Question

If x=3cott-2cos^3t, y=3sint-2sin^3t , find d^2y/dx^2.

Answer 1

Question:

$\sf If \ x=3 \cos t-2 \cos^3 t \ , y = 3\sin t- 2\sin^3 t$

$\sf Find \ \dfrac{d^2y}{dx^2}$

Answer:

$\sf \dfrac{d^2y}{dx^2} =\dfrac{-1}{(3\sin^3 t.\cos 2t)}$

Step-by-step explanation:

Given :

$\sf x = 3 \cos t - 2 \cos^3 t$

Diff. w.r.t. t :

$\sf \dfrac{dx}{dt} = \dfrac{d}{dt}(3 \cos t - 2 \cos^3 t)$

Applying power and rule rule :

$\sf \dfrac{dx}{dt} =- 3 \sin t - 2 ( 3. \cos^2 t ( - \sin t ))$

$\sf \dfrac{dx}{dt} = 6\cos^2 t \sin t - 3 \sin t$

Also given :

$\sf y= 3 \sin t- 2 \sin^3 t$

Diff w.r.t. t

$\sf \dfrac{dy}{dt} = \dfrac{d}{dt}(3 \sin t- 2 \sin^3 t)$

Again applying power and chain rule :

$\sf \dfrac{dy}{dt} =3 \cos t - 2 ( 3 \sin^2 t. \cos t)$

$\sf \dfrac{dy}{dt} =3 \cos t - 6 \sin^2 t. \cos t$

Now as we do in parametric function :

$\sf \dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt} }{\dfrac{dx}{dt} }$

Putting value of above here we get :

$\sf \dfrac{dy}{dx} =\dfrac{3 \cos t - 6 \sin^2 t. \cos t}{6\cos^2 t \sin t - 3 \sin t}$

$\sf \dfrac{dy}{dx} =\dfrac{3 \cos t(1 - 2 \sin^2 t)}{3\sin t(2 \cos^2 t-1)}$

Using multiple angle formula :

$\sf 2 \cos^2 x-1= \cos 2 x$

$\sf 1 - 2 \sin^2 x= \cos 2 x$

$\sf \dfrac{dy}{dx} =\dfrac{ \cos t(\cos 2 t)}{\sin t( \cos 2t)}$

$\sf \dfrac{dy}{dx} = \cot t$

Now second order derivative :

$\sf \dfrac{d^2y}{dx^2} = \dfrac{d}{dx}(\cot t )$

Diff. w.r.t. x

$\sf \dfrac{d^2y}{dx^2} = -\csc^2 .\dfrac{dt}{dx}$

We already have value of derivative 'x' w.r.t. 't' :

$\sf \dfrac{d^2y}{dx^2} = -\csc^2 .\dfrac{1}{(6 \cos^2 t. \sin t - 3 \sin t )}$

$\sf \dfrac{d^2y}{dx^2} =\dfrac{ -\csc^2 }{(3\sin t(2 \cos^2 t- 1)}$

$\sf \dfrac{d^2y}{dx^2} =\dfrac{-1}{(3\sin t.\cos 2t). \sin^2 t}$

$\sf \dfrac{d^2y}{dx^2} =\dfrac{-1}{(3\sin^3 t.\cos 2t)}$

Hence we get required answer.