Question

differentiate krna h.......n


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jldi se ans dedo yrrr.....​

Answer 1

Answer:

$\large\boxed{\sf{(\frac{b}{a},t=\frac{\pi}{4})\:\:(0,t=\frac{\pi}{3})}}$

Step-by-step explanation:

It's being given that,

$\sf{x = a \sin2t(1 + \cos2t)} \\ \\ \sf{= > x = a \sin2t + a \sin(2t) \cos(2t) \times \frac{2}{2} } \\ \\ \sf{= > x = a \sin(2t) + \frac{a}{2} \sin(4t) } \\ \\ \sf{ = > \frac{dx}{dt} = a \cos(2t) \times 2 + \frac{a}{2} \cos(4t) \times 4 }\\ \\ \sf{= > \frac{dx}{dt} = 2a \cos(2t) + 2a \cos(4t) \: \: \: \: \: \: \: ............(1)}$

Also, it's being given that,

$\sf{y = b \cos2t(1 - \cos2t)} \\ \\ \sf{= > y = b \cos2t - b { \cos }^{2} 2t} \\ \\ \sf{ = > \frac{dy}{dt} = b( - \sin2t) \times 2 - b(2 \cos2t) ( - \sin2t) \times 2} \\ \\ \sf{ = > \frac{dy}{dt} = 2b \sin(4t) - 2b \sin(2t) \: \: \: \: \: ...........(2)}$

From $\sf{(1)}$ and $\sf{(2)}$,

$\sf{= > \dfrac{dy}{dx} = \dfrac{ \frac{dy}{dt} }{ \frac{dx}{dt} } } \\ \\ \sf{ = > \frac{dy}{dx} = \frac{2b \sin(4t) - 2b \sin(2t) }{2a \cos(2t) + 2a \cos(4t) }} \\ \\ \sf{ = > \frac{dy}{dx} = \frac{b \sin(4t) - b \sin(2t) }{a \cos(2t) + a \cos(4t) } }$

Now, when t = π/4

$\sf{= > \frac{dy}{dx} = \frac{b \sin(\pi) - b \sin( \frac{\pi}{2} ) }{a \cos( \frac{\pi}{2} ) + a \cos(\pi) } }\\ \\ \sf{= > \frac{dy}{dx} = \frac{ - b}{ - a} = \frac{b}{a} \: \: \: \: \: \: \: \: (t = \frac{\pi}{4} )}$

Now, when t = π/3

$\sf{= > \dfrac{dy}{dx} = \dfrac{b \sin( \frac{4\pi}{3} ) - b \sin( \frac{2\pi}{3} ) }{a \cos( \frac{2\pi}{3} ) + a \cos( \frac{4\pi}{3} ) } } \\ \\ \sf{ = > \dfrac{dy}{dx} = \dfrac{ - \frac{ \sqrt{3} b}{2} + \frac{ \sqrt{3}b }{2} }{a \cos( \frac{2\pi}{3} ) + a \cos( \frac{4\pi}{3} ) } }\\ \\ \sf{= > \frac{dy}{dx} = \frac{0}{a \cos( \frac{2\pi}{3} ) + a \cos( \frac{4\pi}{3} ) } } \\ \\ \sf{ = > \frac{dy}{dx} = 0 \: \: \: \: \: \: \: \: (t = \frac{\pi}{3} )}$

Answer 2

Answer:

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