Question

Find the point(s) on the curve of y=e^x at which the curvature is largest.

Answer 1

@SolomonZelman @goformit100 @EclipsedStar

Answer 2

\[k(x)=\frac{ \left| y'' \right| }{ [1+(y')^2]^{3/2} }\]

Answer 3

y'=e^x
y"=e^x

Answer 4

@SithsAndGiggles

Answer 5

\[k(x)=\frac{ \left| e^x \right| }{ (1+e ^{2x})^{3/2} }\]

Answer 6

You want to find \(x\) that maximize \(k(x)\), so take the derivative. Note: \(e^x>0\) for all \(x\), so \(|e^x|=e^x\). This is your typical optimization problem. Quotient rule gives
\[\begin{align*}k'(x)&=\frac{\dfrac{d}{dx}[e^x](1+e^{2x})^{3/2}-e^x\dfrac{d}{dx}[(1+e^{2x})^{3/2}]}{(1+e^{2x})^3}\\\\
&=\frac{e^x(1+e^{2x})^{3/2}-\dfrac{3}{2}e^x(1+e^{2x})^{1/2}\dfrac{d}{dx}[1+e^{2x}]}{(1+e^{2x})^3}\\\\
&=\frac{e^x(1+e^{2x})^{3/2}-3e^{3x}(1+e^{2x})^{1/2}}{(1+e^{2x})^3}\\\\
&=\frac{e^x(1+e^{2x})^{1/2}\left((1+e^{2x})-3e^{2x}\right)}{(1+e^{2x})^3}\\\\
&=\frac{e^x\left(1-2e^{2x}\right)}{(1+e^{2x})^{5/2}}
\end{align*}\]
Find the critical points. Where is \(k'(x)=0\) ?

Answer 7

Got it! Thank you!