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A spherical balloon of radius
3cm is being pumped by air at a rate of 9 cubic
cm per second. What is the rate of change of the radius?
(V=(4)/(3)pir^(3))

A spherical balloon of radius $3 \mathrm{~cm}$ is being pumped by air at a rate of $9$ cubic $\mathrm{cm}$ per second. What is the rate of change of the radius? $\left(V=\frac{4}{3} \pi r^{3}\right)$

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Find the rate of change of one variable when rate of change of other variable is given

Full solution

Q. A spherical balloon of radius

3 \mathrm{~cm}

is being pumped by air at a rate of

9

cubic

\mathrm{cm}

per second. What is the rate of change of the radius?

\left(V=\frac{4}{3} \pi r^{3}\right)

Given: Given: $\newline$ Volume of a sphere $V$ = $\frac{4}{3}\pi r^3$ $\newline$ Rate of change of volume $\frac{dV}{dt}$ = $9 \, \text{cm}^3/s$ $\newline$ Radius $r$ = $3 \, \text{cm}$ $\newline$ We need to find the rate of change of the radius, $\frac{dr}{dt}$ . $\newline$ First, let's write down the formula for the volume of a sphere: $\newline$ $V = \frac{4}{3}\pi r^3$
Differentiate Volume Equation: Differentiate both sides of the volume equation with respect to time $t$ to find the relationship between the rates of change of volume and radius. $\frac{dV}{dt} = 4\pi r^2\left(\frac{dr}{dt}\right)$
Substitute Values: Substitute the given rate of change of volume and the radius into the differentiated equation. $\newline$ $9 = 4\pi(3)^2\frac{dr}{dt}$ $\newline$ $9 = 4\pi(9)\frac{dr}{dt}$ $\newline$ $9 = 36\pi\frac{dr}{dt}$
Solve for $\frac{dr}{dt}$ : Solve for $\frac{dr}{dt}$ to find the rate of change of the radius. $\newline$ $\frac{dr}{dt} = \frac{9}{36\pi}$ $\newline$ $\frac{dr}{dt} = \frac{1}{4\pi}$
Calculate Numerical Value: Calculate the numerical value of $\frac{dr}{dt}$ .
$\frac{dr}{dt} = \frac{1}{4\pi}$
$\frac{dr}{dt} \approx \frac{1}{4 \times 3.14159}$
$\frac{dr}{dt} \approx \frac{1}{12.56636}$
$\frac{dr}{dt} \approx 0.079577$

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