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

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

Key Concepts

Unit Vector

A unit vector is a vector with a magnitude of one and is used to denote a direction without scaling the size of measurements. In the context of finding a directional derivative, converting a given vector to a unit vector is crucial, since the directional derivative measures the rate of change per unit length in the specified direction.

Maximum Increase and Rate of Change

The maximum rate of increase of a function at a point is given by the magnitude of the gradient, and this maximum is achieved in the direction of the gradient vector. Understanding this concept helps in identifying the direction in which the function increases most rapidly and in quantifying how steep that increase is at a given point.

Directional Derivative

The directional derivative represents the rate at which a function changes as one moves in a specified direction. It is computed as the dot product of the gradient vector and a unit vector in the direction of interest. This concept generalizes the idea of partial derivatives, which only consider changes along the coordinate axes, to any arbitrary direction in the domain.

Gradient Vector

The gradient vector of a multivariable function is a vector composed of its partial derivatives with respect to all variables. It points in the direction of the steepest ascent of the function and its magnitude quantifies the rate of increase in that direction. The process often involves computing the partial derivatives and then evaluating them at a specific point if needed.

Transcript

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

00:17 We know that the gradient of f, that is del f, is equal to fx, fy, fz.

00:39 That is the partial derivative of f with respect to x, then partial derivative of f with respect to x, to y and then partial derivative of f with respect to z so this is equal to fx is 2x divided by 1 plus x square plus y square plus z squared as f is given to be natural log of 1 plus x square plus y square plus z square its derivative is first we find the derivative of a length that is one divided by x squared plus y square plus z square and then by chain rule it will be multiplied by the derivative of one plus x square plus y square plus z square which is two x so it is two x divided by 1 plus x square plus y square plus z square.

01:53 Similarly, fy is 2y divided by 1 plus x square plus y square plus z square and f z is 2 z divided by 1 plus x 2x square plus y square plus z square.

02:52 Now we will write it in the form of unit vectors i j so, this will be, first we can take common 2 divided by 1 plus x square plus y square plus x squared plus x squared plus z squared from all the three components.

03:35 So it will be 2 divided by 1 plus x square plus y square plus z square and multiplied by x, y, z.

03:45 So that can be written as xi plus yj plus z k.

04:16 This is the required gradient.

04:23 Now we have to evaluate this at the given point 1 -1 minus 1.

04:31 That is we have to find del f at 1 -1 -1 -1.

04:46 So it will be equal to 2 divided by 1 plus x square will also be 1, y square will also be 1 and z squared that is minus 1 square will also be 1.

05:05 So denominator will be 1 plus 1 plus 1 plus 1 that is 4 multiplied by x i that is 1 i.

05:21 So i plus 1 j that is j plus minus 1 into k that is minus k so this is equal to 1 divided by 2 i plus j minus k now the next question asked is to find the unit vector in the direction of maximum increase of f at the given point we know that the direction of maximum increase of f is in the direction of the gradient of f, that is del f.

06:40 That is, we have to find the unit vector of del f.

06:46 Unit vector of del f can be found as del f divided by magnitude of del f.

07:03 We will first find the magnitude of del f, which is a single.

07:13 Equal to under the root the coefficients of ijk with their squares, that is squares of the coefficients of ijk...

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