A particle moves in the Euclidean plane. At time t (taking all real values) its coordinates are x = t^{3} - t and y = t^{4} + t. Show that its velocity has a maximum at t = 0, and that its path has an inflection at t = 0.

**Solution**

The speed squared is (dx/dt)^{2} + (dy/dt)^{2} = 16t^{6} + 9t^{4} + 8t^{3} - 6t^{2} + 2. Let this be f(t). We have f '(t) = 12t(8t^{4} + 3t^{2} + 2t - 1). So f '(t) = 0 at t = 0. Also for t small (positive or negative), 8t^{4} + 3t^{2} + 2t - 1 is close to -1 and hence negative, so f '(t) is positive for t just less than 0 and negative for t just greater than 0. Hence f(t) has a maximum at t = 0. Hence the speed does also.

The gradient dy/dx = (4t^{3} + 1)/(3t^{2} - 1). Let this be g(t). Then g'(t) = 6t(2t^{3} - 2t - 1)/(3t^{2} - 1)^{2}. Hence g'(0) = 0. Also g'(t) is positive for t just less than 0 and negative for t just greater than 0, so it is a point of inflection.

*Comment. This seems to be an entirely routine exercise in differentiation.*

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002