Q. 24 - Calculus

Question

In Exercises 19–26, write down an equation that relates the two quantities described. Then use implicit differentiation to obtain a relationship between the rates at which the quantities change over time. The volume V and height h of a cone with a fixed radius of 3 units. 

Step-by-Step Solution

Answer

Equation relating the quantities: $V=3\mathrm{\pi h}$

Implicit differentiation: $\frac{dV}{dt}=3\pi \frac{dh}{dt}$

Step 1. Given information

Volume of the cone $=V$

Radius of the cone $=3$

Height of the cone $=h$

Step 2. Equation relating the quantities

Volume of the cone is given by,

$$\begin{gathered} V=\frac{\pi r^{2}h}{3} \\ \Rightarrow V=\frac{\pi \left(3^2\right)h}{3} \\ \Rightarrow V=3\pi h \end{gathered}$$

Step 3. Implicit differentiation with respect to time

From the above step,

$$\begin{gathered} V=3\pi h \\ \mathrm{Differentiating} \mathrm{on} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get}, \\ \frac{dV}{dt}=3\pi \frac{dh}{dt} \end{gathered}$$