Question

Gas is being pumped in spherical balloon at the rate of 30$f{{t}^{3}}/\min$,then the rate at which radius of balloon increases when it reaches the value of 15ft, is
(a)$\dfrac{1}{30\pi }ft/\min$
(b)$\dfrac{1}{15\pi }ft/\min$
(c)$\dfrac{1}{20}ft/\min$
(d) $\dfrac{1}{25}ft/\min$

Answer 1

To solve this problem, what we will do is firstly, we will write the data given in question, in equation form. Then we will find the derivative of volume of sphere $\dfrac{4}{3}\pi {{r}^{3}}$ with respect to r. After that we will solve the two equations for r = 15 and hence obtain the rate of change of radius of balloon.

Complete step-by-step answer:
Let r be the radius of balloon and V be the volume of balloon at time t min.
Mow, in question t is given that Gas is being pumped in spherical balloon at the rate of 30$f{{t}^{3}}/\min$, so this means volume of 30$f{{t}^{3}}$ is being added in balloon per minute.
So, we are provide with rate of change in volume of balloon with respect to time, which means we have the value of $\dfrac{dV}{dt}$ as derivative of any function with respect to time show the rate of change in function with respect to time.
So, $\dfrac{dV}{dt}=30f{{t}^{3}}/\min$
Now, we know that the volume of a spherical ball is equal to $\dfrac{4}{3}\pi {{r}^{3}}$ , where r is the radius of the balloon.
So, $V=\dfrac{4}{3}\pi {{r}^{3}}$
Differentiating$\dfrac{4}{3}\pi {{r}^{3}}$, with respect to radius t, we get
$\dfrac{dV}{dt}=\dfrac{4}{3}\times 3\times \pi {{r}^{2}}\times \dfrac{dr}{dt}$, as $\dfrac{d}{dt}({{x}^{n}})=n{{x}^{n-1}}$
On solving, we get
$\dfrac{dV}{dt}=4\pi {{r}^{2}}\dfrac{dr}{dt}$
Also, we have $\dfrac{dV}{dt}=30f{{t}^{3}}/\min$from above
So, $4\pi {{r}^{2}}\dfrac{dr}{dt}=30f{{t}^{3}}/\min$
Now, in question it is given that r = 15ft, so
Putting r = 15 in $4\pi {{r}^{2}}\dfrac{dr}{dt}=30f{{t}^{3}}/\min$, we get
$4\pi {{(15ft)}^{2}}\dfrac{dr}{dt}=30f{{t}^{3}}/\min$
$900\pi f{{t}^{2}}\dfrac{dr}{dt}=30f{{t}^{3}}/\min$
On solving, we get
$\dfrac{dr}{dt}=\dfrac{1}{30\pi }ft/\min$

So, the correct answer is “Option A”.

Note: Always remember that derivative of any function with respect to time show the rate of change in function with respect to time and $\dfrac{d}{dt}({{x}^{n}})=n{{x}^{n-1}}$. Also, that volume of spherical ball is equal to $\dfrac{4}{3}\pi {{r}^{3}}$ , where r is the radius of the ball. Always put units after the final numerical answer. Try to avoid calculation mistakes.