I understand that using a unit vector (of a vector say $\vec{a}$ ) and computing the directional derivative gives the slope (or rate of change of the function) in the direction of the vector.
I have three questions :
- If I use the vector itself rather than it's unit vector what will I get when I compute it's dot product with the gradient of the function? It wouldn't give the slope of the curve(formed by the slicing of the function with the plane containing the vector $\vec{a}$ ), would it?
Note: The function is scalar.
Also going by it's formal definition:
$\displaystyle \nabla _{\mathbf {v}}{f}({\mathbf {x}})=\lim _{h\rightarrow 0}{\frac {f({\mathbf {x}}+h{\mathbf {v}})-f({\mathbf {x}})}{h}}$
$\mathbf {v}$ is a vector
Quoting from Wikipedia
This definition is valid in a broad range of contexts, for example where the norm of a vector (and hence a unit vector) is undefined.
- What does that mean?
Also quoting from Wikipedia:
If the function f is differentiable at x, then the directional derivative exists along any vector v, and one has
$\displaystyle \nabla _{\mathbf {v} }{f}({\mathbf {x} })=\nabla f({\mathbf {x} })\cdot {\mathbf {v} }$
Intuitively, the directional derivative of f at a point x represents the rate of change of f, in the direction of v with respect to time, when moving past x.
- Why is it mentioned with respect to time isn't it with respect to the change in x (or/and y ) in the direction of the vector ?