Question

Gradient of a distance function Let \((a, b)\) be a given point in \(\mathbb{R}^{2}\), and let \(d=f(x, y)\) be the distance between \((a, b)\) and the variable point \((x, y)\) a. Show that the graph of \(f\) is a cone. b. Show that the gradient of \(f\) at any point other than \((a, b)\) is a unit vector. c. Interpret the direction and magnitude of \(\nabla f\).

Short answer

Answer: The gradient of the distance function is given by \(\nabla f(x, y) = \frac{1}{d}((x - a), (y - b))\), where d is the distance between points (a, b) and (x, y). The direction of the gradient vector represents the straight line from the fixed point (a, b) to the variable point (x, y), while its magnitude is always equal to 1, indicating a constant rate of change in the direction of the gradient.

Step-by-step solution

Distance Function

Let (a, b) and (x, y) be two points in \(\mathbb{R}^{2}\). We can find the distance between these two points using the Euclidean distance formula, the distance function d=f(x, y) can be given as: $$ f(x, y) = d = \sqrt{(x - a)^2 + (y - b)^2} $$

Square the Distance Function

To simplify the problem, let's square the distance function, which will get rid of the square root: $$ f^2(x, y) = (x - a)^2 + (y - b)^2 $$

Show the Graph is a Cone

To show that the graph of the squared distance function is a cone, we will rewrite it in the form of the standard equation of a cone: $$ z^2 = (x - a)^2 + (y - b)^2 $$ Here, z = f, which is the distance. This is the equation of a right circular cone with its vertex at the point (a, b, 0) and is symmetrical about the z-axis.

Gradient of the Squared Distance Function

Next, let's find the gradient of the squared distance function, f²(x,y). The gradient is a vector that contains the partial derivatives with respect to x and y: $$ \nabla f²(x, y) = \left(\frac{\partial f²(x, y)}{\partial x}, \frac{\partial f²(x, y)}{\partial y}\right) = \left(2(x - a), 2(y - b)\right) $$

Gradient of the Distance Function

To find the gradient of the distance function itself, we can use the chain rule by multiplying the gradient of the squared distance by the derivative of the square root function \(\frac{d\sqrt{f^2}}{df}\) : $$ \nabla f(x, y) = \frac{1}{2 \sqrt{f^2(x, y)}}\nabla f^2(x, y) = \frac{1}{2f(x, y)}(2(x - a), 2(y - b)) = \frac{1}{d}((x - a), (y - b)) $$

Show the Gradient is a Unit Vector

To show that the gradient is a unit vector, we have to prove that the magnitude of the gradient vector is equal to 1 for any point (x, y) different from (a, b): $$ ||\nabla f(x, y)|| = \frac{1}{d}\sqrt{(x - a)^2 + (y - b)^2} = \frac{d}{d} = 1 $$ This proves that the gradient of the distance function is a unit vector at any point (x, y) different from (a, b).

Interpret the Direction and Magnitude of the Gradient

The direction of the gradient vector is given by the vector from the fixed point (a, b) to the variable point (x, y). This means that the gradient always points in the direction of the straight line that joins the two points. The magnitude of the gradient vector is always equal to 1. This indicates that the gradient is a unit vector, and thus the rate of change of the distance function in the direction of the gradient is constant. This constant rate is independent of the location of the variable point (x, y) or the fixed point (a, b).

Key concepts

Euclidean Distance

The Euclidean distance is the length of the straight line joining two points in Euclidean space. In a two-dimensional plane, the distance between points \textbf{(a, b)} and \textbf{(x, y)} is calculated using the formula:
\[ f(x, y) = \sqrt{(x - a)^2 + (y - b)^2} \]
This formula is rooted in the Pythagorean theorem and represents the most common and intuitive way of measuring the distance between points on a plane. Students embarking on physical or data science disciplines learn this foundational concept early, as it is essential for understanding a multitude of more complex theories and applications.

Partial Derivatives

Partial derivatives are the rates at which a multivariable function changes as only one of its variables changes while the others stay fixed. For a function like the squared distance \textbf{f²(x, y)}, the partial derivatives are represented as:
\[ \frac{\partial f²(x, y)}{\partial x} \text{ and } \frac{\partial f²(x, y)}{\partial y} \]
When you calculate these, you treat \textbf{y} as a constant when differentiating with respect to \textbf{x} and vice versa. Understanding this selective differentiation process is crucial for advancing in calculus, optimization problems, and many fields where functions have numerous variables, such as economics, physics, and engineering.

Unit Vector

A unit vector is a vector whose length or magnitude is 1. It represents direction without specifying magnitude and is often used to normalize other vectors. In the context of the exercise, the gradient of the distance function, after simplification, is shown to be a unit vector:
\[ ||abla f(x, y)|| = 1 \]
This is significant in physics for representing a vector direction, such as velocity or force, where only the orientation is of interest, not the quantity.

Chain Rule

The chain rule is a formula for calculating the derivative of a composite function. In our context, it was used to relate the derivatives of the squared distance function and the square root function to find the gradient (\textbf{abla f(x, y)}). The chain rule is expressed as:
\[ \frac{d}{dx}[\sqrt{f²(x, y)}] = \frac{1}{2 \sqrt{f²(x, y)}} \cdot \frac{d}{dx}[f²(x, y)] \]
Understanding the chain rule is essential for solving calculus problems involving composite functions. It is a tool that links simple derivatives to form solutions for more complex functions, allowing for a deeper understanding of how variables and their changes interconnect.