Rates of Change

Question

A container is being filled with water. After t seconds the depth of water, x cm, in the container is increasing at the rate of 1.44tcm/s.

Given that the container is empty when t=0, use calculus to find

the value of t when x=18.

Answer 1

We are given the rate of change of the water depth. You write this, mathematically, as:

 

dx/dt = 1.44 t

 

Multiply both sides by dt:

 

dx = 1.44 t dt

 

Integrate both sides:

 

x = 0.72 t² + C

 

We know that x=0 when t=0, so C=0.

 

Then plug in x=18 and solve for t:

 

18 = 0.72 t²

12.96 = t²

t = +/-3.6

 

The negative answer is illogical, so the answer is t = 3.6 s