# directional derivative, derivation of

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

$x={x}_{0}+at;y={y}_{0}+bt$

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

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

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

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

Title | directional derivative^{}, derivation of |
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Canonical name | DirectionalDerivativeDerivationOf |

Date of creation | 2013-03-22 15:25:22 |

Last modified on | 2013-03-22 15:25:22 |

Owner | apmc (9183) |

Last modified by | apmc (9183) |

Numerical id | 7 |

Author | apmc (9183) |

Entry type | Derivation |

Classification | msc 26B12 |

Classification | msc 26B10 |