Question

Find the equations of tangent and normal to the curves
$x = a \: { \sin}^{3} theta \: and \: y = a \: { \cos }^{3} theta \: at \: theta = \frac{\pi}{4}$
​

Answer 1

GiveN :

• $x\ =\ a \sin^3 \theta$
• $y\ =\ a \cos^3 \theta$

To FinD :

• Equation of Tangent and Equation of normal

SolutioN :

Differentiate and y wrt. theta

$\sf{\implies\ \dfrac{dy}{d \theta}\ =\ (a \cos^2 \theta) (- \sin \theta)}$

$\sf{\implies\ \dfrac{dy}{d \theta}\ =\ -a \cos^2 \theta \sin \theta}$

$\sf{\implies\ \dfrac{dy}{d \theta}_{\bigg( At\ \theta\ =\ \dfrac{\pi}{4} \bigg) }\ =\ -a \cos^2 \bigg(\dfrac{\pi}{4} \bigg) sin \bigg(\dfrac{\pi}{4} \bigg) }$

$\sf{\implies{ \dfrac{dy}{d \theta}_{(At\ \theta\ =\ \dfrac{\pi}{4})}\ =\ -a \bigg( \dfrac{1}{\sqrt{2}} \bigg)^2 \bigg( \dfrac{1}{\sqrt{2}} \bigg)}}$

$\sf{\implies \dfrac{dy}{d \theta}_{(At\ \theta\ =\ \dfrac{\pi}{4})}\ =\ \dfrac{-a}{2 \sqrt{2}} }$

___________________

Differentiate x wrt. Theta

See Attached Picture

__________________

We know that,

$\implies \sf{\dfrac{dy}{dx}\ = \dfrac{\dfrac{dy}{d \theta}}{\dfrac{dx}{d \theta}} }$

Substituting Values ..

$\implies \sf{ \dfrac{dy}{dx}\ =\ \frac{\dfrac{-a}{2 \sqrt{2}}}{\frac{a}{2 \sqrt{2}}}}$

$\implies \sf{ \dfrac{dy}{dx}\ =\ -1}$

_____________________

So, We get :

• Equation of Tangent $\sf{x\ +\ y\ =\ \dfrac{a}{\sqrt{2}}}$

• Equation of Normal $\sf{y\ -\ x\ =\ 0}$