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Equation with two second order variables

  • Thread starter dunkdie
  • Start date
  • #1
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Hi all,

I've come across this forum when searching for solution to my problem and I found that the community is extremely helpful :)

I was working on my college project when I came across the equation below.

Homework Statement


Z = -0.0006*X^2 + 0.0004*X -0.0008*Y^2 - 0.0006*Y + 0.956

Suppose that I know the value of Z (as example Z=0.15), which method can I use to find the value of X and Y?


Homework Equations



The equation above was derived when I was trying to chart a surface graph from these two equations:

Z = -0.0006*X^2 + 0.0004*X + 0.956

and

Z = -0.0008*Y^2 - 0.0006*Y + 0.956

However I dunno the correctness of this method that I used to come up with surface graph Z = -0.0006*X^2 + 0.0004*X -0.0008*Y^2 - 0.0006*Y + 0.956
The picture of this surface graph is attached.


The Attempt at a Solution



I tried using differential method to solve the equation but so far no luck :( Hope somebody would come up with some method or explanation regarding the matter. Many thanks.


Regards,
John
 

Attachments

Answers and Replies

  • #2
33,519
5,196
For a given value of z, your equation determines an ellipse. Complete the square in the x and y terms to get the ellipse into standard form.

However, I'm not so sure about this equation:
Z = -0.0006*X^2 + 0.0004*X -0.0008*Y^2 - 0.0006*Y + 0.956

You said it came about from these equations:
Z = -0.0006*X^2 + 0.0004*X + 0.956
Z = -0.0008*Y^2 - 0.0006*Y + 0.956

Presumably you set the two right sides equal, which would give you

-0.0006*X^2 + 0.0004*X + 0.956 = -0.0008*Y^2 - 0.0006*Y + 0.956
or, equivalently, -0.0006*X^2 + 0.0004*X = -0.0008*Y^2 - 0.0006*Y

Notice that there is no z and no 0.956.

The equation above represents a hyperbola that is the intersection of the to parabolic cylinders Z = -0.0006*X^2 + 0.0004*X + 0.956 and Z = -0.0008*Y^2 - 0.0006*Y + 0.956.
 

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