Math Ellipse Question

seiferseph

How would i do this question?

A canal with cross-section a semi-ellipse of width 20 m is 5 m deep at the centre. Find the equation for the ellipse, and use it to find the depth 2 m from the edge.

I think i have a way to get it using major axis length 20 and minor axis length 10, but what would the specific points be (where is the centre?) thanks.

codyg1985

You could put the center of the ellipse on the origin here. It should not matter in terms of the geometry of the ellipse itself.

Ouabache

You major and minor axes sound fine.

It sounds like there are no constraints for defining your coordinate system.
How about calling the midpoint across your canal, at surface level (0,0).
[you're not obligated to do this, you can still solve this even if you defined the near or far edge or your canal as (0,0)]

Do you already know the general formula for an ellipse?
(hint: take a look at equ. 11 at this URL---> http://mathworld.wolfram.com/Ellipse.html

How far from the centre would 2m from either edge be?
Just plug in your given information into the equation for an ellipse and you can find the depth at 2m from either edge.

seiferseph

for the equation i got

x^2/100 + y^2/25 = 1

then to solve for the depth 2 m from the edge, so i substitute 8 for x? i got +/- 3. what does that mean? the depth would be 3 m deep at 2 m in from either side, right?

codyg1985

The numbers you did get indicate that it is that depth at either side of the ellipse.

HallsofIvy

I just want to expand on codyg1985 and Ouabache's point that you can ASSUME the center of the ellipse is at (0,0).


Canals don't have coordinate systems attached! You are free to set up a coordinate system yourself- in this case it is simplest to choose your coordinate system so that (0,0) is at the center of the ellipse.

In fact you DON'T have to use meters as your units. Let's do this: take half the length of the horizontal axis as your unit of length in the x-direction and half the length of the vertical axis as your unit of length in the y-direction.
What? different units along the x and y axes? Yep, that's perfectly valid. Of course, it messes up the geometry a bit! Now that the semi-axes along the x and y axes are both one, this ellipse (in this rather peculiar coordinate system) has equation x²+ y²= 1- the equation of a circle (you have to kinda "squash" your eyeballs to make them fit this coordinate system!).
(At least I am choosing the coordinate axes along the axes of the ellipse. It would be legitimate to choose them some other angle but that would really complicate the calculations. Even I am not that crazy!)
So how do we "use it to find the depth 2 m from the edge."?

Our coordinate unit in the x-direction corresponds to half the horizontal width of the canal- 10 m. As seiferseph calculated, that is 8 m from the center and so 0.8 "x- units". Put x= 0.8 in x²+ y²= 1 to get (0.8)²+ y²= 0.64 + y² = 1 so y²= 1- 0.64= 0.36 and so
y= 0.6 (do you see the "3-4-5 right triangle" there?) At this point the y-coordinate is 0.6 of a unit. But a unit in the y direction is the depth of the canal- 5 m. 0.5 of 5 m is 3 m just as seiferseph got before!

seiferseph

Thats interesting HallsofIvy, and thanks to everyone.