Name: Finding the Angle Between Two Following Curves
Uploaded: 2019-07-19
Duration: 11 min 3 s
Description: In this video, you will learn how to find the angle between two following curves, through a simple method. Watch this video and learn more now.

Question

Find the equation of tangent and normal to the curve

x = 3 \cos θ - \cos^{3} θ, y = 3 \sin θ - \sin^{3} θ .

Accepted Answer

Given that $x = 3 \cos θ - \cos^{3} θ, y = 3 \sin θ - \sin^{3} θ .$

Differentiating $x$ and $y$ with respect to $θ$ we get

$\frac{dx}{dθ} = - 3 \sin θ - 3 \cos^{2} θ (- \sin θ) = - 3 \sin θ + 3 \cos^{2} θ \sin θ$

And $\frac{dy}{dθ} = 3 \cos θ - 3 \sin^{2} θ (\cos θ) = 3 \cos θ - 3 \sin^{2} θ \cos θ$

We know that $\frac{dy}{dx} = \frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}$

$= \frac{3 \cos θ - 3 \sin^{2} θ \cos θ}{3 \cos θ - 3 \sin^{2} θ \cos θ} = \frac{3 \cos θ (1 - \sin^{2} θ)}{- 3 \sin θ (1 - \cos^{2} θ)} = \frac{- \cos θ \cos^{2} θ}{\sin θ \sin^{2} θ} = - \frac{\cos^{3} θ}{\sin^{3} θ}$

The slope of tangent $(m) = {(\frac{dy}{dx})}_{θ = θ} = - \frac{\cos^{3} θ}{\sin^{3} θ}$

So the equation of the tangent is $y - y_{1} = m (x - x_{1})$

$∴ y - (3 \sin θ - \sin^{3} θ) = - \frac{\cos^{3} θ}{\sin^{3} θ} (x - (3 \cos θ - \cos^{3} θ))$

$\Rightarrow {ysin}^{3} θ - 3 \sin^{4} θ + \sin^{6} θ = - {xcos}^{3} θ + 3 \cos^{4} θ - \cos^{6} θ$

$\Rightarrow {xcos}^{3} θ + {ysin}^{3} θ + \sin^{6} θ + \cos^{6} θ = 3 (\cos^{4} θ + \sin^{4} θ)$

$\Rightarrow {xcos}^{3} θ + {ysin}^{3} θ + {(\sin^{2} θ + \cos^{2} θ)}^{3} + 3 \sin^{2} \cos^{2} θ (\sin^{2} θ + \cos^{2} θ) = 3 ({(\sin^{2} θ + \cos^{2} θ)}^{2} + 2 \cos^{2} {θsin}^{2} θ)$

$\Rightarrow {xcos}^{3} θ + {ysin}^{3} θ + 1 + 3 \sin^{2} \cos^{2} θ = 3 (1 + 2 \cos^{2} {θsin}^{2} θ)$

$\Rightarrow {xcos}^{3} θ + {ysin}^{3} θ = 2 + 3 \cos^{2} {θsin}^{2} θ$

Equation of the normal is $y - y_{1} = - \frac{1}{m} (x - x_{1})$

$∴ y - (3 \sin θ - \sin^{3} θ) = \frac{\sin^{3} θ}{\cos^{3} θ} (x - (3 \cos θ - \cos^{3} θ))$

$\Rightarrow {ycos}^{3} θ - 3 \sin {θcos}^{3} θ + \cos^{3} {θsin}^{3} θ = {xsin}^{3} θ - 3 \cos θ \sin^{3} θ + \sin^{3} θ \cos^{3} θ$

$\Rightarrow {xsin}^{3} θ - {ycos}^{3} θ - 3 \cos θ \sin θ (\sin^{2} θ - \cos^{2} θ) = 0$

Find the equation of tangent and normal to the curve $x = 3 \cos θ - \cos^{3} θ, y = 3 \sin θ - \sin^{3} θ .$

Important Questions on Tangents and Normals

Learn and Prepare for any exam you want !

Find the equation of tangent and normal to the curve x=3cos θ-cos3 θ, y=3sin θ-sin3 θ.

Important Questions on Tangents and Normals

Learn and Prepare for any exam you want !

Find the equation of tangent and normal to the curve $x = 3 \cos θ - \cos^{3} θ, y = 3 \sin θ - \sin^{3} θ .$