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October 6th, 2017, 03:44 AM   #1
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Question A question regarding how to define a plane

Today I took a math test and one of the questions was:

"Prove that 4x + 3y = 7 is a tangent plane to x^2 + 2xy + y = 4 and state the tangent point."

This question confused me, since the question didn't state that x or y were functions of any variables. Therefore I'm thinking that both of the equations are in fact not planes (3 dimensions), but graphs (2 dimensions).

I'm also thinking that there is no way to get a (x,y,z) point, since the functions only consist of 2 variables.

Am I thinking completely wrong or is the question incorrectly stated?

Also, my first language is not English so forgive me if I am using the wrong terms.

Thanks!
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October 6th, 2017, 05:55 AM   #2
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Quote:
Originally Posted by EliasR View Post
Today I took a math test and one of the questions was:

"Prove that 4x + 3y = 7 is a tangent plane to x^2 + 2xy + y = 4 and state the tangent point."

This question confused me, since the question didn't state that x or y were functions of any variables. Therefore I'm thinking that both of the equations are in fact not planes (3 dimensions), but graphs (2 dimensions).
If you draw a set of 3D Cartesian coordinates, you can draw the flat plane described by $\displaystyle 4x + 3y = 7$ and the curved surface $\displaystyle x^2 + 2xy + y = 4$ just fine... it's just that there's just no z-dependence on the plane or surface. Therefore, if your Cartesian axes has the z-axis pointing "up", the plane and the surface will look vertical in the diagram.

The problem is basically a 2D problem since you don't have to worry about the z-coordinate, but there's nothing wrong with using those equations to describes planes or surfaces in 3D space.

You can get 3D coordinates without any problem, but in a problem with no z-dependence, the z-component of any coordinate you find will probably be 0.

Last edited by Benit13; October 6th, 2017 at 05:57 AM.
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October 6th, 2017, 06:08 AM   #3
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Quote:
Originally Posted by EliasR View Post
Today I took a math test and one of the questions was:

"Prove that 4x + 3y = 7 is a tangent plane to x^2 + 2xy + y = 4 and state the tangent point."

This question confused me, since the question didn't state that x or y were functions of any variables. Therefore I'm thinking that both of the equations are in fact not planes (3 dimensions), but graphs (2 dimensions).
I'm not sure what you mean by that. x and y don't have to be functions of any other variables. It is interesting that there is no "z" in either equation. We could write that with 4x+ 3y= 7 as a line in the xy-plane, x^2+ 2xy+ y= 4 as a conic section. However, since the word "plane" is used, I imagine we are expected to think of those as extending out of the xy-plane parallel to the z- axis.

Quote:
I'm also thinking that there is no way to get a (x,y,z) point, since the functions only consist of 2 variables.

Am I thinking completely wrong or is the question incorrectly stated?

Also, my first language is not English so forgive me if I am using the wrong terms.

Thanks!
You are correct that, if this is to be in three dimensions, then the point of contact is a line, along the length of the cylinder, not a point.

You can solve this as "Show that 4x+ 3y= 7 is the tangent line to x^2+ 2xy+ y= 4" in the xy-plane. That (x, y) point, with z any number, would be the line of tangency of the plane and surface.

If it were me, I would double check to make certain there was not supposed to be a "z" in at least one of those equations.
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October 6th, 2017, 06:29 AM   #4
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Thank you for your replies.

There was no "z" in the equation, so I was thrown off by that. I guess I just have a tricky teacher that hands out this problem on a test

At least now I know that I should have treated it as equations in the xy-plane.

The extra tricky thing is that we learned a formula for getting a tangent plane and it contained z and f derived with respect to x and y.
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October 6th, 2017, 08:18 AM   #5
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Note that x = y = 1 satisfies both equations.
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October 6th, 2017, 09:47 AM   #6
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Note that x = y = 1 satisfies both equations.
Yes, I at least got that right and realized that it was the point of tangens (don't now how to express it in English really). Hopefully I get one point for getting that right.
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