যদি x = a cos t, y = b sin t হল, তাহলে \(\rm \dfrac{d^2y}{dx^2}\) এর মান কত হবে?

ধারণা:

অবকলনের শৃঙ্খল সূত্র:

• \(\rm \dfrac{d}{dx}\ f[g(x)]=\dfrac{d}{d\ g(x)}f[g(x)]\times \dfrac{d}{dx}g(x)\)
• \(\rm \dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}\)

ত্রিকোণমিতিক অপেক্ষকের অবকলন:

• \(\rm \dfrac{d}{dx}\sin x=\cos x\ \ \ \ \ \ \ \ \ \ \ \ \ \dfrac{d}{dx}\cos x=-\sin x\\ \dfrac{d}{dx}\tan x=\sec^2x\ \ \ \ \ \ \ \ \ \ \ \dfrac{d}{dx}\cot x=-\csc^2 x\\ \dfrac{d}{dx}\sec x=\tan x\sec x\ \ \ \ \dfrac{d}{dx}\csc x=-\cot x\csc x\)

গণনা:

x = a cos t

⇒ \(\rm \dfrac{dx}{dt}=a\dfrac{d}{dt}\cos t=-a\sin t\)

y = b sin t

⇒ \(\rm \dfrac{dy}{dt}=b\dfrac{d}{dt}\sin t=b\cos t\)

এখন, \(\rm \dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}\)

= \(\rm b\cos t\times \dfrac{1}{-a\sin t}\)

= \(\rm -\dfrac{b}{a}\times\cot t\)

এবং, \(\rm \dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)\)

= \(\rm \left (-\dfrac{b}{a}\right)\left(\dfrac{d}{dx}\cot t\right)\)

= \(\rm -\left(\dfrac{b}{a}\right)\times\left(\dfrac{d}{dt}\cot t\right)\times\dfrac{dt}{dx}\)

= \(\rm \left(-\dfrac{b}{a}\right)\times( -\csc^2t)\times\dfrac{1}{-a\sin t}\)

= \(\rm \dfrac{b}{a^2}\times\dfrac{-1}{\sin^3t}\)

= \(\rm \dfrac{b}{a^2}\times\dfrac{-1}{\left(\tfrac{y}{b}\right)^3}\)

= \(\rm \dfrac{-b}{a^2}\times{\left(\dfrac{b}{y}\right)^3}\)

= \(-\rm \dfrac{b^4}{a^2y^3}\).