II-17 Find the directional derivative of the function at the...

II-17 Find the directional derivative of the function at the given point in the direction of the vector $\mathbf{v}$ .
$$f(x, y)=\ln \left(x^{2}+y^{2}\right), \quad(2,1), \quad \mathbf{v}=\langle- 1,2\rangle$$

Key Concepts

Directional Derivative

The directional derivative measures the rate at which the function changes at a given point in a specified direction. It is computed as the dot product of the gradient vector at that point and a unit vector in the specified direction, combining both the magnitude and direction of change.

Gradient Vector

The gradient vector of a function consists of its partial derivatives with respect to each variable, arranged as a vector. It points in the direction of the greatest rate of increase of the function and its magnitude represents the rate of the steepest ascent.

Unit Vector in a Given Direction

A unit vector in a specified direction is obtained by normalizing the direction vector, which means dividing the vector by its magnitude. This ensures that the direction is preserved while its length is scaled to one, making it suitable for use in computing directional derivatives.

Dot Product

The dot product is an algebraic operation that takes two vectors and returns a scalar. It is used in the context of directional derivatives to combine the gradient vector and a unit direction vector, effectively projecting the gradient onto the given direction to determine the rate of change of the function.

Transcript

00:01 So for this problem, we are given that we have a function f of xy is equal to the natural log of x squared plus y squared.

00:07 We have a point 2 comma 1.

00:09 We want to find the directional derivative in the direction of v here, which is negative 1 comma 2 in that direction.

00:16 To use our directional derivative formula, we need two things.

00:20 The gradient of our function f at that point, which we can calculate in a little bit.

00:25 And then we also need this direction, you.

00:28 However, this directional you must be, a unit vector.

00:32 The magnitude of this vector must be one.

00:35 Otherwise, we cannot use this formula.

00:38 So we see here that this is not going to have the magnitude of one.

00:42 You see that the components are greater than ones that can't have a magnitude of less than one.

00:47 And so because of this, to create a unit vector, what we do is we take the vector, we take the vector and we divide it by the magnitude.

00:56 So in this case, we have negative 1, 2, and we divide it.

01:01 By the magnitude, which is just equal to the sum of the squares of the components and the square root of that.

01:07 So we get negative 1 squared plus 2 squared and the square of that.

01:14 And so we get negative 1 comma 2 divided by the square root of 5.

01:19 And this is our unit vector.

01:22 We can now utilize this unit vector for our problem.

01:26 We must always look at this before we even start to use this formula.

01:30 The next thing we want to do is compute the gradient of our vector v, of our function b.

01:39 So this means we need to compute the derivative with respect to x and the derivative with respect to y...

Please give Ace some feedback