A balloon, which always remains spherical, has a variable diameter
\begin{align} \frac{3}{2}(2x+1)\end{align}
Find the rate of change of its volume with respect to x.
The volume of a sphere (V) with radius (r) is given by,
\begin{align} V=\frac{4}{3}\pi r^3 \end{align}
It is given that:
\begin{align} Diameter =\frac{3}{2}(2x+1) \end{align}
\begin{align} \Rightarrow r =\frac{3}{4}(2x+1) \end{align}
\begin{align} \therefore V =\frac{4}{3}\pi(\frac{3}{4})^3(2x+1)^3=\frac{9}{16}\pi\times(2x+1)^3 \end{align}
Hence, the rate of change of volume with respect to x is as
\begin{align} \frac{dV}{dx}=\frac{9}{16}\pi\frac{d}{dx}(2x+1)^3=\frac{9}{16}\pi\times3(2x+1)^2 \times2=\frac{27}{8}\pi(2x+1)^2\end{align}
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Welcome to the NCERT Solutions for Class 12 Mathematics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 13: A balloon, which always remains spherical, has a variable diameter egin{align} frac{3}{2}(2x+1....
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