Surfaces and Planes in Three-Dimensional Spaces

Question

A roof will be constructed to fit the surface given by the function:

$$ z = f(x,y) = \frac 52 + {1\over 200} (9x^2 - 4y^2) \tag{1} $$

On the domain $[-6, 6] \times [-6, 6]$, with $x$, $y$, and $z$ all measured in metres. 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 (See figure).

Recall that a vertical plane has the form:

$$ ax + by = c \tag{2} $$

and that, generally, cross-sections of a surface $z = f(x,y)$ are the lines of intersections with vertical planes. Usually, the vertical planes are either taken as $x$ is constant, or $y$ is constant. For the surface, $z$, this gives the cross-sections:

$$f(x,c) = \frac52 + {1\over200}(9x^2 - 4c^2)$$ $$ f(c,y) = \frac52 + {1\over200}(9c^2 - 4y^2)$$

which are parabolas. Since the joists are cut from straight pieces of timber, these are not helpful.

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a) Which values of $a$, $b$, and $c$ can be chosen in equation $(2)$ such that the cross-sections of $(1)$ are straight lines in three-dimensional space? Give the general equation of the cross section in the form:

$$\mathbf r(\lambda) = \mathbf r_0 +\lambda \mathbf v$$

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b) Each joist $J_k$, $k = 1, \dots, 9$ 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. Specifically, the $P_k$ are given by:

$$ \begin{align} P_1 &= (-6, 3, f(-6, 3)) \\ P_2 &= (-6, 0, f(-6, 0)) \\ P_3 &= (-6, -3, f(-6, -3)) \\ P_4 &= (-6, -6, f(-6, -6)) \\ P_5 &= (-4, -6, f(-4, -6)) \\ P_6 &= (-2, -6, f(-2, -6)) \\ P_7 &= (0, -6, f(0, -6)) \\ P_8 &= (2, -6, f(2, -6)) \\ P_9 &= (4, -6, f(4, -6)) \end{align} $$

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 $[-6, 6] \times [-6, 6]$.

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c) Calculate the length of each joist. (Give the answer correct to two decimal places).

Answer 1

I'm going to give a hint for the first part to get you started. You should know that it's not in good form too ask so many questions in the same post. This forum does not like posts that are copied homework questions. Here's a quick guide on how to ask a good question

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Our function has the general form $$ f(x,y) = Ax^2 - By^2 + C $$ where $A,B,C$ are known, positive constants

Since a vertical plane is only given in terms of $x$ and $y$, it's a good idea to put one in terms, of the other, so we can rewrite $(2)$ as $y = px + q $ where $p = -a/b$, $q = c/b$

Plugging this in gives $$ z = f(x,px+q) = C + Ax^2 - B(px+q)^2 = (A-Bp^2)x^2 - 2Bpqx + (C-Bq^2) $$

You want this cross-section to be linear, which means the $x^2$ coefficients must be zero, therefore we have $$ A - Bp^2 = 0 $$

This gives you a value for $p$, which is the "slope" of the vertical plane. There are no constraints on $q$

As for the parametric form, notice that everything is already in terms of $x$, so one possible parametrization is $(x,y,z) = (\lambda,p\lambda+q,f(\lambda,p\lambda+q))$

If you have one specific point on this cross section, it's easy to find the value of $q$ (since $(2)$ is always valid and you already know what $p$ is). There are actually two solutions passing the same point, since you have two different values for $p$.