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

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. $$f(x, y, z)=\frac{x-z}{y-z} ; P(3,2,-1) ;\left\langle\frac{1}{3}, \frac{2}{3},-\frac{2}{3}\right\rangle$$

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.
$$f(x, y, z)=\frac{x-z}{y-z} ; P(3,2,-1) ;\left\langle\frac{1}{3}, \frac{2}{3},-\frac{2}{3}\right\rangle$$

Key Concepts

Gradient

The gradient of a scalar function is a vector that consists of its partial derivatives with respect to each variable. It points in the direction of the greatest rate of increase of the function, and its magnitude gives the rate of increase in that direction. Evaluating the gradient at a specific point allows us to understand how the function is changing locally around that point.

Directional Derivative

The directional derivative measures the rate at which a function changes at a point in a specified direction. It is computed by taking the dot product of the gradient and a unit vector that points in the direction of interest. This concept quantifies how the function increases or decreases when moving in that particular direction.

Unit Vector

A unit vector is a vector with a magnitude of one that indicates direction. In the context of directional derivatives, it is essential to normalize any given direction vector so that it properly represents only the direction and does not affect the computed rate of change due to its length.

Maximum Rate of Change

The maximum rate of change of a function at a point is achieved in the direction of the gradient. The magnitude of the gradient at that point represents this maximum rate. This concept is critical when determining the steepest ascent or descent on the surface described by the function.

Transcript

00:05 In this problem, it is asked to compute the gradient of the given function f and then evaluate it at the given point p.

00:18 Gradient of f, that is del f, is equal to fx, f, f, z, that is partial derivative of f with respect to x, partial derivative of f with respect to y, and partial derivative of f with respect to y, and partial derivative of f, fx, with respect to z.

00:55 The given function f is x minus z divided by y minus z so fx will be 1 divided by y minus z f y will be minus x minus x divided by y minus z square and f x will be x minus y divided by y minus z square this can be written as we can take one divided by y minus z common outside so this will be i hat minus x minus z divided by y minus z square no we have taken y minus z common outside so this will be only y minus z j head plus x minus y divided by y minus z k hat so this is our required gradient now we have to evaluate it at the given point p 3 2 minus 1 so we will substitute x equal to 3 y equal to 2 and z equal to minus 1 so that delf at 3 2 minus 1 is equal to 1 divided by y minus z, that is 1 divided by 2 minus 1.

04:19 So 1 divided by 2 plus 1, that is 3.

04:26 I had minus x minus z means 3 minus minus 1.

04:36 So 3 plus 1, 4 divided by y minus z is 3.

04:44 J head plus x minus y means 3 minus 2 that is 1 divided by y minus z is 3 that is 1 divided by 3 k hat that is del f at the given point is 1 divided by 3 common outside multiplied by i hat minus 4 divided by 3 j hat plus 1 divided by 3 j hat plus 1 divided by 3 k hat now, the next question asked is to find the unit vector in the direction of maximum increase of f at p.

05:39 As we know, the direction of maximum increase of f is in the direction of del f.

05:48 That is, we have to find the unit vector in the direction of del f.

05:53 So, unit vector is equal to del f divided by the magnitude of del f.

06:04 To find this, first we will find the value of magnitude of del f.

06:14 That is magnitude of del f is equal to under the root, coefficient of i head, that is 1 by 3, it's square.

06:35 So it is 1 by 9.

06:37 Coefficient of j head is minus 4 by 9 so its square is 16 by 81 plus coefficient of k head is 1 by 9 and so its square is 1 by 81 so this is equal to under the root 26 by 81 and so this is this is equal to 26 by root 26 by 9.

07:37 This is the magnitude of del f.

07:41 So now we will find the required unit vector.

07:46 Our del f is 1 by 3 i minus 4 by 3 j 1 by 3k divided by magnitude of del f we have found to be root 26 divided by by 9.

08:43 So this will be 3 ones are 3 and 3 3 3s are 9.

08:51 So this is equal to 3 divided by root 26 i head minus 4 by 3 j head plus 1 by 3 k head...

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