Level curves

  • Thread starter cscott
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  • #1
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Homework Statement



I need to sketch level curves of [itex]T(x, y) = 50(1 + x^2 + 3y^2)^{-1}[/itex] and [itex] V(x, y) = \sqrt{1 - 9x^2 -4y^2}[/itex]

The Attempt at a Solution



Is it correct that they are ellipses?

ie [tex] 1 = \frac{9}{1 - c^2} x^2 + \frac{4}{1 - c^2}y^2[/itex]

for V(x, y) = c = constant
I feel so rusty going back to school :s
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
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That's for V? Setting [itex]V(x,y)= \sqrt{1- 9x^2- 4y^2}= c[/itex], then [itex]1- 9x^2- 4y^2= c^2[/itex], [itex]9x^2+ 4y^2= 1-c^2[/itex],
[tex]\frac{9}{1- c^2}x^2+ \frac{4}{1-c^2}y^2= 1[/tex]
just as you say. Yes, that's an ellipse. It might be easier to recognise if you wrote it
[tex]\frac{x^2}{\left(\frac{\sqrt{1-c^2}}{3}\right)^2}+ \frac{y^2}{\left(\frac{\sqrt{1-c^2}}{2}\right)^2}= 1[/tex]
an ellipse with center at (0,0) and semi-axes of length
[tex]\frac{\sqrt{1-c^2}}{3} [/tex]
and
[tex]\frac{\sqrt{1-c^2}}{2}[/tex]

Similarly, [itex]T(x,y)= 50(1+ x^2+ 3y^2)^{-1}= c[/itex] gives [itex]c(1+ x^2+ 3y^2)= 50[/itex] so [itex]1+ x^2+ 3y^2= 50/c[/itex], [itex]x^2+ 3y^2= (50/c- 1)[/itex]. Now divide both sides by 50/c- 1:
[tex]\frac{x^2}{50/c-1}+ \frac{y^2}{\frac{50/c-1}{3}}= 1[/tex]
again, an ellipse with center at (0,0), semi-axes of length
[tex]\sqrt{50/c- 1}[/tex]
and
[tex]\sqrt{\frac{50/c- 1}{3}}[/tex]
 
  • #3
782
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That does make it easier to understand.

Thanks for your help.
 

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