So I have this question here which say:
Show that the curve $\vec{r}^{\,}(t)=\left<\sin(t),2\cos(t),\cos(t)\right>,
t \in \mathbb{R}$, lies at the intersection between an ellipsoid and a plane. Find the curvacture at an arbitrary point on the curve and the point(s) on the curve at which the curvacture is a maximum.
Here's the solution:
I'm confused by many things. I understand the parametrization, but how was $x^2+\frac{1}{2}\left(\frac{y^2}{4}+z^2 \right)$ obtained?
For the part of the solution that says "Notice this is not the only ellipsoid…", how were those ellipsoids obtained?
Next, how or why did the solution randomly get $y-2z=0$ from?
Even if that somehow make sense, how does that "prove" that this parametric curve lies at the intersection between an ellipsoid and a plane? I guess I'm just stuck on the general idea of this.
I understand how they got the maximum so no need to explain that part to me.
If someone could clear up the intersection portion, that would be great, thanks! I'm just really trying to understand this concept.
Best Answer
To see this, you could reason as follows: you know that the equation of an ellipsoid is $ax^2 + by^2 + cz^2 = d$ for some numbers $a,b,c,d$ all positive. Now in the case of your parametrized curve, which you want to put on an ellipsoid, you have $x^2 = \text{sin}^2(t)$, $y^2 = 4\text{cos}^2(t)$ and $z^2 = \text{cos}^2(t)$. How can you find the $a,b,c,d$ of an appropriate ellipsoid in this case? Well you’re essentially trying to find an identity that these squared trig functions satisfy. The basic identity involving squared trig functions is $\text{sin}^2(t) + \text{cos}^2(t) = 1$ so you should probably try and use that. If you choose $a,b,c,d$ to be $1,1,-3,1$ then it will look exactly like that identity, but you need them all to be positive to get an ellipsoid. So instead you could take $1,1/8,1/2,1$ and then you get the ellipsoid $x^2 + (1/8)y^2 + (1/2)z^2 = 1$.
Then to find a plane you do something similar - you want an equation of the form $ax + by + cz = d$ (no squares this time) for some different numbers $a,b,c,d$. Since both $y$ and $z$ are multiples of $\text{cos}(t)$ (the multiples being 2 and 1 respectively) a natural first choice is $a,b,c,d$ should be $0,1,-2,0$, giving you the plane $y-2z=0$.