for what values of theta (0 < theta < 2pi) do maximum r values occur on the graph the polar equation r = 4-3 sin 3 theta
To determine the values of theta for which the maximum r values occur on the graph of the polar equation r = 4-3sin(3θ), we need to find the critical points of r.
The first step is to find the derivative of r with respect to θ. Let's call this derivative dr/dθ.
We have:
r = 4 - 3sin(3θ)
Differentiate both sides with respect to θ to get:
dr/dθ = -9cos(3θ)
To find the critical points, we need to solve the equation dr/dθ = 0.
-9cos(3θ) = 0
Now, we need to solve this equation for theta. Since cosine function equals zero at π/2 and 3π/2 for one period, we can solve for 3θ:
3θ = π/2 + kπ or 3π/2 + kπ, where k is an integer.
Dividing both sides by 3, we get:
θ = (π/6 + kπ/3) or (π/2 + kπ/3)
Since we are interested in the values of theta between 0 and 2π, we need to find the values of theta that satisfy 0 < θ < 2π.
For k = 0:
θ = π/6, π/2
For k = 1:
θ = π/3, 2π/3
For k = 2:
θ = π/2, 5π/6
Note that for higher values of k, we would get duplicate solutions.
Therefore, the values of theta (0 < theta < 2pi) for which the maximum r values occur on the graph of r = 4 - 3sin(3θ) are:
θ = π/6, π/3, π/2, 2π/3, 5π/6
To find the values of θ for which the maximum r values occur on the graph of the polar equation r = 4 - 3sin(3θ), we need to analyze the behavior of the function.
First, let's consider the range of the function sin(3θ). The sine function oscillates between -1 and 1, so sin(3θ) will also oscillate between -1 and 1.
Now, let's focus on the term 4 - 3sin(3θ). As sin(3θ) oscillates, 3sin(3θ) will oscillate between -3 and 3. Subtracting that from 4 will result in a range of values between 1 and 7. Therefore, the equation r = 4 - 3sin(3θ) represents a spiral-like graph that extends from r = 1 to r = 7.
To determine the values of θ for which the maximum values of r occur, we need to find the values of θ where sin(3θ) is either -1 or 1.
When sin(3θ) = -1, we have:
-1 = sin(3θ)
3θ = arcsin(-1)
3θ = -π/2 + 2πn, where n is an integer
Solving for θ, we get:
θ = (-π/2 + 2πn)/3
Similarly, when sin(3θ) = 1, we have:
1 = sin(3θ)
3θ = π/2 + 2πm, where m is an integer
Solving for θ, we get:
θ = (π/2 + 2πm)/3
Thus, the values of θ for which the maximum r values occur on the graph of the polar equation r = 4 - 3sin(3θ) are given by the solutions to the equations:
θ = (-π/2 + 2πn)/3 and θ = (π/2 + 2πm)/3
where n and m are integers.