/*! 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 111 Show that the equation \(r=a \co... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 111

Show that the equation \(r=a \cos \theta+b \sin \theta\) where \(a\) and \(b\) are real numbers, describes a circle. Find the center and radius of the circle.

Short Answer

Expert verified
Answer: The circle is centered at the origin (0, 0) and has a radius of \(\sqrt{a^2 + b^2}\).

Step by step solution

01

Convert the polar equation to Cartesian coordinates

We have the polar equation \(r = a\cos\theta + b\sin\theta\). To convert this to the Cartesian coordinate system, we'll use the relationships \(x = r\cos\theta\) and \(y = r\sin\theta\). First, let's get the expression for \(r\): \[r=\frac{x}{\cos\theta} = a\cos\theta + b\sin\theta.\] Now, multiply both sides of the equation by \(\cos\theta\) to eliminate the denominator: \[x=a\cos^2\theta + b\sin\theta\cos\theta.\] Next, let's write the expression for \(r\) in terms of \(y\) and \(\sin \theta\): \[r=\frac{y}{\sin\theta} = a\cos\theta + b\sin\theta.\] Now, multiply both sides of the equation by \(\sin\theta\) to eliminate the denominator: \[y=a\cos\theta\sin\theta+b\sin^2\theta.\]
02

Combine the two equations to eliminate theta

We have the following equations: \[x=a\cos^2\theta + b\sin\theta\cos\theta\] \[y=a\cos\theta\sin\theta+b\sin^2\theta.\] Now, square both equations and add them to eliminate \(\sin\theta\cos\theta\) term. \[x^2 = a^2\cos^2\theta + 2ab\sin\theta\cos\theta + b^2\sin^2\theta\] \[y^2 = a^2\cos^2\theta\sin^2\theta + 2ab\sin^2\theta\cos^2\theta + b^2\sin^4\theta\] Adding the equations, we get \[x^2 + y^2 = (a^2+b^2)(\cos^2\theta+\sin^2\theta).\]
03

Simplify and rewrite in standard form

Using the identity \(\cos^2\theta+\sin^2\theta = 1\), we have \[x^2+y^2=a^2+b^2.\] This is a standard equation for a circle centered at the origin with radius \(\sqrt{a^2+b^2}\). So, the given polar equation represents a circle.
04

Find the center and radius of the circle

The standard equation for a circle is \((x-h)^2+(y-k)^2=r^2\), where \((h,k)\) is the center and \(r\) is the radius. Comparing our equation with the standard equation, we can see that the center of the circle is \((h,k)=(0,0)\). The radius of the circle is obtained by taking the square root of both sides of the equation: \[r = \sqrt{a^2 + b^2}.\] In conclusion, the given polar equation represents a circle centered at the origin with radius \(\sqrt{a^2 + b^2}\).

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