Direction of maximum increase and rate of change

[Hiche]

Homework Statement

Homework Equations

..

The Attempt at a Solution

We find $\nabla f(1,2)$ at point P: the answer is $(-4, 13)$. Now, we know that the directional derivative of f is given by: $\nabla f(1, 2) |u| \cos\theta$
where $|u| = 1$ and $\cos\theta=1$ (since the direction is maximized at $\theta\ = 0$ or $\cos\theta=1$)
So, letting $u$ be the direction of maximum increase:

$u =$ $\frac {(-4, 13)}{\sqrt{185}}$

Is that true (for the first part)? And how can I start with the second? Any help is appreciated. Oh, and $u$ in the first part is irrelevant of the $u$ in the second part of the question.

 

[mighty2000]

I would first verify the accuracy of the given answer for the gradient at point P. I would use the given function to calculate the gradient at (1,2) and compare it to the given answer of (-4,13). If they match, then the first part is correct.

For the second part, we are looking for the direction of maximum increase, which is the direction in which the directional derivative is maximized. This can be found by maximizing the dot product between the gradient and the unit vector in the direction of interest.

So, using the given gradient at point P, we can set up the following equation:

\nabla f(1,2) \cdot u = |\nabla f(1,2)| \cdot |u| \cdot \cos\theta

Since we want to maximize the directional derivative, we can set \cos\theta = 1 to get:

\nabla f(1,2) \cdot u = |\nabla f(1,2)| \cdot |u|

We know that |\nabla f(1,2)| = \sqrt{185} from the first part, so we can substitute that in:

(-4,13) \cdot u = \sqrt{185} \cdot |u|

Solving for u, we get:

u = \frac{(-4,13)}{\sqrt{185}}

Therefore, the direction of maximum increase at point P is in the direction of u = \frac{(-4,13)}{\sqrt{185}}.