Question

The rate of change of the volume V of water in a tank with respect to time t is directly proportional to the cubed root of the volume. Write a differential equation that describes the relationship. Display keyboard shortcuts for Rich Content Editor

Answer 1

Answer:

The differential equation becomes -

$\frac{dV}{dt} = k\sqrt[3]{V}$ i.e. $\frac{dV}{dt} = kV^{\frac{1}{3} }$

Step-by-step explanation:

Given - The rate of change of the volume V of water in a tank with respect to time t is directly proportional to the cubed root of the volume.

To find - Write a differential equation that describes the relationship.

Proof -

Rate of change of volume V with respect to time t is represented by $\frac{dV}{dt}$

Now,

Given that,

The rate of change of the volume V of water in a tank with respect to time t is directly proportional to the cubed root of the volume.

⇒$\frac{dV}{dt}$ ∝ $\sqrt[3]{V}$

Now,

We know that, when we have to remove the Proportionality sign , we just put a constant sign.

Let k be any constant.

So,

The differential equation becomes -

$\frac{dV}{dt} = k\sqrt[3]{V}$ i.e. $\frac{dV}{dt} = kV^{\frac{1}{3} }$