Question

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.)

Answer 1

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

Answer 2

Part A)

`\displaystyle V=(1)/(12)\pi h^3`

Part B)

`\displaystyle (dV)/(dt)=(1)/(4)\pi h^2(dh)/(dt)`

Part C)

`\displaystyle (dh)/(dt)\approx0.4\text{ feet per second}`

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:

`\displaystyle V=(1)/(3)\pi r^2h`

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

`\displaystyle (10)/(5)=(h)/(r)`

Therefore:

`\displaystyle 2=(h)/(r)`

So:

`\displaystyle r=(h)/(2)`

Substituting this into our formula yields:

`\displaystyle V=(1)/(3)\pi \big( (h)/(2) \big)^2h`

Simplify:

`\displaystyle V=(1)/(12)\pi h^3`

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]`

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]`

Implicitly differentiate:

`\displaystyle (dV)/(dt)=(1)/(12)\pi(3h^2(dh)/(dt))`

Simplify. So, our derivative is:

`\displaystyle (dV)/(dt)=(1)/(4)\pi h^2(dh)/(dt)`

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:

`\displaystyle 5=(1)/(4)\pi(4)^2(dh)/(dt)`

Simplify:

`\displaystyle 5=4\pi(dh)/(dt)`

Therefore:

`\displaystyle (dh)/(dt)=(5)/(4\pi)\approx0.3978\approx 0.40`

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