In Exercises 64–70, graph each polar equation. Be sure to test fo...

Question

In Exercises 64–70, graph each polar equation. Be sure to test for symmetry.

r = 2 + cos θ

Accepted Answer

Test the polar equation for symmetry and sketch its graph where we have R equals 13 plus cosine theta. We're also given a coordinate plane here and polar cornets to graph our functional. Now to solve this, we first need to test all of the symmetry. For this equation. We have three types of symmetry to test our first being polar symmetry. We can test polo symmetry by replacing our angle data with negative data and seeing if we get the original function back out. So for polar, we have R equals 13 plus cosine negative data. No, we have cosine negative data. We know cosine is an even function. This tells us if we take cosine of a negative angle, it will be the same as taking cosine of a positive angle, meaning we have R equals 13 plus cosine data. If we use this rule, this is the same as our original function. So we confirm we have polar symmetry. Now we can check our next symmetry which is symmetry over the line data equals pi divided by two. To check this one, we take both our radius R and our angle data can turn the negative. We wanna see if we get the original function back out again. So we will get negative R equals 13 plus cosine negative data by are even functional. We get 13 plus cosign data and divide by negative one to get R equals negative 13 minus cosine data. That is not the same as our original function. So we cannot confirm if this has symmetry of the line theta equals pi divided by two. So we can finally check our last symmetry. Pole symmetry. For pulse symmetry, we will just change our radius to be negative. We get negative R equals 13 plus cosine theta divided by negative one gives us R equals negative 13 minus cosine theta which once again is not the same as our original. So we do not have pole symmetry. At least we can't confirm that there's pole symmetry. Now, we can go ahead and graph our function. We just need to find values from zero to pi because we know this has polar symmetry that will have a reflection. So we can make input and output values of data to get R zero and divided by six pi divided by three, divided by two, two pi divided by three five P divided by six and pi if we were to plug in zero, we get four five divided by six, you get 13.9 pi over 3, 13.5 I divided by 2 13, two pi divided by 3, 12.5 55 divided by 12.1 and pie, we just get 12th. Now we can plot these on our graph. We have zero go all the way to pi divided by 6 13.9 five, divided by 3, 13.5. I divided by two 13, two pi divided by 3 12.5. Pi pi divided by 6 12.1 and pi 12, you can draw a line or a curve connecting the points. And because we have polar symmetry, we will reflect this vertically. So we reflect it, we end up with a circle and we can go ahead and a straw the other half of the circle like. So it's a little rough but this is the graph of our function. OK. I hope to help you solve the problem. Thank you for watching. Goodbye.

In Exercises 64–70, graph each polar equation. Be sure to test for symmetry. r = 2 + cos θ

Key Concepts

Polar Coordinates

Symmetry in Polar Graphs

Graphing Polar Equations

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