91Ó°ÊÓ

The gradient is a vector that points in the direction of the greatest rate of increase of a function. It combines all the partial derivatives of a function into a vector form. For a multivariable function like \[f(x, y)\], the gradient, denoted as \(abla f\), has components consisting of the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\).
This means that if you want to determine how a function is changing at a particular point, the gradient gives the most direct line of steepest ascent.
In the given exercise, we calculate the gradient of the function \(f(x, y) = \frac{x}{x^{2}+y^{2}}\) by finding its partial derivatives and then assembling them into the vector form:

• \(\frac{\partial f}{\partial x} = \frac{y^2 - x^2}{(x^2 + y^2)^2}\)
• \(\frac{\partial f}{\partial y} = -\frac{2xy}{(x^2 + y^2)^2}\)

Thus, the gradient \(abla f\) is \(\left\langle \frac{y^2 - x^2}{(x^2 + y^2)^2}, -\frac{2xy}{(x^2 + y^2)^2} \right\rangle\). By evaluating it at the point \((1, 2)\), we find the specific change in function at that location.

Partial derivatives are an essential concept in understanding the gradient because they represent how a function changes as only one of its variables changes while the others are held constant.
For example, with \(f(x, y)\), partial derivatives like \(\frac{\partial f}{\partial x}\) reveal how the function changes if \(x\) changes but \(y\) remains the same.
The same logic applies to \(\frac{\partial f}{\partial y}\).
Understanding the role of partial derivatives, you can break down complex multivariable functions and comprehend their behavior one variable at a time.

• In our exercise, we calculated the partial derivative with respect to \(x\) as \(\frac{y^2 - x^2}{(x^2 + y^2)^2}\).
• The partial derivative with respect to \(y\) was \(-\frac{2xy}{(x^2 + y^2)^2}\).

These were then used to form the components of the gradient, helping us understand the function's rate of change in each direction.

A unit vector is a vector that has a magnitude of 1, and it is useful for defining direction without affecting magnitude.
To find a unit vector in a given direction, you divide each component of a vector by the vector's magnitude.
In directional derivatives, we use a unit vector to specify the direction in which we want to look at a function's rate of change.
For our problem, the given vector was \(\mathbf{v} = \langle 3, 5 \rangle\).

• First, we calculated its magnitude: \(\|\mathbf{v}\| = \sqrt{3^2 + 5^2} = \sqrt{34}\).
• Then, the unit vector \(\mathbf{u}\) was found by dividing each component by this magnitude: \(\mathbf{u} = \left\langle \frac{3}{\sqrt{34}}, \frac{5}{\sqrt{34}} \right\rangle\).

This unit vector is crucial for calculating the directional derivative across a specified direction.

The dot product is an operation that takes two equal-length sequences of numbers (usually vectors) and returns a single number.
In the context of directional derivatives, it measures how much one vector "projects" onto another.
The dot product between two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is calculated as \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + \ldots + a_n b_n\).

• In this exercise, the dot product was used to compute the directional derivative. This is done by taking the dot product of the gradient \(abla f(1, 2) = \left\langle \frac{3}{25}, -\frac{4}{25} \right\rangle\) with the unit vector \(\mathbf{u} = \left\langle \frac{3}{\sqrt{34}}, \frac{5}{\sqrt{34}} \right\rangle\).

Calculating the dot product here results in \(-\frac{11}{25\sqrt{34}}\), representing the directional derivative, thus determining how rapidly the function changes at point \( (1, 2) \) in the direction specified by \(\mathbf{v}\).