/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 A line segment \(L\) of length \... [FREE SOLUTION] | 91Ó°ÊÓ

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A line segment \(L\) of length \(2 a\) has its two end points on the \(x\) -and \(y\) -axes, respectively. The point \(P\) is on \(L\) and is such that \(O P\) is perpendicular to \(L .\) Show that the set of points \(P\) satisfying this condition is a four-leaved rose by finding its polar equation.

Short Answer

Expert verified
The locus of point P is a four-leaved rose with polar equation \( r = \frac{\sqrt{2}a}{\cos\theta} \).

Step by step solution

01

Understand the Problem

We have a line segment \( L \) with endpoints on the \( x \)- and \( y \)-axes, such that \(L \) has a length \( 2a \). A point \( P \) lying on \( L \) is defined such that \( OP \) (the distance from origin to \( P \)) is perpendicular to \( L \). We need to find the locus of \( P \) and show it is a four-leaved rose in polar coordinates.
02

Define the Equation of L in Cartesian Coordinates

Assuming \( P \) has coordinates \((x, y)\), the endpoints of \( L \) are \((2a, 0)\) and \((0, 2a)\). The equation of line \( L \) can be written as \( \frac{x}{2a} + \frac{y}{2a} = 1 \) or \( x + y = 2a \).
03

Polar Coordinates Setup

Express the coordinates of \( P \) in polar form. If \( P \) has Cartesian coordinates \((r \cos(\theta), r \sin(\theta))\), then we substitute these into the line equation: \( r \cos(\theta) + r \sin(\theta) = 2a \).
04

Condition for OP Perpendicular to L

If \( OP \) is perpendicular to \( L \), the dot product of vector \( OP \) and the gradient direction of \( L \) must be zero. The gradient of \( L \) can be considered as \( (1, 1) \). Hence, \((r \cos(\theta), r \sin(\theta)) \cdot (1, 1) = r(\cos(\theta) + \sin(\theta)) = 0\).
05

Relate Polar Equation of P

From the equation \( r(\cos(\theta) + \sin(\theta)) = 2a \), isolate \( r \): \[ r = \frac{2a}{\cos(\theta) + \sin(\theta)}. \]To reveal the four-leaved rose, use the identity \( \cos(\theta) + \sin(\theta) = \sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right) \).
06

Simplify to Standard Rose Equation

Rewriting the polar equation with the identity, \[ r = \frac{2a}{\sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right)} \approx \frac{\sqrt{2}a}{\cos\left(\theta - \frac{\pi}{4}\right)}. \]Introduce the substitution \( \theta - \frac{\pi}{4} \rightarrow \theta' \), the equation becomes \( r = \sqrt{2}a \sec(\theta') \) which forms a rose with four petals (or leaves) when considering symmetry.

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Key Concepts

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

Four-Leaved Rose
The four-leaved rose is a fascinating curve in polar coordinate geometry. It's named for its resemblance to a flower with four distinct petals, making it aesthetically appealing and mathematically intriguing. This shape is peculiar because it's defined by a polar equation rather than the regular Cartesian coordinates. When graphed, it creates a symmetrical and repetitive pattern that crosses the polar origin. The standard form of its polar equation is often expressed as \(r = a \cos(2\theta)\) or \(r = a \sin(2\theta)\). These equations highlight the four-leaved structure, thanks to the multiplier 2 inside the trigonometric function.In our exercise, the transformation and substitution, through trigonometric identities, allow us to appreciate this beautiful pattern starting from basic geometric conditions. Understanding this concept is key to leveraging polar equations for sketching and analyzing intricate curves found in more advanced mathematics.
Cartesian Coordinates
In mathematics, Cartesian coordinates are used to define a plane through two perpendicular lines called axes. Named after René Descartes, Cartesian coordinates are the most familiar coordinate system where each point is described by an ordered pair \((x, y)\). The sign of these coordinates dictates their position in one of the four quadrants around the origin. For our exercise, we initially express the line segment by its equation in Cartesian form as \(x + y = 2a\). This equation signifies all points \((x, y)\) that lie on the line between the x-axis and y-axis intersection. By setting up an understanding of such a representation, we can convert these coordinates into a polar form, which assists further in solving problems involving complex curve structures like our four-leaved rose.
Polar Equation
Polar equations redefine relationships between two variables using a radius and angle, rather than two intersecting lines. Instead of identifying a point's position through x and y coordinates, polar equations use \(r\) (radius from the origin) and \(\theta\) (angle from the positive x-axis). In polar coordinates, the starting expression \(r(\cos(\theta) + \sin(\theta)) = 2a\) paves the way for discovering complex curves. By simplifying this expression, and using trigonometric identities, we reveal the polar equation of a four-leaved rose: \(r = \frac{2a}{\sqrt{2}\cos(\theta - \frac{\pi}{4})}\).Polar equations are pivotal for transforming mathematical problems into graphically powerful solutions, allowing exploration of symmetry, rotational patterns, and unique geometrical forms.

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Most popular questions from this chapter

In order to graph a polar equation such as \(r=f(t)\) using a parametric equation grapher, you must replace this equation by \(x=f(t) \cos t\) and \(y=f(t) \sin t .\) These equations can be obtained by multiplying \(r=f(t)\) by \(\cos t\) and \(\sin t\), respectively. Confirm the discussions of conics in the text by graphing \(r=4 e /(1+e \cos t)\) for \(e=0.1,0.5,0.9,1,1.1\) and \(1.3\) on \([-\pi, \pi]\).

The position of a comet with a highly eccentric elliptical orbit \((e\) very near 1\()\) is measured with respect to a fixed polar axis (sun is at a focus but the polar axis is not an axis of the ellipse) at two times, giving the two points \((4, \pi / 2)\) and \((3, \pi / 4)\) of the orbit. Here distances are measured in astronomical units \((1 \mathrm{AU} \approx 93\) million miles). For the part of the orbit near the sun, assume that \(e=1\), so the orbit is given by $$r=\frac{d}{1+\cos \left(\theta-\theta_{0}\right)}$$ (a) The two points give two conditions for \(d\) and \(\theta_{0}\). Use them to show that \(4.24 \cos \theta_{0}-3.76 \sin \theta_{0}-2=0\) (b) Solve for \(\theta_{0}\) using Newton's Method. (c) How close does the comet get to the sun?

Show that the polar equation of the circle with center \((c, \alpha)\) and radius \(a\) is \(r^{2}+c^{2}-2 r c \cos (\theta-\alpha)=a^{2} .\)

Consider the two circles \(r=2 a \sin \theta\) and \(r=2 b \cos \theta\), with \(a\) and \(b\) positive. (a) Find the area of the region inside both circles. (b) Show that the two circles intersect at right angles.

Using the same axes, draw the conics \(y=\) \(\pm\left(a x^{2}+1\right)^{1 / 2}\) for \(-2 \leq x \leq 2\) and \(-2 \leq y \leq 2\) using \(a=\) \(-2,-1,-0.5,-0.1,0,0.1,0.6,1 .\) Make a conjecture about how the shape of the figure depends on \(a\).

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