A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm
Solution and Explanation
The volume of a sphere \((V)\) with radius \((r)\) is given by,
\(v=\frac{4}{3}πr^3\)
∴Rate of change of volume \((V)\) with respect to time \((t)\) is given by
\(\frac{dv}{dt}=\frac{dv}{dr}.\frac{dr}{dt}\) [By chain rule]
\(=\frac{dv}{dt}(\frac{4}{3}πr^3).\frac{dr}{dt}\)
\(=4πr^2\frac{dr}{dt}\)
It is given that \(\frac{dv}{dt}=900cm^3/s\)
\(∴ 900=4πr^2.\frac{dr}{dt}\)
\(\frac{dr}{dt}=\frac{900}{4πr^2}=\frac{225}{r^2}\)
Therefore, when radius=15 cm
\(\frac{dr}{dt}=\frac{225}{\pi(15)^2}=\frac{1}{\pi}\)
Hence, the rate at which the radius of the balloon increases when the radius is \(15cm\) is \(\frac{1}{π} cm/s.\)
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Concepts Used:
Rate of Change of Quantities
The rate of change of quantities can be expressed in the form of derivatives. Rate of change of one quantity with respect to another is one of the major applications of derivatives. The rate of change of a function with respect to another quantity can also be done using chain rule.
If some other quantity ‘y’ causes some change in a quantity of certain ‘x’, in view of the fact that an equation of the form y = f(x) gets always satisfied, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
This is also called the Average Rate of Change.
If the rate of change of a function is to be defined at a specific point i.e. a specific value of ‘x’, it is known as the Instantaneous Rate of Change of the function at that point. From the definition of the derivative of a function at a point, we have
From this, it is to be concluded that the instantaneous Rate of Change of the function is represented by the derivative of a function. From the rate of change formula, it represents the case when Δx → 0. Thus, the rate of change of ‘y’ with respect to ‘x’ at x = x0 = (dy/dx)x = x0