Find all points of intersection of the polar curves r= 2 + 4cos theta and r= 6cos theta.
Please explain in detail.
well, where does
2 + 4cosθ = 6cosθ ?
2 = 2cosθ
cosθ = 1
θ = 0,2pi,...
To find the points of intersection between the polar curves \(r = 2 + 4\cos\theta\) and \(r = 6\cos\theta\), we need to determine the values of \(\theta\) for which both equations are satisfied.
We can start by setting the two equations equal to each other:
\(2 + 4\cos\theta = 6\cos\theta\)
Next, rearrange the equation to isolate \(\cos\theta\):
\(2 = 2\cos\theta\)
Divide both sides of the equation by 2:
\(1 = \cos\theta\)
Now, recall that \(\cos\theta\) represents the x-coordinate of a point on the unit circle. The range of \(\theta\) for which \(\cos\theta = 1\) is when \(\theta = 0\) and when \(\theta = 2\pi\).
So, the two polar curves intersect at two points: one point when \(\theta = 0\) and another point when \(\theta = 2\pi\).
To find the corresponding values of \(r\) for each \(\theta\), substitute \(\theta\) back into one of the original equations. Let's use the equation \(r = 2 + 4\cos\theta\) to find the values of \(r\):
For \(\theta = 0\):
\(r = 2 + 4\cos(0) = 2 + 4(1) = 6\)
Therefore, the point of intersection when \(\theta = 0\) is (\(6, 0^\circ\)).
For \(\theta = 2\pi\):
\(r = 2 + 4\cos(2\pi) = 2 + 4(1) = 6\)
Therefore, the point of intersection when \(\theta = 2\pi\) is (\(6, 360^\circ\)).
In conclusion, the polar curves \(r = 2 + 4\cos\theta\) and \(r = 6\cos\theta\) intersect at two points: (\(6, 0^\circ\)) and (\(6, 360^\circ\)).