Question

A cylindrical tank standing upright (with one circular base on the ground) has a radius of 20 cm. How fast does the water level in the tank drop when the water is being drained at 25 cm³/sec?

a) 0.01 cm/sec
b) 0.02 cm/sec
c) 0.03 cm/sec
d) 0.04 cm/sec

Answer 1

The water level in the tank drops at a rate of approximately 0.01 cm/sec.

Step-by-step explanation:

To find how fast the water level in the tank drops, we can use the formula for volume of a cylinder: V = πr²h, where V is the volume, r is the radius, and h is the height. Since the radius of the tank is 20 cm, the volume can be expressed as V = π(20)²h. Taking the derivative of both sides with respect to time, we get dV/dt = π(20)²(dh/dt). We are given that dV/dt = -25 cm³/sec (negative sign because the volume is decreasing), so we can solve for dh/dt:

-25 = π(20)²(dh/dt)

dh/dt = -25 / (π(20)²) ≈ -0.01 cm/sec

Therefore, the water level in the tank is dropping at a rate of approximately 0.01 cm/sec.