Problem 29 - Algebra and Trigonometry with Analytic Geometry

Question

A thin flat plate is situated in an \(x y\)-plane such that the density \(d\) (in \(\mathrm{lb} / \mathrm{ft}^{2}\) ) at the point \(P(x, y)\) is inversely proportional to the square of the distance from the origin. What is the effect on the density at \(P\) if the \(x\) - and \(y\)-coordinates are each multiplied by \(\frac{1}{3} ?\)

Step-by-Step Solution

Answer

The density increases by 9 times when the coordinates are scaled by \(\frac{1}{3}\).

Step 1: Understanding the density equation

The problem states that the density \(d\) is inversely proportional to the square of the distance from the origin \((0,0)\). The distance of a point \((x, y)\) from the origin is given by the formula \(r = \sqrt{x^2 + y^2}\). Since density is inversely proportional to the square of this distance, we can define \(d\) as \(d = \frac{k}{(x^2 + y^2)}\), where \(k\) is a constant of proportionality.

Step 2: Initial density expression at P(x, y)

Initially, at \(P(x, y)\), the density is given by the equation \(d = \frac{k}{(x^2 + y^2)}\). Here, the denominator \((x^2 + y^2)\) represents the square of the distance of point \(P\) from the origin.

Step 3: Calculating the new coordinates

If the coordinates \(x\) and \(y\) are each multiplied by \(\frac{1}{3}\), the new coordinates of the point become \(\left(\frac{x}{3}, \frac{y}{3}\right)\).

Step 4: Substituting the new coordinates into the density equation

Substitute the new coordinates into the density formula: \(d' = \frac{k}{\left(\left(\frac{x}{3}\right)^2 + \left(\frac{y}{3}\right)^2\right)} = \frac{k}{\left(\frac{x^2}{9} + \frac{y^2}{9}\right)}\).

Step 5: Simplifying the new density equation

Combine the terms in the denominator: \(d' = \frac{k}{\left(\frac{x^2 + y^2}{9}\right)} = \frac{k \cdot 9}{x^2 + y^2}\). Notice that this expression can be rewritten as \(d' = 9 \cdot \frac{k}{x^2 + y^2}\).

Step 6: Comparing original and new densities

The original density was \(d = \frac{k}{x^2 + y^2}\). The new density \(d' = 9d\). This shows that the density at point \(P\) when the coordinates are scaled down is 9 times the original density.

Key Concepts

Density Function

In math and physics, a density function often serves to show the distribution of some quantity across a specified domain. In this context, the density at any point on a plate can be thought of as how much mass is present per unit area at that point. This specific problem involves a density that varies inversely as the square of the distance from the origin.
This means:

• As you move further from the origin, the density decreases.
• The distance between each point \( P(x, y) \) and the origin \( (0, 0) \) is vital in determining density.

The equation capturing this is \( d = \frac{k}{x^2 + y^2} \), emphasizing inverse proportionality. Inverse proportionality implies that as one quantity increases, the other decreases. When grappling with density problems like this, it's critical to understand how altering distances affects the overall density, as demonstrated by the given exercise.

Distance Formula

Understanding the relationship between two points in a plane is key, and the distance formula provides a handy measuring tool. It helps us calculate how far apart two locations are, specifically from a central point like an origin. For two points in a plane:

• The origin is \( (0, 0) \).
• The target point is \( (x, y) \).

Using the distance formula, the distance \( r \) from \( (0, 0) \) to \( (x, y) \) is captured by \( r = \sqrt{x^2 + y^2} \). This formula draws on the Pythagorean theorem, using a right triangle to find distances in a plane. By determining \( x^2 + y^2 \), we are actually finding the squared distance in cartesian coordinates. Knowing this helps guide you through boundary changes to \( (x, y) \), especially when transformations are applied.

Coordinate Transformation

A coordinate transformation allows you to see how changes in position affect other elements, like density, in this educational exercise. By multiplying each coordinate by a fraction like \( \frac{1}{3} \), the transformation becomes focused not just on mere geometric shifts, but how such movements impact related mathematical values. Here's what happens:

• The original point changes from \( (x, y) \) to \( \left( \frac{x}{3}, \frac{y}{3} \right) \).
• This rescaling affects the distance of the point from the origin.
• Thus, as the distance changes, the inverse squared function for the density outputs different densities.

Gradually swapping out for transformed coordinates is useful, as this is frequent in physics and mathematics for simplifying problems or rectifying perspectives.

Proportionality Constant

A proportionality constant serves the role of a balancing factor. It's the number outside a direct or inverse proportionality equation that allows comparisons. In this problem, we denote it by \( k \), a constant unchanging regardless of the situation.
In inverse relationships, such as this with density taking the form \( d = \frac{k}{x^2 + y^2} \), \( k \) scales the effect of inverse proportion. Consider this:

• It connects two variables that don't change together at a fixed rate.
• Helps in appreciating density's rate of change with distance squared.
• Remains constant even when the \( x \) and \( y \) coordinates are transformed.

This is important because, in many mathematical experiences, constants like \( k \) simplify our understanding and allow us to make predictions by maintaining consistency.