Problem on finding a directional derivative
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Find the directional derivative of $f(x,y) = 10 - x^2 -2y^2$ at the point $(x,y) = (1,-1)$ and in the direction of $-2 \mathbf{i} + 4 \mathbf{j}$.
Solution
Recall that
Directional derivative and gradient
The directional derivative of $f(x,y)$ in the direction of the vector $\mathbf{v}$ can be computed by
$$D_\mathbf{v} f = \nabla f(x,y) \cdot \frac{\mathbf{v}}{|\mathbf{v}|}$$
To use this formula, we need to compute $\bfv$ and the gradient $\nabla f(1,-1)$.
Length of a vector
We identify that $\bfv = -2 \bfi + 4 \bfj$, which has length $$|\bfv| = \sqrt{(-2)^2 + 4^2} = \sqrt{20}.$$
Definition of the gradient
The gradient is \begin{align}
\nabla f(x,y) &= -2 x \ \mathbf{i} - 4 y \ \mathbf{j},\\
\nabla f(1,-1) &= -2 \ \mathbf{i} + 4 \ \mathbf{j}.
\end{align}
Direct computation of dot product
The directional derivative is thus $$(-2 \bfi + 4 \bfj) \cdot \frac{-2 \bfi + 4 \bfj}{\sqrt{20}} = \sqrt{20}.$$
Related topics
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$\nabla f(x,y) = \partial_x f(x,y) \mathbf{i} + \partial_y f(x,y) \mathbf{j}$
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