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Two curves are orthogonal if their tangent lines are perpendicular at each point of intersection. Show that the given families of curves are orthogonal trajectories of each other; that is, every curve in one family is orthogonal to every curve in the other family. Sketch both families of curves on the same axes. \( x^2 + y^2 = r^2, ax + by = 0 \)

Short Answer

Expert verified
These curves are orthogonal trajectories as their tangent slopes multiply to -1.

Step by step solution

01

Find the Derivative of First Family

Let's begin with the family of circles described by the equation \( x^2 + y^2 = r^2 \). Differentiating both sides with respect to \( x \), we get \( 2x + 2y \frac{dy}{dx} = 0 \). Simplifying, the derivative is \( \frac{dy}{dx} = -\frac{x}{y} \). This represents the slope of the tangent for the circles.
02

Find the Derivative of Second Family

Now consider the family of lines given by the equation \( ax + by = 0 \). Differentiating both sides with respect to \( x \) yields \( a + b \frac{dy}{dx} = 0 \). Therefore, the derivative is \( \frac{dy}{dx} = -\frac{a}{b} \). This is the slope of the tangent line to the straight line.
03

Check for Orthogonality

Two curves are orthogonal if the product of their slopes is -1. For circles, the slope is \( -\frac{x}{y} \) and for lines, it is \( -\frac{a}{b} \). Calculate the product: \( \left(-\frac{x}{y}\right) \times \left(-\frac{a}{b}\right) = \frac{ax}{by} \). Now, for orthogonal trajectories, this value should be -1. Since \( ax + by = 0 \), the slope product \( \frac{ax}{by} \) indeed equals -1, confirming that they are orthogonal.
04

Sketching the Families of Curves

Sketch the circles centered at the origin with varying radii and the lines passing through the origin. Each line is a diameter of each corresponding circle, and since they intersect at the origin at right angles, they demonstrate the orthogonality. Therefore, these sketches are effective in showing that the families \( x^2 + y^2 = r^2 \) and \( ax + by = 0 \) are orthogonal trajectories.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Derivative
The derivative is a crucial concept in calculus that helps us understand how a function changes at any given point. It tells us the rate of change, or the slope, of the tangent line to the curve at a specific point. In this exercise, the derivatives of the given curves are found to determine the slopes of their tangent lines.
For the family of circles with the equation \( x^2 + y^2 = r^2 \), we differentiate it with respect to \( x \) to find the slope of the tangent. The derivative \( \frac{dy}{dx} = -\frac{x}{y} \) gives us the slope at any point on the circle. Similarly, for the line \( ax + by = 0 \), differentiating gives us \( \frac{dy}{dx} = -\frac{a}{b} \), representing the slope of the tangent to the line.
Understanding these derivatives helps us evaluate the relationship between the two curves by examining their slopes, which is key to finding orthogonal trajectories.
Tangent Lines and Their Slopes
Tangent lines touch curves at exactly one point and are essential for studying the behavior of curves at that specific point. They provide crucial information about the direction and steepness of the curve where they meet.
In our exercise, the slope of the tangent to the circle is \( -\frac{x}{y} \), which varies depending on the position on the circle. For the line, the slope is constant, \( -\frac{a}{b} \).
  • Tangent to the circle: showcases how the slope varies across different points.
  • Tangent to the line: offers a consistent slope across its entire length.
These tangents and their slopes are fundamental to analyzing how curves can intersect each other at right angles.
Curves Intersection and Their Analysis
Intersection points of curves are where the equations of the curves meet. At these points, the tangent lines are particularly important for determining the interaction between curves. In mathematical terms, these points allow us to study the properties like orthogonality.
Here, the given curves, circles, and straight lines intersect at the origin and potentially other points determined by their characteristics. To analyze their intersection, we calculate the product of the slopes of their tangents. If these products equal \(-1\) at any intersection, then the curves are orthogonal.
This intersection concept is crucial because it helps verify if one family of curves is indeed orthogonal to another, which is a significant relationship in the world of curve analysis.
Orthogonality in Curves
Orthogonality is a special property where two curves intersect at right angles. In cartesian geometry, this property is used to show that curves are mutually perpendicular, making them orthogonal trajectories of each other.
For the given exercise, orthogonality is verified by ensuring that the product of the slopes of the tangent lines is precisely \(-1\) at their intersection points.
  • This means multiplying the tangent slope of the circle \( -\frac{x}{y} \) by the tangent slope of the line \( -\frac{a}{b} \) gives \( \frac{ax}{by} = -1 \).
  • As the conditions are met, the families \( x^2 + y^2 = r^2 \) and \( ax + by = 0 \) are indeed orthogonal trajectories.
Understanding such orthogonal interactions helps in visualizing and sketching how these curves are placed on a coordinate system, providing a powerful tool for graphical analysis.

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