2

Say you had a function

f(x,y)= $5/2 + 1/200(9x^2-4y^2)$

And if you were to make a series of vertical planes over the domain $[-6,6]*[-6,6]$ in the form of $ax+by=c$. These vertical planes would give cross sections that are straight lines in the 3 dimensional space. Visualization of problem.

I'm trying to find values for $a,b$ and $c$ such that I can find a general formula for the cross-sections in the form $r=r0+tv$

So far my approach as been to set $b=1$ in the equation of a plane, rearranging for $y=c-ax$, and substituting that back into the original function, and setting the $x^2$ terms equal each other, such that $9x^2=4a^2x^2$.

This gives a value of $a=3/2$, which is as far as I've been able to go.

I've been stuck on this problem for some time now, so any help/tips would be greatly appreciated!

Thanks!

MichaelHannen
  • 31
  • 1
  • In the formula for $f(x,y),$ is the second term $(1/200)\cdot (9x^2-4y^2)$? [or is the $9x^2-4y^2$ part of the denominator?] – coffeemath Dec 17 '17 at 06:26
  • It's not clear to me how vertical planes (or even tilted planes satisfying some linear equation in $x,y,$ and $z$) would create linear cross-sections of $f$. – actinidia Dec 17 '17 at 06:33
  • The formula for that section of $f(x,y)$ is $(1/200)\cdot (9x^2-4y^2)$. – MichaelHannen Dec 17 '17 at 06:40
  • At first I thought that if you take the cross sections with $x$ or $y$ as a constant, it would produce parabolas. But it has to be possible that there are some values of $a,b$ and $c$ that satisfy a linear cross-section over the small domain. – MichaelHannen Dec 17 '17 at 06:47

0 Answers0