25 POINTS FOR WHOEVER ANSWERS THIS QUESTION PLEASE HELP! An inverted conical tank with a height of 10 ft and a radius of 5 ft is drained through a hole in the vertex at a rate of 5 ft^3/s. What is the rate of change when the water depth is at 4 ft?

a.) Let V be the volume of water in the tank and let h be the depth of the water. Write an equation that relates V and h.

b.)Differentiate both sides of the equation with respect to t. dV/dt = () dh/dt

c.) when the water depth is 4 ft, the rate of change of the water depth is about . (Round to nearest hundredth if needed.)

So just use Cyclander formula. That’s how you solve!

Part A)

Part B)

Part C)

Step-by-step explanation:

Please refer to the attachment.

We know that the conical tank has a height of 10 feet and a radius of 5 feet.

The rate at which water is being drained through a hole in the vertex is 5 cubic feet per second.

Part A)

Recall the volume formula for a cone:

Since this is a cone, we can draw two similar triangles (refer to the attachment). Then, by properties of similar triangles, we can write:

Therefore:

So:

Substituting this into our formula yields:

Simplify:

Part B)

Let’s take the derivative of both sides with respect to t. So:

The left will simply be dV/dt. We can move the coefficient from the right:

Implicitly differentiate:

Simplify. So, our derivative is:

Part C)

We want to find the rate of change of the water depth (in other words, dh/dt) when h=4.

Since our water is being drained at a rate of 5 cubic feet per second, this means that dV/dt=5.

So, substituting, we get:

Simplify:

Therefore:

So, the change of the water depth is about 0.4 feet per second.

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