Question

A balloon, which always remains spherical, has a variable radius. Find the rate at which its volume is increasing with the radius when radius is $10\text{ }cm$ .

Answer 1

Hint: Volume of sphere is changes with respect to radius of sphere. For that we differentiate the volume of the sphere with respect to radius and substitute values for the final answer. The volume of the sphere increases as the radius of the sphere increases.
Formula used
Volume of sphere $=\dfrac{4}{3}\pi {{r}^{3}}$
V$=\dfrac{4}{3}\pi {{r}^{3}}$ & $\text{used }\pi =3\cdot 14$ in this question.

Complete step-by-step answer:
 The volume of sphere is$\dfrac{4}{3}\pi {{r}^{3}}$
Where r is the radius of sphere
As radius of sphere increase, volume also increase
So differentiate with respect to radius
V=$\dfrac{4}{3}\pi {{r}^{3}}$
\[\begin{align}
  &\Rightarrow \dfrac{dv}{dr}=\dfrac{4}{3}\pi \dfrac{d{{r}^{3}}}{dr} \\
 &\Rightarrow \dfrac{dv}{dr}=\dfrac{4}{3}\pi \left( 3{{r}^{2}} \right) \\
 & \Rightarrow \dfrac{dv}{dr}=4\pi {{r}^{2}} \\
\end{align}\]
Volume of sphere change with respect to radius and when radius is $10\text{ }cm$ can change in volume will be
\[\dfrac{dv}{dr}=4\pi {{\left( 10 \right)}^{2}}=4\times 3\cdot 14\times 100=1256\]

Additional information:
We can also find the surface area change with respect to radius. Also we can find changes in various shapes with respect to other variables also like time.

Note: We know that the volume of sphere is $\dfrac{4}{3}\pi {{r}^{3}}$ and we also know that volume of sphere increases as its radius increases. We differentiate the volume of the sphere with respect to radius and we calculate the rate of change of volume with respect to radius at r = 10.