A particle moves in the Euclidean plane. At time t (taking all real values) its coordinates are x = t3 - t and y = t4 + 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 = 16t6 + 9t4 + 8t3 - 6t2 + 2. Let this be f(t). We have f '(t) = 12t(8t4 + 3t2 + 2t - 1). So f '(t) = 0 at t = 0. Also for t small (positive or negative), 8t4 + 3t2 + 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 = (4t3 + 1)/(3t2 - 1). Let this be g(t). Then g'(t) = 6t(2t3 - 2t - 1)/(3t2 - 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