I am having difficulty finding the sum of maximum and minimum curvature of the ellipse $9(x-1)^2 + y^2 = 9$. I know that I am supposed to parametrize the ellipse as $f(x(t), y(t))$, with $x(t) = 1 + cos(t)$, then find $y(t)$ and sketch the ellipse to find where the curvature is greatest and least. I managed to sketch the graph, but I am at a loss as to how to continue from there.
Anyone knows how should this problem be solved? Much appreciated!
Given a parametric curve $(x(t), y(t))$, the (signed) curvature is given by \begin{equation*} \kappa(t) = \frac{x' y'' - y' x''}{((x')^2 + (y')^2)^{3/2}}. \end{equation*}
The ellipse $9(x-1)^2 + y^2 = 9$ can be rewritten in standard form as $(x-1)^2 + \frac{y^{2}}{9} = 1$. Then parametrize it as \begin{align*} x(t) & = 1 + \cos{t}\\ y(t) & = 3\sin{t}. \end{align*}
Then compute \begin{align*} \kappa(t) & = \frac{-\sin{t}\left(-3\sin{t}\right) - 3\cos{t}(-\cos{t})}{\left(\sin^{2}{t} + 9\cos^{2}{t}\right)^{3/2}}\\ & = \frac{3}{\left(\sin^{2}{t} + 9\cos^{2}{t}\right)^{3/2}}\\ & = \frac{3}{\left(1 + 8\cos^{2}{t}\right)^{3/2}}. \end{align*}
Thus $\kappa(t)$ has minimum and maximum values of $\frac{3}{9^{3/2}} = \frac{3}{27} = \frac{1}{9}$ and $3$, respectively. These occur when $t = 0, \pi$ and $t = \frac{\pi}{2}, \frac{3\pi}{2}$, respectively.
In order to arrive at the formula above for curvature (above is the signed curvature, though it never changes sign so it doesn't matter), we can start from the definition: \begin{equation*} \kappa = \| \frac{d \mathbf{T}}{ds} \| \end{equation*} where $s$ is the arc length parameter and $\mathbf{T}(s)$ is the unit tangent vector to the curve.
Assuming we have a two-dimensional parametrization, $\gamma(t) = \langle x(t), y(t) \rangle$, we have \begin{equation*} \mathbf{T}(t) = \frac{\langle x'(t), y'(t)\rangle}{\sqrt{ (x'(t))^{2} + (y'(t))^{2} }} = \frac{1}{\|\gamma'(t)\|}\langle x'(t), y'(t) \rangle. \end{equation*}
Now consider that the arc length from time $0$ to $t$ is given by \begin{equation*} s(t) = \int_{0}^{t}\|\gamma'(u)\|\,du. \end{equation*} By the Fundamental Theorem of Calculus, we have $\frac{ds}{dt} = s'(t) = \|\gamma'(t)\|$.
Note that \begin{equation*} \frac{d \mathbf{T}}{ds} = \frac{d\mathbf{T}}{dt} \frac{dt}{ds} = \frac{\mathbf{T}'(t)}{\frac{ds}{dt}} = \frac{\mathbf{T}'(t)}{\|\gamma'(t)\|}. \end{equation*}
Now we compute \begin{align*} \mathbf{T}'(t) &= \left\langle \frac{x''}{\|\gamma'(t)\|} - \frac{1}{2}\frac{x'\cdot 2 \cdot (x' x'' + y' y'')}{\|\gamma'\|^{3}}, \, \frac{y''}{\|\gamma'(t)\|} - \frac{1}{2}\frac{y'\cdot 2 \cdot (x' x'' + y' y'')}{\|\gamma'\|^{3}} \right\rangle\\ & = \frac{1}{\|\gamma'(t)\|^{3}} \langle x'' \left( (x')^{2} + (y')^{2} \right) - (x')^{2}x'' - y'x'y'', \, y''\left( (x')^{2} + (y')^{2} \right) - (y')^{2}y'' - y'x'x'' \rangle\\ & = \frac{1}{\|\gamma'(t)\|^{3}} \langle y'(x''y' - x'y''), \, x'(y''x' - x''y') \rangle\\ & = \frac{x''y' - x'y''}{\|\gamma'(t)\|^{3}}\langle y',\, -x' \rangle. \end{align*}
Thus \begin{align*} \kappa & = \left\| \frac{d\mathbf{T}}{ds} \right\|\\ & = \left\|\frac{\mathbf{T}'(t)}{\|\gamma'(t)\|} \right\|\\ & = \left\|\frac{x''y' - x'y''}{\|\gamma'(t)\|^{4}}\langle y',\, -x' \rangle \right\|\\ & = \frac{|x''y' - x'y''|}{\|\gamma'(t)\|^{3}}\\ & = \frac{|x''y' - x'y''|}{((x')^{2} + (y')^{2})^{3/2}}. \end{align*}