Q 10. - Calculus

Question

If the volume and radius of a cone are functions of time, what is the relationship between the rate of change of the volume of the cone and the rate of change of the radius of the cone? 

Step-by-Step Solution

Answer

The required relationship between the rate of change of the volume of the cone and the rate of change of the radius of the cone is $V^{\text{'}}\left(t\right)=\frac{2\pi h}{3}r\left(t\right)r^{\text{'}}\left(t\right)$.

Step 1. Given information and formulas

The volume of cone is $V=\frac{1}{3}\pi r^{2}h$

And its given that the volume and radius of cone are functions of time.

Therefore, $V=V\left(t\right)$ and $r=r\left(t\right)$

Step 2. Explanation

Now $V\left(t\right)=\frac{1}{3}\pi {\left(r\left(t\right)\right)}^{2}h$

Differentiate both sides with respect to $t$

$\Rightarrow V^{\text{'}}\left(t\right)=\frac{\pi h}{3}\left\{2r\left(t\right)r^{\text{'}}\left(t\right)\right\}$

$\Rightarrow V^{\text{'}}\left(t\right)=\frac{2\pi h}{3}r\left(t\right)r^{\text{'}}\left(t\right)$