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Chapter 6: Problem 71

Convert each polar equation to a rectangular equation. Then use a rectangular coordinate system to graph the rectangular equation. $$r=6 \cos \theta+4 \sin \theta$$

Short Answer

Expert verified
The polar equation \( r = 6 \cos \theta + 4 \sin \theta \) converts to the rectangular equation \( (x-3)^2 + (y-2)^2 = 13 \), which represents a circle centered at (3,2) with radius \(\sqrt{13}\).

Step by step solution

01

Conversion from Polar to Rectangular Form

Use the identities for x and y to convert the equation from polar to rectangular: \( r = 6 \cos \theta + 4 \sin \theta \), multiply all parts of the equation by \(r\) (which equals \(\sqrt{x^2 +y^2}\)) to get: \( r^2 = r\cdot6 \cos \theta + r\cdot4 \sin \theta \), substituting \( x = r \cos\theta \), \( y = r \sin\theta \), \( r^2 = x^2 + y^2 \) to the equation to give: \( x^2 + y^2 = 6x + 4y \)
02

Rearranging the Rectangular Equation

For convenience, let's rearrange the rectangular equation as \( (x-3)^2 + (y-2)^2 = 13 \). This is done by completing the square.
03

Graphing in Rectangular Coordinate System

Last step is to graph our equation. The graph represents a circle in the plane with center at (3,2) and radius \(\sqrt{13}\). We can use these values to plot the circle on a rectangular coordinate system.

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Most popular questions from this chapter

Draw two vectors, \(\mathbf{v}\) and \(\mathbf{w},\) with the same initial point. Show the vector projection of \(\mathbf{v}\) onto \(\mathbf{w}\) in your diagram. Then describe how you identified this vector.

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