupees received from the sale of $x$ units of a product is given by $R(x) = 3x^2 + 36x + 5$. The marginal revenue, when $x = 15$ is . . . . . . .

$A$ $10 \, cm$ long rod $AB$ moves with its ends on two mutually perpendicular straight lines $OX$ and $OY$. If the end $A$ is moving at the rate of $2 \, cm/sec$,then when the distance of $A$ from $O$ is $8 \, cm$,the rate at which the end $B$ is moving is:

If a circular iron sheet of radius $30 \, cm$ is heated such that its area increases at the uniform rate of $6\pi \, cm^2/hr$,then the rate (in $cm/hr$) at which the radius of the circular sheet increases is

The radius of a right circular cylinder increases at the rate of $0.1 \text{ cm/min}$,and the height decreases at the rate of $0.2 \text{ cm/min}$. The rate of change of the volume of the cylinder,in $\text{cm}^3/\text{min}$,when the radius is $2 \text{ cm}$ and the height is $3 \text{ cm}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module