Paper 3, Section I, A

Vector Calculus | Part IA, 2015

The smooth curve C\mathcal{C} in R3\mathbb{R}^{3} is given in parametrised form by the function x(u)\mathbf{x}(u). Let ss denote arc length measured along the curve.

(a) Express the tangent t\mathbf{t} in terms of the derivative x=dx/du\mathbf{x}^{\prime}=d \mathbf{x} / d u, and show that du/ds=x1d u / d s=\left|\mathbf{x}^{\prime}\right|^{-1}.

(b) Find an expression for dt/dsd \mathbf{t} / d s in terms of derivatives of x\mathbf{x} with respect to uu, and show that the curvature κ\kappa is given by

κ=x×xx3\kappa=\frac{\left|\mathbf{x}^{\prime} \times \mathbf{x}^{\prime \prime}\right|}{\left|\mathbf{x}^{\prime}\right|^{3}}

[Hint: You may find the identity (xx)x(xx)x=x×(x×x)\left(\mathbf{x}^{\prime} \cdot \mathbf{x}^{\prime \prime}\right) \mathbf{x}^{\prime}-\left(\mathbf{x}^{\prime} \cdot \mathbf{x}^{\prime}\right) \mathbf{x}^{\prime \prime}=\mathbf{x}^{\prime} \times\left(\mathbf{x}^{\prime} \times \mathbf{x}^{\prime \prime}\right) helpful.]

(c) For the curve

x(u)=(ucosuusinu0)\mathbf{x}(u)=\left(\begin{array}{c} u \cos u \\ u \sin u \\ 0 \end{array}\right)

with u0u \geqslant 0, find the curvature as a function of uu.

Typos? Please submit corrections to this page on GitHub.