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प्रश्न
Find \[\frac{dy}{dx}\] at \[t = \frac{2\pi}{3}\] when x = 10 (t – sin t) and y = 12 (1 – cos t).
उत्तर
x = 10(t – sint) and y = 12(1 – cos t)
\[\frac{dx}{dt} = 10 - 10\cos t\]
\[\frac{dy}{dt} = 0 + 12\sin t = 12\sin t\]
\[ \therefore \frac{dy}{dx} = \frac{12\sin t}{10 - 10\cos t}\]
\[\text { We have to find the value of } \frac{dy}{dx} at \ t = \frac{2\pi}{3}\]
\[\frac{dy}{dx} = \frac{12\sin t}{10 - 10\cos t}\]
\[ = \frac{12\sin\frac{2\pi}{3}}{10 - 10\cos\frac{2\pi}{3}}\]
\[ = \frac{12 \times \frac{\sqrt{3}}{2}}{10 - 10 \times \frac{- 1}{2}}\]
\[ = \frac{6\sqrt{3}}{15}\]
\[ = \frac{2\sqrt{3}}{5}\]
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