For what values of theta 0<theta < 2pie (also equal to these) do maximum r -values occur on the graph of the polar equation r= 2-4sin3 theta ? Note that a maximum r-value occurs at a point that is the max distance from the pole?
I know the max is 6 but what would be the radians around the unit circle?
To find the values of theta for which maximum r-values occur on the graph of the polar equation r = 2 - 4sin(3theta), we need to find when the derivative of r with respect to theta equals zero.
1. Take the derivative of r with respect to theta using the chain rule:
dr/dtheta = d(2 - 4sin(3theta))/dtheta.
2. Simplify the derivative:
dr/dtheta = -12cos(3theta).
3. Set the derivative equal to zero:
-12cos(3theta) = 0.
4. Solve for theta:
cos(3theta) = 0.
To find the values of theta, we need to consider the unit circle and where the cosine function is equal to zero. On the unit circle, the cosine function is equal to zero at π/2 and 3π/2.
Since we have 3theta, we can set up the following equation to find the values of theta:
3theta = π/2, 3π/2.
5. Solve for theta:
theta = π/6, π/2.
Now, we need to consider the given range 0 < theta < 2π.
In this range, both π/6 and π/2 lie.
Therefore, the values of theta for which maximum r-values occur on the graph of the polar equation r = 2 - 4sin(3theta) are:
θ = π/6 and θ = π/2 (in radians).
To find the values of theta for which the maximum r-values occur on the graph of the polar equation r = 2 - 4sin(3theta), we need to determine when the sine function has its maximum value.
The maximum value of the sine function is 1, which occurs at theta = pi/2 and at theta = 3pi/2. Since theta ranges from 0 to 2pi, we need to check if either of these values lies within this range.
For theta = pi/2, we have 0 < pi/2 < 2pi, so this value satisfies the condition.
For theta = 3pi/2, we have 0 < 3pi/2 < 2pi, so this value also satisfies the condition.
Therefore, the maximum r-values occur at theta = pi/2 and theta = 3pi/2, which correspond to 90 degrees and 270 degrees, respectively, on the unit circle.
first off, that's pi not "pie".
r=2-4sin3θ
Since sin3θ has a minimum of -1, that makes max r=6.
So, when is sin3θ = -1? When 3θ = 3π/2
That is, when θ=π/2
A glance at the graph makes this clear.
http://www.wolframalpha.com/input/?i=r%3D2-4sin3%CE%B8
Of course, since sin3θ has a period of 2π/3, r=6 when θ=pi/2+2kπ/3
That is, when θ=π/2, 7π/6, 11π/6