# To find the rate at which the radius of the balloon increases when the radius is 15 cm. A balloon which always remains spherical is being inflated by pumping in $900c{{m}^{3}}$ of gas per second.

Answer

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Hint: We can write the expression for volume of a balloon and from there we find the rate of change. Volume of a sphere is given by $\dfrac{4}{3}\pi {{r}^{3}}$ . To find Rate of change we use derivatives, so we can write the rate of change of volume with time by differentiating volume with respect to time.

Complete step-by-step answer:

First of all we collect the information which is given in the question which is,

Balloon always remains spherical $\Rightarrow $ Volume of balloon $=\dfrac{4}{3}\pi {{r}^{3}}$

Balloon is being inflated $\Rightarrow \dfrac{dV}{dt}=900c{{m}^{3}}$ which is positive as we are pumping air in and the volume is increasing.

Now we need to find out the rate at which the radius of the balloon increases. For this we will write-

$V=\dfrac{4}{3}\pi {{r}^{3}}$

Differentiating both sides with respect to time we have,

$\begin{align}

& \dfrac{dV}{dt}=\left( \dfrac{4}{3}\pi \right)\dfrac{d}{dt}({{r}^{3}}) \\

& =\left( \dfrac{4}{3}\pi \right)(3{{r}^{2}})\dfrac{dr}{dt} \\

\end{align}$

On further simplification we have,

$\dfrac{dV}{dt}=4\pi {{r}^{2}}\dfrac{dr}{dt}$

We are given $\dfrac{dV}{dt}=900c{{m}^{3}}$ and we need to find out $\dfrac{dr}{dt}$ when r= 15 cm.

Therefore we write,

$\dfrac{dr}{dt}=\dfrac{1}{4\pi {{r}^{2}}}\dfrac{dV}{dt}$

Substituting r=15 and $\dfrac{dV}{dt}=900c{{m}^{3}}$ we have,

$\dfrac{dr}{dt}=\dfrac{1}{4\pi {{(15)}^{2}}}.900$ cm/s

On simplifying we have,

$\dfrac{dr}{dt}=\dfrac{1}{\pi }cm/s$

Hence, the answer is $\dfrac{1}{\pi }cm/s$ .

Note:

There are a number of things to keep in mind. First of all while differentiating we must take care about with respect to what we are differentiating. If we differentiate with respect to ‘r’ we will not get anything useful. And we should also keep in mind about the units, which will also help us in understanding with respect to what we need to differentiate. We should also keep in mind the sign of rate of change as rate of change may be increasing as well as decreasing. Increasing rate of change will have positive signs and decreasing rate of change will have negative signs.

Complete step-by-step answer:

First of all we collect the information which is given in the question which is,

Balloon always remains spherical $\Rightarrow $ Volume of balloon $=\dfrac{4}{3}\pi {{r}^{3}}$

Balloon is being inflated $\Rightarrow \dfrac{dV}{dt}=900c{{m}^{3}}$ which is positive as we are pumping air in and the volume is increasing.

Now we need to find out the rate at which the radius of the balloon increases. For this we will write-

$V=\dfrac{4}{3}\pi {{r}^{3}}$

Differentiating both sides with respect to time we have,

$\begin{align}

& \dfrac{dV}{dt}=\left( \dfrac{4}{3}\pi \right)\dfrac{d}{dt}({{r}^{3}}) \\

& =\left( \dfrac{4}{3}\pi \right)(3{{r}^{2}})\dfrac{dr}{dt} \\

\end{align}$

On further simplification we have,

$\dfrac{dV}{dt}=4\pi {{r}^{2}}\dfrac{dr}{dt}$

We are given $\dfrac{dV}{dt}=900c{{m}^{3}}$ and we need to find out $\dfrac{dr}{dt}$ when r= 15 cm.

Therefore we write,

$\dfrac{dr}{dt}=\dfrac{1}{4\pi {{r}^{2}}}\dfrac{dV}{dt}$

Substituting r=15 and $\dfrac{dV}{dt}=900c{{m}^{3}}$ we have,

$\dfrac{dr}{dt}=\dfrac{1}{4\pi {{(15)}^{2}}}.900$ cm/s

On simplifying we have,

$\dfrac{dr}{dt}=\dfrac{1}{\pi }cm/s$

Hence, the answer is $\dfrac{1}{\pi }cm/s$ .

Note:

There are a number of things to keep in mind. First of all while differentiating we must take care about with respect to what we are differentiating. If we differentiate with respect to ‘r’ we will not get anything useful. And we should also keep in mind about the units, which will also help us in understanding with respect to what we need to differentiate. We should also keep in mind the sign of rate of change as rate of change may be increasing as well as decreasing. Increasing rate of change will have positive signs and decreasing rate of change will have negative signs.

Last updated date: 21st Sep 2023

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