can you explain how you find the eccentricity of r=4/-2-costheta?
Certainly! To find the eccentricity of a polar equation in the form r = f(θ), where f(θ) is some function of θ, we can use the formula:
eccentricity (e) = √(1 + f'(θ)²) / |f(θ)|
In the given equation r = 4 / (-2 - cosθ), we need to find the value of f(θ) and calculate the derivative f'(θ).
First, let's simplify the equation:
r = 4 / (-2 - cosθ)
Multiply both sides by (-2 - cosθ) to eliminate the denominator:
r(-2 - cosθ) = 4
Distribute the r:
-2r - rcosθ = 4
Now, isolate the cosθ term:
-rcosθ = 4 + 2r
Rearrange the equation:
cosθ = -(4 + 2r) / r
Now we have an expression for cosθ in terms of r.
Next, we will differentiate cosθ with respect to θ to find f'(θ). In this case, f(θ) = cosθ.
Differentiating cosθ with respect to θ gives us:
d(cosθ) / dθ = -sinθ
Now we have f'(θ) = -sinθ.
Substituting f'(θ) and f(θ) into the eccentricity formula, we get:
eccentricity (e) = √(1 + (-sinθ)²) / |cosθ|
To evaluate the eccentricity, we need a specific value of θ. Choose any value that satisfies the domain of the equation.
For example, let's evaluate it at θ = 0:
eccentricity (e) = √(1 + (-sin0)²) / |cos0|
= √(1 + (0)²) / |1|
= √(1 + 0) / 1
= √(1) / 1
= 1 / 1
= 1
Therefore, the eccentricity of the equation r = 4 / (-2 - cosθ) is 1.