Find the rate of change?

Question

Find the rate of change of f, df/dt, along the parametric curve x(t)=2t-1, y(t)=3-2t^2 at t=1.

The answer is -8 and I need to find df/dt(1).

Answer 1

You have to show the fucntion f(x,y) othervise the problem can not be solved. If you know the function this is what you hav to do:

                                     df/dt = (∂f/∂x) (dx/dt) + (∂f/∂y) (dy/dt)    (we used the chain rule)

For t = 1 we have x(1) = 2x1-1=1, y(1) = 3 - 2x1 = 1. Derivatives dx/dt and dy/dt are equal to:

                         dx/dt = 2     dy/dt = -4t.    We also have                dx/dt (1) =2,      dy/dt (1) = -4

Thus                         df/dt (1) = 2 (∂f/∂x) - 4 (∂f/∂y)  

Your steps now are: define the function f = f(x,y), find its derivatives    ∂f/∂x and  ∂f/∂y and calculate them at x = 1 and y = 1 (x, y at t=1).  Good luck!