Solve the problem. Round your answer, if appropriate.
-The volume of a sphere is increasing at a rate of $7 \mathrm {~cm} ^ { 3 } / \mathrm { sec }$. Find the rate of change of its surface area when its volume is $\frac { 4 \pi } { 3 } \mathrm {~cm} ^ { 3 }$. (Do not round your answer.)
A) $14 \mathrm {~cm} ^ { 2 } / \mathrm { sec }$
B) $\frac { 28 } { 3 } \mathrm {~cm} ^ { 2 } / \mathrm { sec }$
C) $14 \pi \mathrm { cm } ^ { 2 } / \mathrm { sec }$
D) $\frac { 1 } { 3 } \mathrm {~cm} ^ { 2 } / \mathrm { sec }$

Question

Quizplus · Accepted Answer

The rate of change of the surface area can be found using the relationship between the volume and surface area of a sphere and the given rate of change of volume. The volume $V$ of a sphere is given by $V = \frac{4}{3}\pi r^3$, and the surface area $A$ is given by $A = 4\pi r^2$. Differentiating both with respect to time ($t$) gives the relationship between the rates of change of volume ($\frac{dV}{dt}$) and radius ($\frac{dr}{dt}$), and subsequently between the radius and the surface area ($\frac{dA}{dt}$). Given $\frac{dV}{dt} = 7 \, \text{cm}^3/\text{sec}$ and $V = \frac{4\pi}{3} \, \text{cm}^3$ (implying $r = 1 \, \text{cm}$), we can find $\frac{dr}{dt}$ and then use it to find $\frac{dA}{dt}$, which results in $14 \, \text{cm}^2/\text{sec}$.

Solve the Problem 7 cm3/sec7 \mathrm {~cm} ^ { 3 } / \mathrm { sec }7 cm3/sec

Solve the Problem $7 \mathrm {~cm} ^ { 3 } / \mathrm { sec }$