Question

Find the points at which the following polar curves have a horizontal or a vertical tangent line. $$r=2+2 \sin \theta$$

Short answer

Question: Determine the points at which the polar curve r = 2+2sin(θ) has horizontal or vertical tangent lines. Answer: To find these points, follow the steps below: Step 1: Convert the polar equation to Cartesian form: x = (2+2sin(θ))cos(θ) y = (2+2sin(θ))sin(θ) Step 2: Compute the derivatives of x and y with respect to θ: dx/dθ = -(2+2sin(θ))sin(θ) + 2cos²(θ) dy/dθ = (2+2sin(θ))cos(θ) + 2sin(θ)cos(θ) Step 3: Compute the slope of the tangent and its reciprocal: dy/dx = (dy/dθ) / (dx/dθ) dx/dy = (dx/dθ) / (dy/dθ) Step 4: Determine the values of θ where the slope is zero or undefined: For horizontal tangents, set dy/dx = 0 and solve for θ. For vertical tangents, set dx/dy = 0 and solve for θ. Step 5: Compute the points in polar coordinates: After solving for θ, replace θ in the polar equation r = 2 + 2sin(θ) to find the corresponding r values. Then, combine the values (r, θ) to find the points in polar coordinates where horizontal or vertical tangents occur.

Step-by-step solution

Convert the polar equation to Cartesian form

Recall the polar to Cartesian conversion formulas: $$x = r \cos \theta$$ $$y = r \sin \theta$$ Given the polar equation: $$r=2+2 \sin \theta$$ Replace r in both formulas: $$x = (2+2 \sin \theta) \cos \theta$$ $$y = (2+2 \sin \theta) \sin \theta$$ Now we have the Cartesian equations.

Compute the derivatives of x and y with respect to θ

Compute dx/dθ and dy/dθ: $$\frac{dx}{d\theta} = - (2+2 \sin \theta) \sin \theta + 2 \cos^2 \theta$$ $$\frac{dy}{d\theta} = (2+2 \sin \theta) \cos \theta + 2 \sin \theta \cos \theta$$

Compute the slope of the tangent and its reciprocal

To find the slope of the tangent, we compute dy/dx by dividing dy/dθ by dx/dθ: $$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$ Compute the reciprocal of the slope, which is dx/dy: $$\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{\frac{dx}{d\theta}}{\frac{dy}{d\theta}}$$

Determine the values of θ where the slope is zero or undefined

A tangent is horizontal when the slope is 0, so set dy/dx equal to 0. A tangent is vertical when the slope is undefined, so set dx/dy equal to 0. Horizontal Tangents: $$\frac{dy}{dx} = 0$$ $$\frac{(2+2 \sin \theta) \cos \theta + 2 \sin \theta \cos \theta}{ - (2+2 \sin \theta) \sin \theta + 2 \cos^2 \theta} = 0$$ The numerator must be 0: $$(2+2 \sin \theta) \cos \theta + 2 \sin \theta \cos \theta = 0$$ Now solve for θ to find the points at which horizontal tangents occur. Vertical Tangents: $$\frac{dx}{dy} = 0$$ $$\frac{- (2+2 \sin \theta) \sin \theta + 2 \cos^2 \theta}{(2+2 \sin \theta) \cos \theta + 2 \sin \theta \cos \theta} = 0$$ The numerator must be 0: $$- (2+2 \sin \theta) \sin \theta + 2 \cos^2 \theta = 0$$ Now solve for θ to find the points at which vertical tangents occur.

Compute the points in polar coordinates

After solving for θ, replace θ in the polar equation: $$r = 2 + 2 \sin \theta$$ to find the corresponding r values, and combine those values (r, θ) to find the points in polar coordinates where horizontal or vertical tangents occur.