We need one more tool to solve for intersections of polar functions. We need to know how to solve a system of two equations.
We can find where two rectangular functions y = f (x) and y = g(x) intersect by setting them equal to each other and solving the equation f (x) = g(x) for x. This is the place (or places) where both equations are true at the same time.
Like the intersection of two rectangular functions, we can find most of the places two polar equations r = f (θ) and r = g(θ) intersect by setting them equal to each other and solving the equation f (θ) = g(θ) for θ.
It's useful to graph the functions before finding where they intersect. We want to know how many intersection points we need to find.
Also, setting equations equal to each other and solving might not catch an intersection point at the origin.
This sort of thing can happen because with polar coordinates there are infinitely many ways to write any point. Pick your favorite and stop there.
When finding where two polar graphs intersect, graph the functions first. Then look at how many intersection points there are and which quadrants they're in. Then we'll know how many points we need to find and roughly where they are.
We can estimate the intersection points from the graph. The graphs of r = cos θ and r = sinθ do look like they hit each other around θ = π / 4. However, we still need to set the functions equal to each other and solve for θ.
Practice:
Find the points at which r = 1 and r = sin θ intersect.
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We have
1 = sinθ
when θ = \frac{π}{2}.
This makes sense when we look at a graph of the two functions: PICTURE graph functions, emphasize intersection
They intersect at
(r,θ) = (1,\frac{π}{2}),
and we can see from the graph that this is the only intersection point.
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Find where r = cos θ and r = sin θ intersect.
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Start by graphing the functions: PICTURE: graph them We can see that they intersect at two places. Set the equations equal: cosθ = sinθ This is true when θ = \frac{π}{4}. Then cosθ = sinθ = \frac{\sqrt2}{2}. One of our points of intersection is (r,θ) = (\frac{\sqrt2}{2}, \frac{π}{4}). We can see from the picture that these graphs also intersect at the origin, but we can't find that by solving the equation cosθ = sinθ because different values of θ make cosθ and sinθ equal to 0. This doesn't matter since the point (0,θ) is the same point no matter what θ is. PICTURE polar funcs 30 We may as well call the point r = 0 and not worry about θ. This point, r = 0, is the second intersection point. | |
- Find all points where r = 1 + cosθ and r = 1 + sin θ intersect.
Answer
- Graph the functions first:PICTURE graph functionsIf we zoom in enough we can see that there are three points of intersection. One of these is the origin, r = 0.PICTURE polar funcs 33Set the functions equal and solve:
1 + cosθ = 1 + sinθcosθ = sinθθ = \frac{π}{4} \text{ or }\frac{5π}{4}
We didn't need to use θ = \frac{5π}{4} when finding the intersection of sin θ and cosθ, but we need it now because we have morepoints of intersection.When θ = \frac{π}{4} we have r = \frac{\sqrt2}{2}, and when θ = \frac{5π}{4} we have r = -\frac{\sqrt2}{2}. The points of intersection arer = 0, (\frac{\sqrt2}{2},\frac{π}{4}), (-\frac{\sqrt2}{2},\frac{5π}{4}).PICTURE: graph functions, label intersections
- Find all points where r = 1 + cos θ and r = -cos θ intersect.
Answer
- First, draw the graph:PICTURE: graph functionsIt looks like we have three points of intersection again, and one of them is r = 0.PICTURE graph functions, emphasize intersection pointsTo find the other two points we need to set the functions equal and solve for θ.
1 + cosθ = -cosθ1 = -2cosθ-\frac{1}{2}& = &cosθθ& = &\frac{2π}{3} \text{ or } \frac{4π}{3}
At these values of θ,r = -cosθ = \frac{1}{2}.The points of intersection arer = 0, (\frac{1}{2},\frac{2π}{3} ), (\frac{1}{2},\frac{4π}{3} ).PICTURE graph functions, label points of intersection
- Find all points in the first quadrant where r = 2sin(3θ) and r = 1 intersect.
Answer
- Graph the functions:PICTURE: graph functionsThere are two points of intersection in the first quadrant, we set the functions equal and solve for θ.
1& = &2sin(3θ)\frac{1}{2}& = &sin(3θ)3θ& = &\frac{π}{6}\text{ or }\frac{5π}{6}θ& = &\frac{π}{18}\text{ or }\frac{5π}{18}
The points of intersection are(1, \frac{π}{18} ), (1,\frac{5π}{18} ).PICTURE: graph functions, label intersections