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\author{Kyle Spaans}
\date{June 1, 2009}
\title{Calculus 3 Lecture Notes}
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\section*{Lecture 12 -- Directional Derivatives}
Recalling the definition of a directional derivative, if $f$ has continuous
partials at $\vec{a}$, then
\[ D_{\vec{u}}f(\vec{a}) = \nabla f(\vec{a}) \cdot \vec{u} \]
\paragraph*{Example 1}
Given $f(x,y) = 2x^2 + 3y^2$ calculate the rate of change of $f(x,y)$ at point
$(2,1)$ in direction of point $(5,5)$. We need the gradient and the direction.
We can use the formula given above. We can see by inspection that the partials
are continuous. So we compute the gradient and the direction:
$\nabla f = (f_x, f_y) = (4x, 6y)$ so $\nalba f(2,1) = (8,6)$. We find the
direction by finding the vector from the wanted point to the direction point.
Let $\vec{U} = (3,4) = (5,5) - (2,1)$, $\|\vec{U}\| = 5$ so the unit vector
$\displaystyle vec{u} = \frac{\vec{U}}{\|\vec{U}\|} = (\frac{3}{5},
\frac{4}{5})$
We can plug these in now to get the answer $\frac{48}{5}$.
We can get the same answer using the earlier equation with limits.
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