Radius of Curvature of a Parametric Plane Curve

Intuition

The osculating circle is the circle that best hugs the curve at a point: it matches the curve's position, direction, and bending. Its radius is the radius of curvature $R$. A sharp turn means small $R$ (large curvature $\kappa$); a nearly straight stretch means huge $R$.

Result / Method

For a plane curve $\vec r(t)=x(t)\,\hat i + y(t)\,\hat j$: $\kappa=\frac{|x'y''-y'x''|}{\big((x')^2+(y')^2\big)^{3/2}},\qquad R=\frac1\kappa.$ Steps: (1) compute $x',y',x'',y''$; (2) plug into the numerator (a $2\times2$ determinant of $\vec r'$ and $\vec r''$) and denominator (speed cubed); (3) invert for $R$. For an explicit $y=f(x)$ use $R=(1+(y')^2)^{3/2}/|y''|$.

Worked Example

For $\vec r(t)=\sinh t\,\hat i+\cosh t\,\hat j$ at $t=\ln 2$: $x'=\cosh t$, $y'=\sinh t$, $x''=\sinh t$, $y''=\cosh t$. Numerator $x'y''-y'x''=\cosh^2 t-\sinh^2 t=1$. With $\sinh(\ln2)=\tfrac34,\ \cosh(\ln2)=\tfrac54$, speed$^2=(\tfrac54)^2+(\tfrac34)^2=\tfrac{17}{8}$, so $R=(\tfrac{17}{8})^{3/2}=\tfrac{17\sqrt{34}}{32}\approx3.10$.

Pitfall: Don't confuse $R$ with $\kappa$ — the question almost always wants $R=1/\kappa$. Also simplify $a^{3/2}$ carefully: $8^{3/2}=16\sqrt2$, and any $\sqrt2$ you rationalize out of the denominator must reappear in the numerator (a common dropped factor in answer keys).