#### To determine

**To find:** The rate of change of the water level of the cylindrical tank whose radius is 5 m and being filled with water at a rate of 3m3/min.

#### Answer

The rate of change of the height of the water level in the cylindrical tank is 325πm/min.

#### Explanation

**Given:**

The radius of the cylindrical tank is 5 m.

The rate at which the cylindrical tank being filled with water is 3m3/min.

**Formula used:**

(1) Chain rule: dydx=dydu⋅dudx.

(2) Volume of cylinder with radius *r* and height *h*: V=πr2h.

**Calculation:**

Let *V* be the volume of the cylindrical tank, *r* be the radius of the cylindrical tank and *h* be the height of the water level in a cylindrical tank.

The volume and height of the water in a cylindrical tank increases when the time *t* increases.

That is, the volume and height of the water level depends on the time variable *t*.

Thus, the volume *V* and the height *h* is a function of the time variable *t*.

Since the cylindrical tank being filled with water at a rate of 3m^{3}/min, dVdt=3m3/min.

Obtain the derivative dhdt when the radius is 5 m.

First, substitute 5 for *r* in the volume of cylinder.

V=π(5)2h=π(25)h=25πh

Differentiate *V* with respect to the time variable *t*,

dVdt=ddt(25πh) =(25π)ddt(h) =(25π)dhdt

Thus, dhdt=125πdVdt.

Substitute dVdt=3 in dhdt.

dhdt=125π⋅(3) =325πm/min

Therefore, the rate of change of the height of the water level in the cylindrical tank is, dhdt=325πm/min.