Find the point(s) of intersection for the polar curves with equations r = 6 cos θ and r = 4 − 2 cos θ. (10 points)
A) pi over 6 and negative pi over 6
B) pi over 3 and negative pi over 3
C) pi over 2 and negative pi over 2
D) 0, π
just plug and chug
6 cosθ = 4 - 2cosθ
6cosθ = 6
cosθ = 1
so, what do you think?
I got 0, pi, is this right? I got this because cos 0 = 1 and cos pi = -1
but we want cosθ = 1, not -1.
I think you made a mistake
6 cosθ = 4 - 2cosθ
You wrote: 6cosθ = 6
But, 6 cos theta = 4 - 2cos theta is equal to 8 cos theta = 4
which means cos theta = 4/8 0r cos theta = 1/2
Please check if I am right or if I made a mistake somewhere
You are correct; Good catch. I got my 2's and 4's mixed up ...
To find the points of intersection for the polar curves with equations r = 6 cos θ and r = 4 - 2 cos θ, we need to set the two equations equal to each other and solve for θ.
First, set r = r:
6 cos θ = 4 - 2 cos θ
Now, combine like terms:
8 cos θ = 4
Next, divide both sides by 8:
cos θ = 1/2
To find the values of θ that satisfy this equation, we can use the unit circle or the inverse cosine function. Since we are looking for θ in the range between 0 and 2π, we can make use of the inverse cosine function. Inverse cosine of 1/2 gives us the principal value of π/3.
So, θ = π/3 is one of the solutions.
To find the other solution, we need to consider the symmetry of the cosine function. Since cosine is an even function, cos(θ) = cos(-θ). Therefore, if θ = π/3 is a solution, then θ = -π/3 is also a solution.
The points of intersection for these equations are when θ = π/3 and θ = -π/3.
Therefore, the correct answer is:
B) π/3 and -π/3