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Calculus/Find intesection points of polar graphs

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Expert: - 3/24/2011


Question
When I graph r(1) = cos(2x) and r(2)=sin(2x) I see two intersections in each quadrant. When I solve sin(2x)=cos(2x) I get only the first point (ie pi/8, 5pi/8 ...)which means the graphs arrive at those points at different x-angles... How do I algebraically find those other points? Is there a method that is taught to find those points.
Thanks,
Phillip

Answer

Polar
Questioner: Phillip
Country: United States
Category: Calculus
Private: No
Subject: polar graphs - Find intesection points
Question: When I graph r(1) = cos(2x) and r(2)=sin(2x) I see two intersections in each quadrant. When I solve sin(2x)=cos(2x) I get only the first point (ie pi/8, 5pi/8 ...)which means the graphs arrive at those points at different x-angles... How do I algebraically find those other points? Is there a method that is taught to find those points.
Thanks,
Phillip
...............................
This is indeed a difficulty with polar coordinates -- a given point in the plane can have many sets of polar coordinates.  

You graphed (see attached) and found 8 points of intersection (I hope), marked  P1,...,P8.

You found that the first point is at pi/8, for example.  You solved like this, I assume:

sin(2t)=cos(2t)   <<< I changed x to t (for theta)

tan(2t) = 1

2t = pi/4 + n pi,  n = 0,1,2,3,....  (and minus, too)

t = pi/8 + n pi/2

Then t = pi/8, 5pi/8, AND 9pi/8, 13pi/8, plus others that are duplicates.  (17pi/8 is the same as pi/8, and so on)

These are P1, P3, P5, P7  in the picture.  What about the even-numbered ones?

Look at the cosine graph, in black.  I have marked the quadrants that contain the actual value of theta (t),  So, at P2, you have one of the two pieces of the quadrant III section of the graph.  So at P2, you have:

On the sine graph, P2 has t = 3pi/8.  sin(2*3pi/8) = sin(3pi/4) = +0.7
P2 is on the sine graph (red) with coordinates (r = 0.7, t = 3pi/8)

On the cosine graph, P2 has t in the second portion of Q III, or t = 11 pi/8
cos(2 * 11pi/8) = cos(11pi/4) = cos(3pi/4) = -0.7
P2 is on the cosine graph (black) with coordinates (- 0.7, t = 11pi/8)

OK, we have found out why they cross there, now how about why?

We note that the values of theta differ by pi, and the r's are +-.

Solve:

sin(2(t+pi)) = - cos(2t)

How do you do that?

Tune in later for the answer, after these words from our sponsor.

Paul Klarreich

Expertise

All topics in first-year calculus including infinite series, max-min and related rate problems. Also trigonometry and complex numbers, theory of equations, exponential and logarithmic functions. I can also try (but not guarantee) to answer questions on Analysis -- sequences, limits, continuity.

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I taught all mathematics subjects from elementary algebra to differential equations at a two-year college in New York City for 25 years.

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(See above.)

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