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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.)

2 Answers

3 votes

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

User Dwstu
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4 votes

Answer:

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.

25 POINTS FOR WHOEVER ANSWERS THIS QUESTION PLEASE HELP! An inverted conical tank-example-1
User Amrish
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