How fast is the height of the water increasing a cylindrical tank?

Question

 a cylindrical tank with radius 5m is being filled with water at a rate of 3m^3/min. The formula for Volume of the cylinder is   V = πr²h  and this a related rate question.

Answer 1

Let V = volume at time t

      h = height at time t

 

Given:  dV/dt = 3

Find:  dh/dt 

 

V = πr²h

 

Differentiate implicitly with respect to t:

 

By the product rule, dV/dt = π[2r(dr/dt)h + r²(dh/dt)]

   

since r remains constant (r = 5), dr/dt = 0

 

So, dV/dt = πr²(dh/dt)

 

Substituting in what we know, 3 = π(5)²(dh/dt)

 

dh/dt = 3/(25π) = 0.038 m/min

Answer 2

When we have related rates, we are dealing with derivatives. We need to find out how fast the height is increasing.  We denote this as

 

dh/dt

 

Given:

r = 5

dV/dt = 3

V = πr²h

 

We can establish a formula using these variables.

 

dV/dt = (dV/dh)*(dh/dt)

 

 

First, we need to find the derivative of V in terms of h.  That is dV/dh.

 

dV/dh = πr²

 

Substitute r=5 into this derivative.

 

dV/dh = π(5)² = 78.54 m³/m = 78.54 m²

 

 

Going back to our formula, we can solve for dh/dt.

 

dh/dt = (dV/dt) / (dV/dh)

 

dh/dt = (3 m³/min) / (78.54 m²)

 

dh/dt = 0.38 m/min