Question

Find the curvature of the curve in R^2 which is given parametrically by
x(t) =from0to t ∫sin u^2 /2 du, y(t) =from 0 to t ∫ cos u^2 /2 du
where t > 0.

Answer 1

Recall that
k = |x' y" - x" y'|/((x')^2 + (y')^2)^(3/2)
Since x and y are expressed as integrals, we just apply the (2nd) Fundamental Theorem of Calculus to get
x' = 1/2 sin(t^2)
x" = t cos(t^2)
y' = 1/2 cos(t^2)
y" = -t sin(t^2)
Now plug that into k, and we get
|(1/2 sin(t^2))(-t sin(t^2))-(t cos(t^2))(1/2 cos(t^2))|/((1/2 sin(t^2))^2+(1/2 cos(t^2))^2)^(3/2)
= (t/2)/(1/4)
= 2t