Respuesta :
The slope of tangent line at point(4, 2pi) is undefined.
For given question,
We have been given a polar equation r = 2 + 2cos(θ)
We need find dy/dx as well as the slope of tangent line at point(4, 2π).
We know that, for polar equation we use,
x = r cos(θ) and y = r sin(θ)
plug the given value of r into these equations we get:
⇒ x = r cos(θ)
⇒ x = (2 + 2cos(θ) ) × cos(θ)
⇒ x = 2(cos(θ) + cos²(θ))
⇒ x = 2cos(θ) + 2cos²(θ)
Similarly,
⇒ y = r sin(θ)
⇒ y = (2 + 2cos(θ) ) × sin(θ)
⇒ y = 2(sin(θ) + sin(θ)cos(θ))
⇒ y = 2sin(θ) + 2sin(θ)cos(θ)
Now we find derivative of x and y with respect to theta.
[tex]\Rightarrow \frac{dx}{d\theta} =-2sin(\theta)+2(-2cos(\theta)sin(\theta))\\\\\Rightarrow \frac{dx}{d\theta} =-2sin(\theta)-2sin(2\theta)[/tex] .............(1)
Similarly,
[tex]\Rightarrow \frac{dy}{d\theta}=2cos(\theta)+2(cos^2(\theta)-sin^2(\theta))\\\\\Rightarrow \frac{dy}{d\theta}=2cos(\theta)+2(cos(2\theta))[/tex] ..............(2)
Now we find dy/dx
⇒ dy/dx = (dy/dθ) / (dx/dθ)
From (1) and (2),
[tex]\Rightarrow \frac{dy}{dx} =\frac{2cos(\theta)+2cos(2\theta)}{-2sin(\theta)-2sin(2\theta)} \\\\\Rightarrow \frac{dy}{dx} =\frac{2(cos(\theta)+cos(2\theta))}{-2(sin(\theta)+sin(2\theta))}\\\\\Rightarrow \frac{dy}{dx} =-\frac{cos(\theta)+cos(2\theta)}{sin(\theta)+sin(2\theta)}[/tex]
We know that The slope of tangent line is given by dy/dx.
So, the slope is: [tex]m =-\frac{cos(\theta)+cos(2\theta)}{sin(\theta)+sin(2\theta)}[/tex]
Now we need to find the slope of tangent line at point(4, 2pi)
Substitute θ = 2π in above slope formula.
[tex]\Rightarrow m =-\frac{cos(2\pi)+cos(2\times 2\pi)}{sin(2\pi)+sin(2\times 2\pi)}\\\\\Rightarrow m=-\frac{1+cos(4\pi)}{0+sin(4\pi)}\\\\\Rightarrow m=-\frac{1+cos(4\pi)}{sin(4\pi)}[/tex]
⇒ m = ∞
The slope of tangent line at point(4, 2pi) is not defined.
This means, the tangent line must be parallel to Y-axis.
Therefore, the slope of tangent line at point(4, 2pi) is undefined.
Learn more about the slope here:
https://brainly.com/question/10785137
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