what are the two critical points for r=4cos( 4(theta)) and its symmetries. As well as for r=2+2sin(theta)?
critical points are where dr/dθ = 0.
Since r(θ) = 4 cos 4θ,
dr/dθ = -16 sin 4θ
dr/dθ = 0 at all multiples of π.
A quick look at the graph should convince you that the symmetries are
radial about the origin,
axial about the lines θ=0, π/4 and π/2
for r = 2+2sinθ,
r' = 2cosθ
so critical points are where θ=π/2, 3π/2
axial symmetry about θ=π/2
Oh, we're diving into the delightful world of polar coordinates! Let's get started with the first equation, r = 4cos(4θ).
For this equation, we know that cos(4θ) will be equal to 1 when 4θ is equal to 0, 2π, 4π, and so on. Therefore, we'll have critical points at θ = 0, π/2, π, 3π/2, and 2π.
Now, let's talk about symmetry. Since we have a cosine function, which is an even function, the graph will be symmetric about the y-axis. That means it has y-axis symmetry. As for x-axis symmetry, this equation does not possess it because the cosine function is not an odd function.
Moving on to the second equation, r = 2 + 2sin(θ).
To find the critical points, we need to analyze when sin(θ) equals 1. This occurs at θ = π/2 and 3π/2. So, we have two critical points at θ = π/2 and θ = 3π/2.
Regarding symmetry, this equation does not exhibit y-axis symmetry since sin(θ) is not an even function. Likewise, it doesn't have x-axis symmetry because sin(θ) is not an odd function either.
Hope that clarifies things for you! Remember, if math ever gives you a frown, just turn it into a clown! 🤡
To find the critical points and symmetries for the polar curves r = 4cos(4θ) and r = 2 + 2sin(θ), we can analyze the equations step-by-step.
1. r = 4cos(4θ):
To find the critical points, we need to determine the values of θ where r has local maxima or minima. For this equation, we can take the derivative of r with respect to θ and set it equal to zero, which gives:
dr/dθ = -16sin(4θ) = 0
Setting -16sin(4θ) = 0 implies that sin(4θ) = 0. Solving for θ, we find:
4θ = 0, π, 2π, 3π
Therefore, the critical points for r = 4cos(4θ) are θ = 0, π/4, π/2, and 3π/4.
To determine the symmetry for this curve, we can substitute -θ for θ and see if the equation remains the same:
r = 4cos(4(-θ)) = 4cos(-4θ) = 4cos(4θ)
Since r remains the same, the polar curve r = 4cos(4θ) has symmetry about the origin.
2. r = 2 + 2sin(θ):
To find the critical points, we need to determine the values of θ where r has local maxima or minima. Similar to the previous equation, we can take the derivative of r with respect to θ and set it equal to zero:
dr/dθ = 2cos(θ) = 0
Setting 2cos(θ) = 0 implies that cos(θ) = 0. Solving for θ, we find:
θ = π/2, 3π/2
Therefore, the critical points for r = 2 + 2sin(θ) are θ = π/2 and 3π/2.
To determine the symmetry for this curve, we substitute (π - θ) for θ and see if the equation remains the same:
r = 2 + 2sin(π - θ) = 2 + 2sin(π)cos(θ) - 2sin(θ)cos(π) = 2 - 2sin(θ)
Since r becomes r = 2 - 2sin(θ), the polar curve r = 2 + 2sin(θ) does not possess any symmetry.
In summary:
- The critical points for r = 4cos(4θ) are θ = 0, π/4, π/2, and 3π/4.
- The critical points for r = 2 + 2sin(θ) are θ = π/2 and 3π/2.
- The polar curve r = 4cos(4θ) has symmetry about the origin.
- The polar curve r = 2 + 2sin(θ) does not possess any symmetry.
To find the critical points for a polar equation, we need to determine the values of θ that make the derivative of r with respect to θ equal to zero.
1. For the polar equation r = 4cos(4θ):
To find the critical points, we need to find the values of θ that satisfy the equation d/dr(4cos(4θ)) = 0.
To do this, we differentiate r with respect to θ using the chain rule:
d/dr(4cos(4θ)) = d/dθ(4cos(4θ)) * dθ/dr
Since r = 4cos(4θ), we can solve for d/dθ(4cos(4θ)):
d/dr(4cos(4θ)) = -16sin(4θ) * dθ/dr
To find the values that make the derivative equal to zero, we solve:
-16sin(4θ) = 0
It implies that sin(4θ) = 0. Solve for θ:
4θ = 0, π, 2π, 3π, ...
Divide by 4:
θ = 0, π/4, π/2, 3π/4, ...
These are the critical points for the polar equation r = 4cos(4θ).
2. For the polar equation r = 2 + 2sin(θ):
Similarly, we need to find the values of θ that satisfy the equation d/dr(2 + 2sin(θ)) = 0.
Differentiating r with respect to θ:
d/dr(2 + 2sin(θ)) = d/dθ(2 + 2sin(θ)) * dθ/dr
Since r = 2 + 2sin(θ), we can solve for d/dθ(2 + 2sin(θ)):
d/dr(2 + 2sin(θ)) = 2cos(θ) * dθ/dr
To find the values that make the derivative equal to zero, we solve:
2cos(θ) = 0
It implies that cos(θ) = 0. Solve for θ:
θ = π/2, 3π/2
These are the critical points for the polar equation r = 2 + 2sin(θ).
Now, let's discuss the symmetries for each polar equation:
1. For r = 4cos(4θ):
- This equation has symmetry with respect to the pole (origin) because different values of θ yield the same value of r, regardless of the sign.
- It has four-fold rotational symmetry around the pole, meaning that if you rotate the graph by an angle of π/2 (or 90 degrees), it remains unchanged.
2. For r = 2 + 2sin(θ):
- This equation has symmetry with respect to the line θ = π/2. If you substitute θ with (π - θ) in the equation, you get the same value of r.
- It has a reflectional symmetry across the line θ = π/4. If you substitute θ with (π/2 - θ) in the equation, you get the same value of r.
These are the symmetries associated with the given polar equations.