Let the directional derivative of a function $f(x,y)$ at a point $P$ in the direction of $\big(\frac{1}{\sqrt{5}}\,i+\frac{2}{\sqrt{5}}\,j\big)$ be $\frac{16}{\sqrt{5}}$ and $\frac{\partial f}{\partial x}$ evaluated at $P$ be 6. What is the directional derivative in the direction of $i - j$?

Looking up numerous examples, I still can't find out what to do with all the information given.


I'm assuming that you are asking for the directional derivative in the new direction evaluated at the same point. So we are given:

$$\nabla f \cdot \hat{u} \Big|_P = \frac{16}{\sqrt{5}}$$

$$\hat{u} = \begin{pmatrix} \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} \end{pmatrix}$$

$$f_x \Big|_P = 6$$

So using the first piece of information:

$$f_x \Big|_P u_1 + f_y \Big|_P u_2 = \frac{16}{\sqrt{5}}$$

$$(6)\Big( \frac{1}{\sqrt{5}} \Big) + f_y \Big|_P \Big( \frac{2}{\sqrt{5}} \Big) = \frac{16}{\sqrt{5}}$$

$$f_y \Big|_P = 5$$

With this we define the new direction:

$$v = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$$

$$\hat{v} = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{pmatrix}$$

and compute the directional derivative:

$$\nabla f \cdot \hat{v} \Big|_P = f_x \Big|_P v_1 + f_y \Big|_P v_2$$

$$= (6) \Big( \frac{1}{\sqrt{2}} \Big) + (5) \Big( -\frac{1}{\sqrt{2}} \Big)$$

$$= \frac{1}{\sqrt{2}}$$

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$\nabla{f}\cdot (\frac{1}{\sqrt5}, \frac{2}{\sqrt5}) = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) = (6)(\frac{1}{\sqrt5}) + (\frac{\partial f}{\partial y})(\frac{2}{\sqrt5}) = \frac{16}{\sqrt5}$

Thus, you can solve for $\frac{\partial f}{\partial y}$

With this, you just need to compute $\nabla{f}\cdot (\frac{1}{\sqrt2},\frac{-1}{\sqrt2})$ at $P$, which is certainly possible, because you now know $\frac{\partial f}{\partial y}$ at $P$

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