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

Answer:

Part A)

$\displaystyle V=\frac{1}{12}\pi h^3$

Part B)

$\displaystyle \frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}$

Part C)

$\displaystyle \frac{dh}{dt}\approx0.4\text{ feet per second}$

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:

$\displaystyle V=\frac{1}{3}\pi r^2h$

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

$\displaystyle \frac{10}{5}=\frac{h}{r}$

Therefore:

$\displaystyle 2=\frac{h}{r}$

So:

$\displaystyle r=\frac{h}{2}$

Substituting this into our formula yields:

$\displaystyle V=\frac{1}{3}\pi \big( \frac{h}{2} \big)^2h$

Simplify:

$\displaystyle V=\frac{1}{12}\pi h^3$

Part B)

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

$\displaystyle\frac{d}{dt}\big[V\big]=\frac{d}{dt}\big[\frac{1}{12}\pi h^3}\big]$

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

$\displaystyle \frac{dV}{dt}=\frac{1}{12}\pi\frac{d}{dt}\big[h^3\big]$

Implicitly differentiate:

$\displaystyle \frac{dV}{dt}=\frac{1}{12}\pi(3h^2\frac{dh}{dt})$

Simplify. So, our derivative is:

$\displaystyle \frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}$

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:

$\displaystyle 5=\frac{1}{4}\pi(4)^2\frac{dh}{dt}$

Simplify:

$\displaystyle 5=4\pi\frac{dh}{dt}$

Therefore:

$\displaystyle \frac{dh}{dt}=\frac{5}{4\pi}\approx0.3978\approx 0.40$

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