Calculus word problem

Question

A concial tank with an upper radius of 4m and a height of 5m drains into a cylindrical tank with a radius of 4m and a height of 5m. If the water level in the conical tank drops at a rate of 0.5m/min, at what rate does the water level in the cylindrical tank rise when the water level in the tank is 3m?

Answer 1

In related rates problems, we are looking at rate of change of all variables with respect to time.

We have a conical tank draining into a cylindrical tank. The rate at which volume of water in the conical tank drops has to be the rate at which volume in the cylindrical tank increases, assuming there is leakage.

Volume of a Cone is given as V₁ = 1/3 Π r² h and the Volume of a cylinder is V₂ = Π R²H .

The ask is here to find the rate of change of level in cylindrical tank dH / dt given the rate at which water level in conical tank is dropping dh / dt. So, we need a relationship between these two. We can first get a relationship between the variables H and h and then we can do differentiation to get an equation in their rates of changes.

Differentiating V₁ = 1/3 Π r² h we get dV₁ / dt = 1/3 Π r² dh/dt . For the cone both r and h are changing as the water level drops. Using similar triangles we see can get h in terms of r as r = 4/5 h . We substitute this into V₁ to reduce it into one variable equation V₁ = 1/3 Π (5/4 h)² h = 25/48 Π h³ . Differentiating that we get dV₁ / dt = 25/16 Π h² dh/dt

Similarly, Differentiating V₂ = Π R² H we get dV2 / dt = Π R² dH/dt . Here R is constant with time as it is a cylinder. Now since both these rates dV₁ / dt and dV2 / dt are equal ,

we get 25/16 Π h² dh/dt = Π R² dH/dt

Simplifying dH / dt = 25/16 h²/ R² dh/dt.

Substituting 0.5 m/min for dh/dt and 3 for h , dH / dt = 25/16 * 9/16 * 0.5 = 0.44 m/s .

I am assuming the level given is the height of water in the conical tank. Patrick has a good question on that.

So, the water level is rising in the cylindrical tank at 0.44 m / second.

I really wanted to record a video answer but there are technical difficulties.