Directional derivative

[gulsen]

Suppose I have a path $$\vec{r}(x,y)$$ and some vector $$\vec{a}(x,y)$$.
Question is: how do I find the tangential and perpendicular component of a along the path r at a given point?

For tangential component, I'd just take the projection of a on r with dot product (I guess this's correct). But what about perpendicular one?

 

[Hootenanny]

HINT: The dot product of two perpendicular vector lines is zero.

~H

 

[kyle8921]

The directional derivative is a measure of how a function changes in a specific direction, and it can be used to find the tangential and perpendicular components of a vector along a path. To find the tangential component of a vector \vec{a} along a path \vec{r}(x,y), you are correct in using the dot product to project \vec{a} onto \vec{r}. This will give you the component of \vec{a} that is parallel to the path \vec{r}.

To find the perpendicular component, you can use the cross product between \vec{a} and the unit tangent vector of the path \vec{T}(x,y). The unit tangent vector is given by \vec{T}=\frac{\vec{r}'(x,y)}{|\vec{r}'(x,y)|}, where \vec{r}'(x,y) is the derivative of \vec{r} with respect to the path parameter. The cross product will give you a vector that is perpendicular to both \vec{a} and \vec{T}, and you can then use the magnitude of this vector as the perpendicular component of \vec{a} along the path \vec{r}.

In summary, to find the tangential component of \vec{a} along the path \vec{r}, you can use the dot product, and to find the perpendicular component, you can use the cross product with the unit tangent vector of the path. These techniques can be used to analyze the directional changes of a function or vector along a specific path, and they are important tools in many fields of science, including physics, engineering, and mathematics.