Find dy/dx when r=2 2cos(theta) , then find slope of tengent line at point(4, 2pi)

Mathematics

1 answer:

Korolek [52]2 years ago

8 0

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.

$\Rightarrow \frac{dx}{d\theta} =-2sin(\theta)+2(-2cos(\theta)sin(\theta))\\\\\Rightarrow \frac{dx}{d\theta} =-2sin(\theta)-2sin(2\theta)$ .............(1)

Similarly,

$\Rightarrow \frac{dy}{d\theta}=2cos(\theta)+2(cos^2(\theta)-sin^2(\theta))\\\\\Rightarrow \frac{dy}{d\theta}=2cos(\theta)+2(cos(2\theta))$ ..............(2)

Now we find dy/dx

⇒ dy/dx = (dy/dθ) / (dx/dθ)

From (1) and (2),

$\Rightarrow \frac{dy}{dx} =\frac{2cos(\theta)+2cos(2\theta)}{-2sin(\theta)-2sin(2\theta)} \\\\\Rightarrow \frac{dy}{dx} =\frac{2(cos(\theta)+cos(2\theta))}{-2(sin(\theta)+sin(2\theta))}\\\\\Rightarrow \frac{dy}{dx} =-\frac{cos(\theta)+cos(2\theta)}{sin(\theta)+sin(2\theta)}$

We know that The slope of tangent line is given by dy/dx.

So, the slope is: $m =-\frac{cos(\theta)+cos(2\theta)}{sin(\theta)+sin(2\theta)}$

Now we need to find the slope of tangent line at point(4, 2pi)

Substitute θ = 2π in above slope formula.

$\Rightarrow m =-\frac{cos(2\pi)+cos(2\times 2\pi)}{sin(2\pi)+sin(2\times 2\pi)}\\\\\Rightarrow m=-\frac{1+cos(4\pi)}{0+sin(4\pi)}\\\\\Rightarrow m=-\frac{1+cos(4\pi)}{sin(4\pi)}$