= Solution
Use $x>0$ as the parameter of the <smooth curve>:
$$
\mathbf r(x)=(x,\log x,0).
$$
Its <velocity vector> and speed are
$$
\mathbf r'(x)=\left(1,\frac1x,0\right),
\qquad
|\mathbf r'(x)|=\frac{\sqrt{x^2+1}}x.
$$
The <unit tangent vector> is therefore
$$
\boxed{\mathbf t(x)=\frac{(x,1,0)}{\sqrt{x^2+1}}}.
$$
The parameter-independent formula for the <curvature of a space curve> gives
$$
\kappa(x)
=\frac{|\mathbf r'(x)\times\mathbf r''(x)|}
{|\mathbf r'(x)|^3}
=\boxed{\frac{x}{(x^2+1)^{3/2}}}.
$$
Differentiating,
$$
\kappa'(x)=\frac{1-2x^2}{(x^2+1)^{5/2}}.
$$
Hence $\kappa$ increases for $0<x<1/\sqrt2$ and decreases for $x>1/\sqrt2$. Its <global maximum> occurs at $x=1/\sqrt2$ and equals
$$
\boxed{\kappa_{\max}=\frac{2}{3\sqrt3}}.
$$
The corresponding point is
$$
\boxed{\left(\frac1{\sqrt2},-\frac12\log2,0\right)}.
$$
Back to article page
