Answer:
Part A)

Part B)

Part C)

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:
![\displaystyle(d)/(dt)\big[V\big]=(d)/(dt)\big[(1)/(12)\pi h^3}\big]](https://img.qammunity.org/2021/formulas/mathematics/high-school/hpbilnkurg96bee6exla5tilgzvft5tz3q.png)
The left will simply be dV/dt. We can move the coefficient from the right:
![\displaystyle (dV)/(dt)=(1)/(12)\pi(d)/(dt)\big[h^3\big]](https://img.qammunity.org/2021/formulas/mathematics/high-school/604tlq5z7cwa75vwyo599ky6014qjk399w.png)
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.