please help!! urgent

Question

Find the points on the given curve where the tangent line is horizontal or vertical:

𝑟 = cos 𝜃 + sin 𝜃.

Answer 1

x = (cosθ + sinθ)cosθ; y = (cosθ + sinθ)sinθ; dy/dx = (dy/dθ)/(dx/dθ) = (cos2θ + sin2θ)/(cos2θ - sin2θ);

Horizontal tangent line: dy/dx = 0; cos2θ + sin2θ = 0; tan2θ = -1; 2θ = 3π/4, θ = 3π/8 or 2θ = 7π/4, θ = 7π/8

x = (1 - √2/2)/2 + 1/2·√2/2 = 1/2; y = 1/2·√2/2 + (1 + √2/2)/2 = (1 + √2)/2.

The same calculation you do for θ = 7π/8 to get 2nd point where dy/dx = 0

Vertical tangent line: dy/dx undefined; cos2θ - sin2θ = 0; tan2θ = 1; 2θ = π/4, θ = π/8; or 2θ = 5π/4, θ = 5π/4

and you evaluate x any coordinate of points where tangent line is vertical

Answer 2

Let x = rcosθ and y = rsinθ

So, x² + y² = r²cos²θ + r²sin²θ = r²(cos²θ + sin²θ) = r²(1) = r²

r = cosθ + sinθ

r² = rcosθ + rsinθ

So, x² + y² = x + y

2x + 2y(dy/dx) = 1 + (dy/dx)

(dy/dx)(2y - 1) = 1 - 2x

dy/dx = (1 - 2x) / (2y - 1)

Vertical tangent when dy/dx is undefined. So, y = 1/2.

x² + 1/4 = x + 1/2

x² - x - 1/4 = 0

Using the quadratic formula, x = (1±√2)/2

So, vertical tangents at ((1+√2)/2, 1/2) and ((1-√2)/2, 1/2)

Horizontal tangent when dy/dx = 0. So, x = 1/2.

1/4 + y² = 1/2 + y

y² - y - 1/4 = 0

Horizontal tangents at (1/2, (1+√2)/2) and (1/2, (1-√2)/2)