Find the equation of the tangent and the normal to the curves at the i

1.	Find the equation of the tangent and the normal to the curves at the indicated points: x = a(θ + sinθ), y = a(1 – cosθ) at θ
Answer» Given as x = a(θ + sinθ), y = a(1 – cosθ) at θ Differentiate with respect θ, to get the slope of the tangent dx/dθ = a(1 + cosθ) dy/dθ = a(sinθ) Dividing dy/dθ and dx/dθ to obtain the slope of tangent dy/dx = sinθ/(1 + cosθ) m(tangent) at θ is sinθ/(1 + cosθ) The normal is perpendicular to tangent therefore, m₁m₂ = – 1 m(normal) at θ is -sinθ/(1 + cosθ) The equation of tangent is given by y – y₁ = m(tangent)(x – x₁) y - a(1 - cosθ) = (sinθ/(1 + cosθ))(x - a(θ + sinθ)) The equation of normal is given by y – y₁ = m(normal)(x – x₁) y - a(1 - cosθ) = ((1 + cosθ)/sinθ)(x - a(θ + sinθ)

Find the equation of the tangent and the normal to the curves at the indicated points: x = a(θ + sinθ), y = a(1 – cosθ) at θ

Answer»

Given as x = a(θ + sinθ), y = a(1 – cosθ) at θ

Differentiate with respect θ, to get the slope of the tangent

dx/dθ = a(1 + cosθ)

dy/dθ = a(sinθ)

Dividing dy/dθ and dx/dθ to obtain the slope of tangent

dy/dx = sinθ/(1 + cosθ)

m(tangent) at θ is sinθ/(1 + cosθ)

The normal is perpendicular to tangent therefore, m₁m₂ = – 1

m(normal) at θ is -sinθ/(1 + cosθ)

The equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y - a(1 - cosθ) = (sinθ/(1 + cosθ))(x - a(θ + sinθ))

The equation of normal is given by y – y₁ = m(normal)(x – x₁)

y - a(1 - cosθ) = ((1 + cosθ)/sinθ)(x - a(θ + sinθ)