Problem on computing a directional derivative
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Compute the directional derivative of $f(x,y) = x^2 + y^2$ at $(1,1)$ in the direction of $2 \mathbf{j}$.
Solution
Recall that:
Directional derivative and gradient
The directional derivative of $f(x,y)$ in the direction of $\mathbf{v}$ is:
$$D_\mathbf{v} f = \nabla f(x,y) \cdot \frac{\mathbf{v}}{|\mathbf{v}|}$$
To compute the directional derivative, we first compute $\nabla f(1,1)$ and $\mathbf{v}/|\mathbf{v}|$.
Recall that
Definition of the gradient
The definition of the gradient of $f$ is $$\nabla f(x,y) = \partial_x f(x,y) \mathbf{i} + \partial_y f(x,y) \mathbf{j}$$
We now compute the gradient of $f$ at $(1,1)$:
\begin{align}
\nabla f(x,y) &= 2 x \ \mathbf{i} + 2 y \ \mathbf{j}\\
\nabla f(1,1) &= 2 \ \mathbf{i} + 2 \ \mathbf{j}\\
\end{align}
To compute $\mathbf{v}/|\mathbf{v}|$, we identify $\mathbf{v} = 2 \mathbf{j}$, which has length $| \mathbf{v} | = 2$. Hence, $$\frac{\mathbf{v}}{|\mathbf{v}|} = \mathbf{j}$$
Putting these computations together, we get
\begin{align}D_\mathbf{v} f &= (2 \mathbf{i} + 2 \mathbf{j}) \cdot \mathbf{j} = 2
\end{align}
The directional derivative is $2$.
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