Find the equation of the tangent to the curve x = θ + sin θ, y = 1 +

1.	Find the equation of the tangent to the curve x = θ + sin θ, y = 1 + cos θ at θ = π/4.
Answer» finding slope of the tangent by differentiating x and y with respect to theta \(\frac{dx}{d\theta}=1+cos\theta\) \(\frac{dy}{d\theta}=-sin\theta\) Dividing both the above equations \(\frac{dy}{dx}=-\frac{sin\theta}{1+cos\theta}\) m at theta ( \(\pi/4\) ) = \(-1+\frac{1}{\sqrt{2}}\) equation of tangent is given by y – y₁= m(tangent)(x – x₁) \(y-1-\frac{1}{\sqrt{2}}\)\(=(-1+\frac{1}{\sqrt{2}})(x-\frac{\pi}{\sqrt{4}}-\frac{1}{\sqrt{2}})\)

Find the equation of the tangent to the curve x = θ + sin θ, y = 1 + cos θ at θ = π/4.

Answer»

finding slope of the tangent by differentiating x and y with respect to theta

\(\frac{dx}{d\theta}=1+cos\theta\)

\(\frac{dy}{d\theta}=-sin\theta\)

Dividing both the above equations

\(\frac{dy}{dx}=-\frac{sin\theta}{1+cos\theta}\)

m at theta ( \(\pi/4\) ) = \(-1+\frac{1}{\sqrt{2}}\)

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

\(y-1-\frac{1}{\sqrt{2}}\)\(=(-1+\frac{1}{\sqrt{2}})(x-\frac{\pi}{\sqrt{4}}-\frac{1}{\sqrt{2}})\)