Find the directional derivative of $f(x,y,z) = x^2yz^3$ along the curve with parametric equations $$\begin{align}x & = e^{-t}, \\ y & = 1 + 2 \sin t, \\ z & = t - \cos t, \end{align}$$ at the point $P$ where $t = 0$.
I know how to find the directional derivative along a vector, but how I can I find it along this curve?
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$\begingroup$It means, along the tangent vector to the curve.
There is a very simple way to do this: substitute $x,y,z$ with the coordinates of the curve, getting: $$ F(t) = f(x(t),y(t),z(t)) = f(e^{-t}, 1+2\sin t, t-\cos t) = ... $$
Now simply take the derivative with respect to $t$!
EDIT Otherwise, take the scalar product between the gradient of $f$, and the tangent vector to the curve at $t=0$.
The tangent vector to the curve is the vector: $$ v= \bigg(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\bigg), $$
so in our case, $(-e^{-t},etc.)$.
If your book (or professor) want you to take a unit tangent vector, divide this by the norm (but I don't think you have to do it, along the curve...ask your teacher).
The gradient is: $$ \nabla f = \bigg(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\bigg), $$
so in our case, $(2xyz^3,etc.)$.
Now take the scalar product of these two vectors, at $t=0$. Which means at the coordinates $(x(0),y(0),z(0))= (1,etc.)$
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