Question: Find the directional derivative of f=x^2·y·z^3 along the curve x = e^-u; y = 2 sin u + 1; z = u - cos u at the point P where u = 0
My working:
At u=0, x=1, y=1, z=-1 so let u = (1,1,-1).
Know that D_u f(x) = ∇f(x) · u = (2x·y·z^3, x^2·z^3, 3x^2·y·z^3) · (1,1,-1) = (2x·y·z^3, x^2·z^3, -3x^2·y·z^2).
At u=0, x=(1,1,-1) so D_u f(x)=(2·1·1·-1, 1·-1, -3·1·1·1)=(-2,-1,-3)
However, I'm not sure if this is correct as I don't know whether I'm meant to substitute u into f to find the derivative at a specific point or not?
Hint:
$$\lim_{h\to0}\frac{f(x(u+h),y(u+h),z(u+h))-f(x(u),y(u),z(u))}h=\lim_{h\to0}\frac{\left(f_x\dfrac{dx}{du}+f_y\dfrac{dy}{du}+f_z\dfrac{dz}{du}\right)h}h \\=\nabla f\cdot\left(\dfrac{dx}{du},\dfrac{dy}{du},\dfrac{dz}{du}\right).$$