Question

Find the equation of tangent to curve x=acost + atsint and y=asint-atcost at any point t​

Answer 1

Explanation:-

$\large\rm { x = a \ \cos \ t + \ a \ \sin \ t}$

$\large\rm { \therefore x = a(t + \sin \ t + \cos \ t)}$

$\large\rm { y = a \ \sin \ t - a \ t \ \cos \ t}$

$\large\rm {\therefore y = a(1+ \sin \ t)^{2}}$

Now,

$\large\rm { \frac{dx}{dt} = 2a \ \cos^{2} t}$

$\large\rm { \frac{dy}{dt} = 2a( \cos \ t + \cos \ t \ \sin \ t}$

$\large\rm { \implies \frac{dy}{dx} = \frac{1 + \sin \ t}{ \cos \ t}}$

$\LARGE\rm { = \frac{ ( \cos \frac{ t }{ 2 } + \sin \frac{ t }{ 2 } ) ^ { 2 } }{ \cos ^ { 2 } \frac{ t }{ 2 } - \sin ^ { 2 } \frac{ t }{ 2 } } }$

$\large\rm { = \tan \bigg ( \frac{ \pi}{4} + \frac{t}{2} \bigg ) }$

So, slope of tangent to the curve = $\rm { = \tan \bigg ( \frac{ \pi}{4} + \frac{t}{2} \bigg ) }$

So, Equation for tangent to the curve

$\large\boxed{\rm{ \psi = \frac{1}{4} ( \pi + 2t) }}$