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Okay, so I missed CalcIII today and I'm struggling a bit here.

$r(t) = (\sin t,\cos t,7\sin t + 4\cos 2t)$

Find the projection of $r(t)$ onto the xz-plane for $−1 \leq x \leq 1$

Answer as an equation using the variables $x$, $y$, and $z$.

P.S. I clepped CalcI and II.... in 2010. Haven't done much since then so I'm just a little out of practice, need to kick my brain back into gear.

UserX
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SpencerThoughts
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  • Don't see what the variable $y$ could usefully do in such an equation. The natural coordinates in the $xz$-plane are $x$ and $z$, as the name suggests. You could add a cameo appearance of $y$ if you like, but it would just stand for $0$. – Marc van Leeuwen Sep 13 '14 at 06:41
  • I completely agree, however this is through WebAssign. Quite possibly the worst aspect of my college going experience thus far. A lot of colleges are requiring teachers to use it because it grades objectively and tracks time and percentages throughout the semester. The problem is you have to type your answer in a specific format for it to grade, which can be a serious issue if you don't know exactly how it wants it typed. The question I asked was exactly how it was asked by WebAssign. Don't get the wrong idea, I only post questions I can't find an explanation for elsewhere, and then only one. – SpencerThoughts Sep 14 '14 at 01:36
  • Well, you might want to give feedback to whoever is responsible for this WebAssign question; it is very sloppily formulated: $r(t)$ is just a single point (on a parametrized curve $r$) and so is its projection $(\sin t,7\sin t+4\cos2t)$; saying "for $-1\leq x\leq1$" is not meaningful for a formula that does not contain $x$ (though $-1\leq\sin t\leq1$ always holds here); as I said $y$ has no role. The question should ask for an equation in $x,z$ such combined with $y=0$ and $-1\leq x\leq1$ it describes the projection image of the curve $r$. You can demand precision from the teachers too. – Marc van Leeuwen Sep 14 '14 at 04:54

3 Answers3

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The question asks you to describe a curve without the parameter $t$ too. That means after you project (by setting $y = 0$), you have to relate $x$ and $z$ directly instead of via $t$. Now note that after setting $y = 0$, you have \begin{align} x & = \sin t \\ z & = 7 \sin t + 4 \cos 2t. \end{align} The problem is expressing $z$ in terms of $x$, which you can do with a bit of trigonometry: $$ z = 7x + 4(1 - 2x^2). $$ The complete answer would have to include the constraint $-1 \le x \le 1$ (because we lost this condition, which was automatic when $x$ was expressed in terms of $\sin$ of something), and the equation $y = 0$.

Tunococ
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Looks kinda like a parabola:

enter image description here

And with good reason. It just so happens that the $z$-component can be rewritten:

$$7 \sin (t)+4 \cos (2 t)=4-8 \sin ^2(t)+7 \sin (t).$$

Thus, the $x$ and $z$ components satisfy the relationship $z=4+7x-8x^2$.

Mark McClure
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The idea is to get rid of the y-component of the vector so that it lies in the xz-plane. I.e. write the vector as $$(\sin t,0,7 \sin t+ 4\cos 2t)$$

zeta
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