x=a(t+sin t),y=a(1-cost) হলে  dy^2/dx^2   এর মান কোনটি?

Explanation:

Another Explanation (5): সমাধান: আমরা জানি, x = a(t + sin t) এবং y = a(1 - cos t) প্রথমে \(\frac{dy}{dx}\) নির্ণয় করি: \(\frac{dx}{dt} = a(1 + cos t)\) \(\frac{dy}{dt} = a sin t\) অতএব, \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a sin t}{a(1 + cos t)} = \frac{sin t}{1 + cos t}\) আমরা জানি, \(sin t = 2 sin(t/2) cos(t/2)\) এবং \(1 + cos t = 2 cos^2(t/2)\) সুতরাং, \(\frac{dy}{dx} = \frac{2 sin(t/2) cos(t/2)}{2 cos^2(t/2)} = \frac{sin(t/2)}{cos(t/2)} = tan(t/2)\) এখন, \(\frac{d^2y}{dx^2}\) নির্ণয় করি: \(\frac{d^2y}{dx^2} = \frac{d}{dx} (\frac{dy}{dx}) = \frac{d}{dx} (tan(t/2)) = \frac{d}{dt} (tan(t/2)) \cdot \frac{dt}{dx}\) আমরা জানি, \(\frac{d}{dt} (tan(t/2)) = \frac{1}{2} sec^2(t/2)\) এবং \(\frac{dx}{dt} = a(1 + cos t) = 2a cos^2(t/2)\) সুতরাং, \(\frac{dt}{dx} = \frac{1}{2a cos^2(t/2)}\) অতএব, \(\frac{d^2y}{dx^2} = \frac{1}{2} sec^2(t/2) \cdot \frac{1}{2a cos^2(t/2)} = \frac{1}{4a cos^4(t/2)}\) এখন, y = a(1 - cos t) দেওয়া আছে। সুতরাং, \(cos t = 1 - \frac{y}{a}\) আমরা জানি, \(cos t = 2 cos^2(t/2) - 1\) সুতরাং, \(2 cos^2(t/2) - 1 = 1 - \frac{y}{a}\) \(2 cos^2(t/2) = 2 - \frac{y}{a} = \frac{2a - y}{a}\) \(cos^2(t/2) = \frac{2a - y}{2a}\) সুতরাং, \(\frac{d^2y}{dx^2} = \frac{1}{4a (\frac{2a - y}{2a})^2} = \frac{1}{4a \cdot \frac{(2a - y)^2}{4a^2}} = \frac{4a^2}{4a(2a - y)^2} = \frac{a}{(2a - y)^2}\) অতএব, \(\frac{d^2y}{dx^2} = \frac{a}{(2a - y)^2}\) ✅