I am given a vector function <2cos(t) + cos(2t), 2sin(t) + sin(2t), 0>

Question

(in other words) : (2cos(t) + cos(2t))i + (2sin(t) + sin(2t)j + 0k

Compute the slope dy/dx and concavity d^2x/(dx^2) at t = pi/3.

I understand that finding the slope of a vector is possible by using dy/dx such that (dy/dt)/(dx/dt) and finding the concavity is by finding the derivative of (dy/dt) then divide (dy/dt)/(dx/dt) but what I'm confused here is when exactly do I plug t in? This part is really confusing to me and I would like some guidance on to actually solve for both the slope and concavity.

Any help is greatly appreciated!

Answer 1

Forgot the closing parenthesis:

(2cos(t) + cos(2t))i + (2sin(t) + sin(2t))j + 0k

Answer 2

Any ideas?