Question

find d^2y / dx^2 if. 1) x=3 cost - 2 cos^3 t & y= 3sint - 2 Sin^3t

Answer 1

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

$\displaystyle\frac{dx}{dt}=-3\;sin\;t-6\;cos^2t(-sint)$

$\displaystyle\frac{dx}{dt}=-3\;sin\;t+6\;cos^2t\;sint$

$\displaystyle\frac{dx}{dt}=3\;sin\;t(2\;cos^2t-1)$

$\text{Using }\boxed{\bf\,cos2A=2\;cos^2A-1}$

$\displaystyle\frac{dx}{dt}=3\;sin\;t\;cos2t$

$\text{and}$

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

$\displaystyle\frac{dy}{dt}=3\;cos\;t-6\;sin^2t(cos\;t)$

$\displaystyle\frac{dy}{dt}=3\;cos\;t(1-2\;sin^2t)$

$\displaystyle\frac{dy}{dt}=3\;cos\;t(1-2\;sin^2t)$

$\text{Using }\boxed{\bf\,cos2A=1-2\;sin^2A}$

$\displaystyle\frac{dy}{dt}=3\;cos\;t\;cos2t$

$\text{Now}$

$\displaystyle\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$\displaystyle\frac{dy}{dx}=\frac{3\;cos\;t\;cos2t}{3\;sin\;t\;cos2t}$

$\displaystyle\frac{dy}{dx}=\frac{cos\;t}{sin\;t}$

$\displaystyle\frac{dy}{dx}=cot\;t$

$\text{Then,}$

$\displaystyle\frac{d^2y}{dx^2}=-cosec^2\;t\;\frac{dt}{dx}$

$\displaystyle\frac{d^2y}{dx^2}=\frac{-cosec^2\;t}{3\;sin\;t\;cos2t}$

$\implies\bf\frac{d^2y}{dx^2}=\frac{-1}{3\;sin^3t\;cos2t}$