অব্যক্ত ও পরামিতিক ফাংশনের অন্তরীকরণ (Differentiation of implicit and Parametric function)

অনুশীলনী \(9.F / Q.3\)-এর প্রশ্নসমুহ
নিচের পরামিতিক সমীকরণ হতে \(\frac{dy}{dx}\) নির্ণয় করঃ
\(Q.3.(i)\) \(x=a(\theta+\sin{\theta}), y=a(1+\cos{\theta})\)
উত্তরঃ \(\frac{dy}{dx}=-\tan{\frac{\theta}{2}}\)

\(Q.3.(ii)\) \(x=a(\cos{\theta}+\theta\sin{\theta}), y=a(\sin{\theta}-\theta\cos{\theta})\) যখন \(\theta=\frac{3\pi}{4}\)
উত্তরঃ \(\frac{dy}{dx}=-1\)

\(Q.3.(iii)\) \(x=\tan^{-1}\sqrt{\frac{1-\cos{\theta}}{1+\cos{\theta}}}\), \(y=\tan^{-1}\frac{\cos{\theta}}{1+\sin{\theta}}\)
উত্তরঃ \(\frac{dy}{dx}=-1\)

\(Q.3.(iv)\) \(x=a\cos{\theta}+b\sin{\theta}\), \(y=a\sin{\theta}-b\cos{\theta}\)
উত্তরঃ \(\frac{dy}{dx}=\frac{a\cos{\theta}+b\sin{\theta}}{b\cos{\theta}-a\sin{\theta}}\)

\(Q.3.(v)\) \(x=a\sec^3{\theta}\), \(y=a\tan^3{\theta}\) যখন \(\theta=\frac{\pi}{4}\)
উত্তরঃ \(\frac{dy}{dx}=\frac{2}{3}\)

\(Q.3.(vi)\) \(\tan{y}=\frac{2t}{1-t^2}\) এবং \(\sin{x}=\frac{2t}{1+t^2}\)
উত্তরঃ \(\frac{dy}{dx}=1\)

\(Q.3.(vii)\) \(x=a\cos^3{\theta}\) এবং \(y=a\sin^3{\theta}\)
উত্তরঃ \(\frac{dy}{dx}=-\tan{\theta}\)

\(Q.3.(viii)\) \(x=e^t\cos{t}\) এবং \(y=e^t\sin{t}\)
উত্তরঃ \(\frac{dy}{dx}=\frac{\sin{t}+\cos{t}}{\cos{t}-\sin{t}}\)

\(Q.3.(ix)\) \(x=a\cos{t}\) এবং \(y=b\sin{t}\)
উত্তরঃ \(\frac{dy}{dx}=-\frac{b}{a}\cot{t}\)

\(Q.3.(x)\) \(x=\sqrt{t}\) এবং \(y=t-\frac{1}{\sqrt{t}}\)
উত্তরঃ \(\frac{dy}{dx}=2\sqrt{t}+\frac{1}{t}\)
\(Q.3.(xi)\) \(x=2\sin{t}\) এবং \(y=\cos{2t}\)
উত্তরঃ \(\frac{dy}{dx}=-2\sin{t}\)

\(Q.3.(xii)\) \(x=\cos^3{\theta}\) এবং \(y=\sin^3{\theta}\)
উত্তরঃ \(\frac{dy}{dx}=-\tan{\theta}\)

\(Q.3.(xiii)\) \(x=a(\cos{\phi}+\phi\sin{\phi}), y=a(\sin{\phi}-\phi\cos{\phi})\)
উত্তরঃ \(\frac{dy}{dx}=\tan{\phi}\)

\(Q.3.(xiv)\) \(x=\frac{3at}{1+t^3}, y=\frac{3at^2}{1+t^3}\)
উত্তরঃ \(\frac{dy}{dx}=\frac{ay-x^2}{y^2-ax}\)

\(Q.3.(xv)\) \(x=a(\theta-\sin{\theta}), y=a(1-\cos{\theta})\)
উত্তরঃ \(\frac{dy}{dx}=\cot{\frac{\theta}{2}}\)

\(Q.3.(xvi)\) \(x=\frac{a\cos{t}}{t}, y=\frac{a\sin{t}}{t}\)
উত্তরঃ \(\frac{dy}{dx}=\frac{\sin{t}-t\cos{t}}{t\sin{t}+\cos{t}}\)

\(Q.3.(xvii)\) \(x=a(\cos{t}+t\sin{t}), y=a(\sin{t}-t\cos{t})\)
উত্তরঃ \(\frac{dy}{dx}=\tan{t}\)

\(Q.3.(xviii)\) \(x=a\cos{\theta}\) এবং \(y=a\sin{\theta}\)
উত্তরঃ \(\frac{dy}{dx}=-\cot{\theta}\)

\(Q.3.(xix)\) \(x=a\sec{\phi}\) এবং \(y=b\tan{\phi}\)
উত্তরঃ \(\frac{dy}{dx}=\frac{b}{a} cosec \ {\phi}\)
\(Q.3.(xx)\) \(x=a(t-\sin{t}), y=a(1+\cos{t})\) হলে, দেখাও যে, \(t=\frac{5\pi}{3}\) যখন \(\frac{dy}{dx}=\sqrt{3}\)
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