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Doing some revision and came across this question. Would rather step along my thought process first rather than just type down the questions at the start if that's okay.

A roof is given by the function $z = f(x, y) = \frac{5}{2} + \frac{1}{200} (9x^2-4y^2)$ and is on the domain

$[-6, 6] × [-6, 6]$

Nine timber joists will be cut, and positioned such that they lie on cross sections of the surface. Shadecloth will be laid on top of the joists to produce a curved roof effect.

So I can see we have a hyperbola here and can understand where the 9 joists would be positioned. Here is what I think this function is looking like, I can see where each joint would go in this picture:

picture

Our vertical plane for cross sections is going to take the form $ax+by=c$ I think and taking x or y as constants give the cross sections.

$z = f(x, y) = \frac{5}{2} + \frac{1}{200} (9x^2-4c^2)$

and

$z = f(x, y) = \frac{5}{2} + \frac{1}{200} (9c^2-4y^2)$

These are parabolas so I don't really know what to do with these, seems kind of useless with the problem.

I'm first asked this by the book

Which values of $a, b, c$ can be chosen in my general equation such that the cross sections of our original equation are straight lines in three-dimenstional space?

Okay, I'd think I'd have to determine the positions of the joists to answer this. That's why I'm scratching my head.

I'll continue here with more info.

Each joist $J_{k}$, $k = 1, ..., 9$ is to have one end attached to the position $P_{k}$ and the other end attached to the position $Q_{k}$ in three-dimensional space. $P_{k}$ is specified in the question.

Determine the other endpoints $Q_{k}$, such that each joist $J_{k}$ lies on the surface $f(x, y)$ and the $x$ and $y$ coordinates of each $Q_{k}$ lie on the boundary of the domain $[−6, 6] × [−6, 6].$

I'm not going to list all 9 points because I'd eventually like to solve this myself but I'll give 4 $P_{k}$ as an example here.

$P_{1} = (-6, 3, f(-6, 3))$

$P_{3} = (-6, -3, f(-6, -3))$

$P_{5} = (-4, -6, f(-4, -6))$

$P_{7} = (0, -6, f(0, -6))$

So once I find the general equation. Which I'll probably do through some sort of simultaneous method maybe? Then I have to find the point on a plane from a vector at each $P_{k}$ 9 different times? I think that's how I go about solving it but it seems so inefficient and I'm also not sure.

Any help appreciated, I'll keep working with mucking with the general equation to find a b and c values in the meantime.

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1 Answer 1

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The way to get the information that you want is to start with the equation of the plane, and plug it into the equation for $z$. This way you get the intersection. All you need is now to have $z$ as a linear function of either $x$ or $y$.

Suppose $a\ne 0$. Then we can write $x=\frac{c-by}{a}$. If we put this expression into $f(x,y)$ we get $$z=\frac{5}{2}+\frac{1}{200}\left(9\frac{c^2-2bcy+b^2y^2}{a^2}-4y^2\right)$$ The coefficient of $y^2$ is $9\frac{b^2}{a^2}-4$. In order to have the intersection as a line, this coefficient must be $0$. You then get $$\frac{b}{a}=\pm\frac{2}{3}$$ You notice that you have two sets of parallel lines. In your figure you show one of these. In order to get the $c$ value, you just need to plug in the value for a point on the line $P_k$. Also note that you can get only ratios of coefficients, since $ax+by=c$ is equivalent to say $2ax+2by=2c$. You can choose $a=1$ and then calculate $b$ and $c$. As a last observation, make sure you cannot get $a=0$. That's easy. Just plug the equation for the plane into the expression for $z$, and show that the coefficient of $y^2$ is not $0$.

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  • $\begingroup$ Hi @Andrei, I was wondering if you could expand further on finding the general equation. So by solving for $x$ in the general equation and plugging that into our surface function we can see that $\frac{b}{a} = +-\frac{2}{3}$. How does this translate to finding an a,b or c value? May I now just let $a$ = 1 and solve for b and c that way? With b being $+-\frac{2}{3}$? Thank you. $\endgroup$
    – 99 Fishing
    Dec 18, 2017 at 6:20
  • $\begingroup$ Yes. That's the idea. Since any $a\ne 0$ is fine as long as $b/a=\pm2/3$, you can choose $a=1$. Or even better, $a=3$. That way $b=\pm2$. To find out $c$, you need to have one more equation. You get that from $P_k$. $\endgroup$
    – Andrei
    Dec 18, 2017 at 6:52
  • $\begingroup$ so it seems like solving for c using any $P_{k}$ gives me differing values that make up vertical intersection planes. Having general values that satisfy the equation of the plane like $a=3$, $b=+-2$, and $c=1.78$ looks to be correct as a plane itself. Is it possible for me to find a way to express that in $r(t) = r_{0} + tv $ form? $\endgroup$
    – 99 Fishing
    Dec 18, 2017 at 7:41
  • $\begingroup$ Sure. Start with $P_k$. we want to calculate a new point on the line. Since the coefficients for $x$ and $y$ are not $0$, we can choose for $R_k$ the $x$ coordinate to be the $x$ coordinate of $P_k$. Then from the equation of the plane you calculate $y$. Then from $f(x,y)$ you get $z$. Now the equation of the line is $P_k+t(R_k-P_k)$ $\endgroup$
    – Andrei
    Dec 18, 2017 at 22:18
  • $\begingroup$ @99Fishing Let me know if my answer was helpful. $\endgroup$
    – Andrei
    Dec 20, 2017 at 20:29

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