Question

Find the values of θ at which there are horizontal tangent lines on the graph of r = 1 + cos θ.

Answer 1

Answer:

θ = π/3 and 5π/3

Step-by-step explanation:

x = r cos θ and y = r sin θ.  Therefore:

dy/dx = (dy/dθ) / (dx/dθ)

dy/dx = (r cos θ + r' sin θ) / (-r sin θ + r' cos θ)

In this case, r = 1 + cos θ and r' = -sin θ.

dy/dx = ((1 + cos θ) cos θ + (-sin θ) sin θ) / (-(1 + cos θ) sin θ + (-sin θ) cos θ)

dy/dx = (cos θ + cos²θ − sin²θ) / (-sin θ − sin θ cos θ − sin θ cos θ)

dy/dx = (cos θ + cos²θ − (1 − cos²θ)) / (-sin θ (1 + 2 cos θ))

dy/dx = (cos θ + 2 cos²θ − 1) / (-sin θ (1 + 2 cos θ))

The tangent line is horizontal when the numerator of dy/dx is 0 and the denominator of dy/dx is not 0.

0 = 2 cos²θ + cos θ − 1

0 = (cos θ + 1) (2 cos θ − 1)

cos θ = -1 or 1/2

θ = π/3, π, 5π/3

0 = -sin θ (1 + 2 cos θ)

sin θ = 0 or cos θ = -1/2

θ = 0, 2π/3, π, 4π/3

Therefore, the answer is θ = π/3 and 5π/3.