Finding the Rate of Change of Water in a Trough

In summary, the conversation discusses a problem involving a trough filled with water and the rate at which the volume of water is changing. The given equation for the volume of water is V = 12x^2 and it is known that water is flowing into the trough at a rate of 60 cm^3/s. The goal is to find the rate at which x, representing the length of the trough, is increasing when x = 10. By using the chain rule and solving for \frac{dx}{dt}, the final answer is found to be 1440x.
  • #1
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Imagine a trough filled with water (I can't put up a picture).
The water in the tank at time t seconds is given by [tex]V = 12x^2[/tex]. Given that water is flowing into the trough at the rate of [tex]60 cm^3/s[/tex], find the rate at which x is increasing when x = 10.

[tex]\frac{dV}{dx} = 24x[/tex]
[tex]\frac{d?}{ds} = 60 cm^3/s[/tex]
Then what do I do?
 
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  • #2
Who's "x" and what does your second equation stand for...?

Daniel.
 
  • #3
If it shouldn't be x, can you tell me what it should be? I don't know what to do for the second equation. I just know its [tex]\frac{d\text{ something}}{dx} = 60 cm^3/s[/tex]. Again, if I'm wrong please tell me what it should be.
 
  • #4
I don't know about "x",what is its point...But:
[tex] \frac{dV}{dt}=24x \frac{dx}{dt} [/tex]

,uding the chain rule...Now plug in the values and solve for the rate of "x"...

Daniel.
 
  • #5
Oh now I know why you asked me about the second equation. I always get mixed up and put ds instead of dt :grumpy: .
So I got 1440x. I know that's not the final answer because there's still the x = 10 part. And I'm confused on that part.
 
  • #6
Well,[itex] \frac{dV}{dt}=60cm^{3}s^{-1} [/itex]...x=10,you must find [itex] \frac{dx}{dt} [/itex]...

Daniel.
 
  • #7
I thought [tex]\frac{dx}{dt} = 60 cm^3/s[/tex] and I've to find [tex]\frac{dV}{dt}[/tex] which is 1440x?
 
  • #8
Nope,it's the other way aroung,the volume is increasing at the rate given in the problem...:wink:

Daniel.
 
  • #9
The point is that
[tex]\frac{dV}{dt}= \frac{dV}{dx}\frac{dx}{dt}[/tex]
(the chain rule).

You are given [itex]\frac{dV}{dt}[/itex] and you can (and did) calculate [itex]\frac{dV}{dx}[/itex]. Put them into that equation and solve for [itex]\frac{dx}{dt}[/itex].
 
  • #10
Oh... Got it! Thank you!
 

1. What is the purpose of finding the rate of change of water in a trough?

The purpose of finding the rate of change of water in a trough is to determine how quickly the water level is changing over a specific period of time. This information can be useful in various applications, such as monitoring water usage or predicting future water levels.

2. How is the rate of change of water in a trough calculated?

The rate of change of water in a trough can be calculated by measuring the change in water level over a specific time period and dividing it by the time interval. This is known as the average rate of change and is expressed in units of distance per time (e.g. meters per hour).

3. What factors can affect the rate of change of water in a trough?

The rate of change of water in a trough can be affected by factors such as the rate of water input or output, the size and shape of the trough, and external factors like temperature or wind. These factors can cause the water level to change at a faster or slower rate.

4. How can the rate of change of water in a trough be used in real-life situations?

The rate of change of water in a trough can be used in various real-life situations, such as monitoring water usage in agriculture or industrial processes, predicting water levels in reservoirs or dams, and managing water resources in areas prone to droughts or floods.

5. What are some limitations of using the rate of change of water in a trough?

Some limitations of using the rate of change of water in a trough include the assumption that the trough has a constant cross-sectional area, neglecting any changes in temperature or evaporation, and not accounting for any leaks or spills that may affect the water level. Additionally, the accuracy of the measurement equipment and the frequency of measurements can also affect the accuracy of the calculated rate of change.

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