Problem on finding a directional derivative
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Let $f(x,y) = x^2/y$. Compute the directional derivative of $f$ at $(1,2)$ in the direction of $\mathbf{i} + 3\ \mathbf{j}$.
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 apply this formula, we find $\bfv$ and compute $\nabla f(x,y)$ at $(1,2)$.
Length of a vector
We identify $\bfv = \bfi + 3 \bfj$, which has length $|\bfv| = \sqrt{10}$.
Definition of the gradient
Now, we compute $\nabla f(x,y)$ and evaluate it at $(1,2)$: $$\begin{align}\nabla f(x,y) &= \frac{2x}{y} \ \mathbf{i} -\frac{x^2}{y^2} \ \mathbf{j} \\ \nabla f(1,2) &= \mathbf{i} - \frac{1}{4} \mathbf{j} \end{align}$$
Applying the formula for directional derivative, we get
Direct computation of dot product
$$D_{(\bfi + 3 \bfj)} f = \left( \bfi - \frac{1}{4} \bfj \right) \cdot \frac{\bfi + 3 \bfj}{\sqrt{10}} = \frac{1}{4 \sqrt{10}}.$$
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