Related Rates: water tank has shape of horizontal cylinder w/ radius 1, length 2...

[Antonino]

A water tank has the shape of a horizontal cylinder with radius 1 and length 2.If water is being pumped into the tank at a rate of 16 m3 per minute, find therate at which the water level is rising when the water is 12 m deep.

Sorry about the sideways picture. I got an answer, but I'm not sure it's correct. I got 1/(6pi)m/min= the derivative of h with respect to t.
May someone please confirm or refute my answer?

 

[tkhunny]

Well, we certainly have problems of scale.

It's 2x2? Are there units on those?

It's 2x2. How does it get to be 12m deep?

Please clear that up and give it another go.

Maybe we're just having an input problem? Try really hard to make sure what appears on the screen is what you intend.

BTW: 1/2m is not 1/2 the 2m Diameter.

 

[Antonino]

Clarification

Well, we certainly have problems of scale.

It's 2x2? Are there units on those?

It's 2x2. How does it get to be 12m deep?

Please clear that up and give it another go.

Maybe we're just having an input problem? Try really hard to make sure what appears on the screen is what you intend.

BTW: 1/2m is not 1/2 the 2m Diameter.

All values are listed with units, meters, and the depth is 0.5 meters. It is a 2 by 2, but I can't make any sense out of that information. I consulted with a friend and he suggested this could not be solved with Calculus AB. Do you know of any clever way to solve this without double integrals?(I'm in AP Calc AB)

 

[tkhunny]

The tank has a 2 m diameter.
The first thing to realize is that at 1/2 m, you can simply ignore the top half the tank.

 

[anthonydalvaro]

Hello Antonio! Your volume formula is written for a vertical cylinder!

the volume for a horizontal cylinder (for these purposes) is different.

When water is filling a vertical cylinder, the volume of the water will be the area of the circle times the height of the water or... (pi)r^2h,

But when water is filling a vertical cylinder, the volume will be the area of the "Segment of water" times the length of the cylinder.

See this webpage https://www.mathsisfun.com/geometry/cylinder-horizontal-volume.html

You can differentiate this volume formula to solve the problem.

 

[lolaberry]

Hello Antonio! Your volume formula is written for a vertical cylinder!

the volume for a horizontal cylinder (for these purposes) is different.

When water is filling a vertical cylinder, the volume of the water will be the area of the circle times the height of the water or... (pi)r^2h,

But when water is filling a vertical cylinder, the volume will be the area of the "Segment of water" times the length of the cylinder.

See this webpage https://www.mathsisfun.com/geometry/cylinder-horizontal-volume.html

You can differentiate this volume formula to solve the problem.

why are you multiplying V by L? shouldn't you be dividing?

 

[limiTS]

I'm getting [imath]\frac{\sqrt{3}}{36}[/imath] for the [imath]\frac{\text{d}h}{\text{d}t}[/imath]

 

[skeeter]

let the circular part of the cylinder be centered at the origin

a representative horizontal cross section of liquid has volume, [imath]dV = 2 \cdot 2\sqrt{1-y^2} \cdot dy[/imath]

[imath]\displaystyle V = 4 \int_{-1}^h \sqrt{1-y^2} \, dy[/imath] where depth of the water is [imath]-1 \le h \le 1[/imath]

[imath]\dfrac{dV}{dt} = 4\sqrt{1-h^2} \cdot \dfrac{dh}{dt}[/imath]

you should get the rising rate in m/min when h = 1/2 m to be the value given in post #7