Question:

A balloon which always remains spherical is being inflated by pumping in $10$ cubic centimeters of gas per second. Find the rate at which the radius of the balloon is increasing when the radius is $15$ cms.

Updated On: May 22, 2024
  • $\frac {1}{90\pi}cm /sec $
  • $\frac {1}{9\pi}cm /sec $
  • $\frac {1}{30\pi}cm /sec $
  • $\frac {1}{\pi}cm /sec $
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The Correct Option is A

Solution and Explanation

We have, $\frac{d V}{d t} =10 cm ^{3} / s $
$r =15 \,cm $
$\therefore V =\frac{4}{3} \pi r^{3} $
$\Rightarrow \frac{d V}{d r}=4 \pi r^{2}=4 \pi(15)^{2}=900 \pi $
Now, $ \frac{d V}{d t} =\frac{d V}{d r} \times \frac{d r}{d t} $
$\Rightarrow 10=900 \pi \times \frac{d r}{d t} $
$\Rightarrow \frac{d r}{d t}=\frac{1}{90 \pi} cm / s$
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Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by 

\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

  • If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
  • Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
  • The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.

Read More: Application of Derivatives