Consider the following functions f, points P, and unit vectors 𝐮 a. Compute the gradient of f an

Consider the following functions f, points P, and unit...

Consider the following functions $f,$ points $P,$ and unit vectors $\mathbf{u}$ a. Compute the gradient of $f$ and evaluate it at $P$. b. Find the unit vector in the direction of maximum increase of $f$ at $P$. c. Find the rate of change of the function in the direction of maximum increase at $P$. d. Find the directional derivative at $P$ in the direction of the given vector. $$f(x, y, z)=\frac{x-z}{y-z} ; P(3,2,-1) ;\left\langle\frac{1}{3}, \frac{2}{3},-\frac{1}{3}\right\rangle$$

Consider the following functions $f,$ points $P,$ and unit vectors $\mathbf{u}$
a. Compute the gradient of $f$ and evaluate it at $P$.
b. Find the unit vector in the direction of maximum increase of $f$ at $P$.
c. Find the rate of change of the function in the direction of maximum increase at $P$.
d. Find the directional derivative at $P$ in the direction of the given vector.
$$f(x, y, z)=\frac{x-z}{y-z} ; P(3,2,-1) ;\left\langle\frac{1}{3}, \frac{2}{3},-\frac{1}{3}\right\rangle$$

Key Concepts

Gradient

The gradient is a vector that represents the collection of all first-order partial derivatives of a multivariable function. It points in the direction of the greatest rate of increase of the function, and its components give the rate of change of the function in the corresponding coordinate directions. Computing the gradient involves taking partial derivatives with respect to each variable and evaluating them if a specific point is given.

Directional Derivative

The directional derivative measures the rate of change of a function in a specified direction. It is computed as the dot product of the gradient vector and a unit vector in the desired direction. This concept generalizes the idea of a derivative to arbitrary directions in the domain of the function.

Unit Vector in the Direction of Maximum Increase

The unit vector in the direction of maximum increase is obtained by normalizing the gradient vector. Since the gradient points in the direction of the steepest ascent, dividing it by its magnitude gives a unit vector that retains the direction but has a length of one, ensuring that the directional derivative correctly reflects the rate of change per unit distance.

Maximum Rate of Change

The maximum rate of change of a function at a point is given by the magnitude of the gradient vector at that point. This value represents how rapidly the function increases in the direction of the steepest ascent, and it is achieved when moving in the direction of the normalized gradient vector.

Transcript

00:01 For this problem, in part a, we'll note that we are given the function f of x, y, y, z, equals x minus z, divided by y minus z, and for part a, we are asked to compute the gradient of f and evaluate it at the point 3 to negative 1.

00:15 So the gradient at a point x, y, z will be, well, we have partial derivative with respect to x in the first component, so that will be just 1 over y minus z in the first component.

00:28 Then it will be the partial derivative with respect to y is the second component, which will be z minus x, that minus x divided by y minus z squared, is the second component.

00:44 And then for the third component, we take the partial derivative with respect to z, which in its most simplified form will be x minus y.

00:53 Oops, it's a little bit sketchy there, x minus y divided by y minus z all squared.

01:00 Evaluating this at the point 3 to negative 1 we'll get that the gradient will be 1 over 3 negative 9 1 over 9 then for part b we are asked to find the unit vector in the direction of maximum increase of f at p so the direction of maximum increase is going to be the same thing as the direction of the gradient but we are asked for the unit vector so i'll say that our unit vector here, we can just take our gradient and divide it by its magnitude.

01:42 Let's see here.

01:44 So we'd take the gradient, and we're dividing by the square root of 1 over 3 squared, so 1 over 9, plus 4 over 9 squared, so that would be plus 16 over 9 squared, would be 81.

01:59 Then we'd have plus 1 over 81, which we can simplify into 1 over 1 over, oops, write this as 1 over the square root of 26 times the vector 3 minus 4 1...

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