# directional derivative, derivation of

Let $f(x,y)$ be a function where $x=x(t)$ and $y=y(t)$. Let $\vec{r}=a\hat{\mathbf{i}}+b\hat{\mathbf{j}}$ be the vector in the desired direction. The line through this vector is given parametrically by:

$\displaystyle x=x_{0}+at;\quad y=y_{0}+bt$

The derivative of $f$ with respect to $t$ is as follows:

$\displaystyle\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}\frac{% dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$

But from the equation of the line, we know that $\frac{dx}{dt}=a$ and $\frac{dy}{dt}=b$ so the derivative becomes:

$\displaystyle\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}a+% \frac{\partial f}{\partial y}b=\nabla f\cdot\vec{r}$

Title directional derivative, derivation of DirectionalDerivativeDerivationOf 2013-03-22 15:25:22 2013-03-22 15:25:22 apmc (9183) apmc (9183) 7 apmc (9183) Derivation msc 26B12 msc 26B10