How Do You Calculate the Depth When a Parabolic Channel is Half Full?

[jrand26]

Homework Statement

The cross-section of a channel is parabolic. It is 3 metres wide at the top and 2 metres deep. Find the depth of water, to the nearest cm, when the channel is half full.

Homework Equations

Nil.

The Attempt at a Solution

I found the function for the graph to be y=x^2-3x. Then I figured that I could just plug (0,0) and (3,0) into the integrand, \frac{\1}{3}x^3-1.5x^2, and then divide by two to get the answer. I'm pretty sure that the problem lies with the method, not my math. I have a feeling that's not the above method does not solve the problem, but I'm not sure why that is so.

For reference, my method gets 225cm while the book has 126cm.

 

[Gib Z]

Well, I would have personally interpreted the graph to be such that the parabola passed through the points (-1.5, 2) (0,0) and (1.5, 2). With those points we can find the equation of the parabola, sub in y=1 and solve for x.

Your equations has a width of 3, but its depth is not 2 as the description states.

 

[jrand26]

Ok thanks, now I'm getting 137cm, and its probably because my function is a bit out of the points. Also, Is my latex correct, I'm supposed to see the stuff right?

 

[Gib Z]

To start tex code, use [ tex ] but to end it its [ /tex ], without the spaces of course.

 

[HallsofIvy]

Homework Statement

The cross-section of a channel is parabolic. It is 3 metres wide at the top and 2 metres deep. Find the depth of water, to the nearest cm, when the channel is half full.

Homework Equations

Nil.

The Attempt at a Solution

I found the function for the graph to be y=x^2-3x. Then I figured that I could just plug (0,0) and (3,0) into the integrand, \frac{\1}{3}x^3-1.5x^2, and then divide by two to get the answer. I'm pretty sure that the problem lies with the method, not my math. I have a feeling that's not the above method does not solve the problem, but I'm not sure why that is so.

For reference, my method gets 225cm while the book has 126cm.

I have corrected the LaTex. You just need "[/tex]" at the end, not "[ tex ]".

The simplest way to do this would be to take the origin of your coordinate system to be at the vertex so that the equation is just y= ax^2. That is, the parabola passes through (0,0), (3/2, 2), and (-3/2, 2) and you determine a by 2= a(3/2)^2.

From that, the "full" volume is \int_{-3/2}^{3/2}2- ax^2 dx and the depth half full, h, is given by a\int_{-3/2}^{3/2} h- ax^2 dx= (1/2)\int_{-3/2}^{3/2}2- ax^2 dx.

 

[Unit]

I agree with Halls of Ivy that you should probably look at the way you're integrating.
Once you've gotten the proper equation for a parabola, I see two methods. .
A) Make a rectangle from the x-axis up to your line y = 2, ending at the points (+1.5, 2) and (-1.5, 2). What is the area of this box?
How does the area given by the definite integral relate to this?
B) More tricky method: Take the inverse of the parabola, and integrate directly the area of water, and use that.