Related Rates: Finding the Rate of Change of Liquid Height in a Draining Tank

In summary: If it specifically asks what the rate of decreasing is, then you should put a negative sign because that is what is happening.
  • #1
gillgill
128
0
am i doing this right?
A cylindrical tank with a radius of 5 ft and a height of 20ft is filled with a certain liquid chemical. A hole is punched in the bottom. At that moment the chemical drains ut of the tank at the rate of 2ft^3/min. At what rate is the height of liquid in the tank changing?

V = 2(pi)rh
dV/dt = [2(pi)r]dh/dt
2 = [2(pi)5]dh/dt
dh/dt = 1/5(pi)

The height of liquid in the tank is changing at the rate is 1/5(pi) /min
 
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  • #2
Looks good to me.
 
  • #3
thanks...
 
  • #4
Except that's not the volume of a cylinder. :)
 
  • #5
[tex] 2\pi rh[/tex] is the area of the curved portion of the cylinder.
 
  • #6
Such a type of textbook questions never include the real effects viz., changing flowrate with respect to height of liquid column in the tank etc. Otherwise, your procedure is ok except that the volume equation of the cylinder as suggested already.
 
  • #7
V = (pi)r^2h


V = 25(pi)h

dV/dt = 25(pi)(dh/dt)
-2 = 25(pi)(dh/dt)
-2/(25)(pi) ft/min = dh/dt

better?
 
  • #8
Much better!
 
  • #9
just for any question like that...when it is asking what the rate of changing is...u need a negative sign...
when it is asking for the rate of decreasing...u don't need a negative sign...is that correct?
 
  • #10
That depends on the context of the question. Usually when something is decreasing you will use a negative sign. For example, if the height is decreasing at a rate of 2 meters per second, you would write that like this.

[tex]\frac{dh}{dt} = -2m/s[/tex]
 
  • #11
but when the question ask what is the rate of decreasing...do u still need a negative sign because we know that it is decreasing already...
 
  • #12
I would put the negative to indicate decreasing. If you did not put the negative sign and applied the positive value, you would have an incorrect answer.
 
  • #13
i mean in the answer...do u need negative signs to indicate decreasing when the question asks for decreasing already...
 
  • #14
I would put, [tex]\frac{dh}{dt} = - \mbox{rate}[/tex] just to make sure there's no mix up.

If the question specifically asks, "what rate is the height decreasing?" then I guess it's acceptable to put a positive answer, although that could be ambiguous.

I would put my first answer just to make clear you know what you are talking about.
 
  • #15
okay...i see...thanks...
 
  • #16
If it gives you find the rate at which h is decreasing, I would report my answer as positive, since when read with the question it makes more sense that way. "h is decreasing at a rate of 3 m/s." Not "h is decreasing at a rate of -3m/s"
 
  • #17
Right, if it is read that way then the negative is implied by the word decreasing. But I was saying that safest way to make sure there was no mix up is to put

[tex]\frac{dh}{dt} = ...[/tex]

where the sign is correctly placed. If you keep the decreasing rate positive because of a wording you could forgot to change it to a negative when working with other things using that value.
 

Related to Related Rates: Finding the Rate of Change of Liquid Height in a Draining Tank

1. What is a "Related Rates tank drain" problem?

A "Related Rates tank drain" problem is a type of math problem that involves finding the rate at which the water level in a tank is changing based on the rate at which the tank is draining. It is a common type of related rates problem that is often used in calculus courses.

2. How do you solve a "Related Rates tank drain" problem?

The first step in solving a "Related Rates tank drain" problem is to draw a diagram and label all the given information. Then, identify the known and unknown variables and write an equation that relates them. Next, take the derivative of the equation with respect to time and substitute in all the known values. Finally, solve for the unknown variable and include appropriate units.

3. What is the formula for solving a "Related Rates tank drain" problem?

The formula for solving a "Related Rates tank drain" problem is dV/dt = A * dh/dt, where dV/dt is the rate of change of volume, A is the cross-sectional area of the tank, and dh/dt is the rate of change of height.

4. What are some tips for solving "Related Rates tank drain" problems?

Some tips for solving "Related Rates tank drain" problems include drawing a clear diagram, labeling all the given information, using appropriate units, and taking the derivative with respect to time.

5. How can "Related Rates tank drain" problems be applied in real life?

"Related Rates tank drain" problems can be applied in real life to determine the rate at which a liquid is being drained from a tank, such as in a swimming pool or a water tower. It can also be used to calculate the rate at which a body of water, like a lake or a pond, is changing due to evaporation or rainfall.

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