Calculus Problem: Finding Work to Empty Trough of Water

  • Thread starter Tranquility13
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    Calculus
In summary, to empty a trough by pumping the water over the top, the weight of the water in the trough is 62 pounds per cubic foot, and it takes 6200 foot-pounds of work to do so.
  • #1
Tranquility13
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Homework Statement


A trough is 5 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of y=x^{2} from x=-1 to x=1 . The trough is full of water. Find the amount of work in foot-pounds required to empty the trough by pumping the water over the top. Note: The weight of water is 62 pounds per cubic foot.


Homework Equations





The Attempt at a Solution

 
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  • #2
Welcome to PF!

Hi Tranquility13! Welcome to PF! :smile:

Show us what you've tried, and where your'e stuck, and then we'll know how to help you! :smile:
 
  • #3
Not really sure how to go about this one. Online calculus course, teachers notes bery confusing.
 
  • #4
… one step at a time …

Tranquility13 said:
A trough is 5 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of y=x^{2} from x=-1 to x=1.

Hi Tranquility13! :smile:

ok … divide the trough into horizontal slices, a distance y above the bottom, and with height dy.

what is the volume of this slice?

what is the weight of the water in it?

how much work is needed to lift that water to the top of the trough (ie to y = 1)?

get that far, and then we'll sort out the integral! :smile:
 
  • #5


tiny-tim said:
Hi Tranquility13! :smile:

ok … divide the trough into horizontal slices, a distance y above the bottom, and with height dy.

what is the volume of this slice?

what is the weight of the water in it?

how much work is needed to lift that water to the top of the trough (ie to y = 1)?



Hi TinyTim (or someone else),
Maybe you can help me finish this problem:
A=l*w=5*2*sqrt(y)
dV=A*dy=5*2*sqrt(y)*dy
dF=p*dV=62*5*2*sqrt(y)*dy
dW=y*dF=y*62*5*2*sqrt(y)*dy

so then the integral is:
int{6200*y*sqrt(y)*dy, from y=0 to y=1}

But this fails to give me the right answer. Hopefully you can shed some light.
Thanks.
 
  • #6
Hi greenandblue!

(are you the same as Tranquility13? … have a square-root anyway: √ :smile:)
greenandblue said:
dW=y*dF=y*62*5*2*√y*dy

Nooo … you have to pump the water over the top! :wink:
 
  • #7
That's what I thought I was doing... well my think was that if it took the force dF for the particular sliver dy, then it would take dF * the height of each sliver (distance = y, from 0, the top, to 1, the bottom), and do that for every sliver dy.

Since the water level is equal to the top of the trough then no further work would be required to get it out.

Of course I know I am off somewhere here, just not sure where...

I'm new to this site, I didn't change my name from T.13, but I certainly appreciate your help!
 
  • #8
greenandblue said:
I'm new to this site, I didn't change my name from T.13, but I certainly appreciate your help!

Hi greenandblue! :smile:

erm … you're really not supposed to do other members' problems for them …

we'd better wait for Tranquility13 to catch up. :smile:
 
  • #9
tiny-tim said:
we'd better wait for Tranquility13 to catch up. :smile:
I completely agree except in this case: T.13 has left her problem untouched since her last post on May 1st, 3.5 months ago. It seems that by now she has either solved the problem or lost interest in it.

I was continuing her question b/c it is similar to one I have, which also deals with emptying a trough of the shape y=x^8 by 4ft long by 1ft tall.

Please reconsider, your advice is helpful. Thanks.
 
  • #10
The way I'd approach the problem is to first think about how you'd find the volume, and then remember that the work needed to pull out the water at each point is proportional to the height of the water at that point.
 
  • #11
… oops!

greenandblue said:
I completely agree except in this case: T.13 has left her problem untouched since her last post on May 1st, 3.5 months ago. It seems that by now she has either solved the problem or lost interest in it.

oops! You're absolutely right … my apologies! :smile:

ok … the distance to the top of the trough isn't y, it' s 1 - y. :wink:
 
  • #12


tiny-tim said:
ok … the distance to the top of the trough isn't y, it' s 1 - y. :wink:
Ok, I got it!, I was considering that before but now it's clear and makes sense. Thanks for your help.
 

1. What is calculus and how is it used to solve problems?

Calculus is a branch of mathematics that deals with the study of continuous change. It is used to solve problems involving rates of change, optimization, and finding areas and volumes of complex shapes.

2. What is the basic concept behind finding work to empty a trough of water?

The basic concept behind finding work to empty a trough of water is to calculate the amount of work required to lift the water from the trough to a certain height. This involves using the principles of force, work, and energy.

3. How do you set up and solve a calculus problem involving finding work to empty a trough of water?

To set up and solve a calculus problem involving finding work to empty a trough of water, you first need to define the variables involved (such as the volume of water, height of the trough, and force of gravity). Then, you can use the formula for work (W = Fd) and integrate it over the desired height to find the total work required.

4. What are some real-life applications of using calculus to solve problems like finding work to empty a trough of water?

Some real-life applications of using calculus to solve problems like finding work to empty a trough of water include determining the amount of energy needed to pump water from a well, calculating the force required to lift objects, and optimizing the design of water systems.

5. What are some common challenges when solving calculus problems like finding work to empty a trough of water?

Some common challenges when solving calculus problems like finding work to empty a trough of water include understanding the problem and setting up the correct equations, dealing with complex integrals and derivatives, and interpreting the results in a real-life context.

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