If a cylindrical tank with radius 5 meters is being filled with water at a rate of 3 cubic meters per minute, how fast is the height of the water increasing?

Question

Emily Ammerman · Accepted Answer

The answer is #(dh)/(dt)=3/(25 pi)m/(min)#. With related rates, we need a function to relate the 2 variables, in this case it is clearly volume and height.  The formula is:
#V=pi r^2 h# There is radius in the formula, but in this problem, radius is constant so it is not a variable.  We can substitute the value in:
#V=pi (5m)^2 h# Since the rate in this problem is time related, we need to implicitly differentiate wrt (with respect to) time:
#(dV)/(dt)=(25 m^2) pi (dh)/(dt)# In the problem, we are given #3(m^3)/min# which is #(dV)/(dt)#.  So we substitute this in:
#(dh)/(dt)=(3m^3)/(min (25m^2) pi)=3/(25 pi)m/(min)# In general
- find a formula to relate the 2 variables
- substitute values to remove the constant variables
- implicitly differentiate wrt time (most often the case)
- substitute the given rate
- and solve for the desired rate.

Cameron Alexander · Answer

undefined To find how fast the height of the water in the cylindrical tank is increasing, you can use related rates.

Given:
- Radius of the cylindrical tank, $ r = 5 $ meters
- Rate of water being filled, $ \frac{dV}{dt} = 3 $ cubic meters per minute

You can use the formula for the volume of a cylinder: $ V = \pi r^2 h $, where $ V $ is the volume, $ r $ is the radius, and $ h $ is the height.

Differentiate the volume formula with respect to time $ t $ to find $ \frac{dh}{dt} $, the rate at which the height of the water is increasing.