Question

${\textbf{\large{Solve it for 100 points }}}$

$x = \frac{ {sin}^{3}t }{ \sqrt{cos2t}}$
$y = \frac{cos {}^{3} t}{ \sqrt{cos2t} }$

Find dy /dx ​

Answer 1

Hii its tom85

your solution is attached above

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Answer 2

Answer:

If x=sin3tcos2t−−−−−√x=sin3⁡tcos⁡2t and y=cos3tcos2t−−−−−√,y=cos3⁡tcos⁡2t, Then dydxdydx in terms of tt

MyTry::MyTry::

Using The

Formula dydx=dydtdxdt.dydx=dydtdxdt.

Now Given

x=sin3tcos2t−−−−−√.

x=sin3⁡tcos⁡2t.

So

dxdt=cos2t−−−−−√⋅3sin2t⋅cost−sin3t⋅12cos2t√⋅−sin2t⋅2tcos2t

dxdt=cos⁡2t⋅3sin2⁡t⋅cos⁡t−sin3⁡t⋅12cos⁡2t⋅−sin⁡2t⋅2tcos⁡2t

So we get

dxdt=cos2t⋅3sin2t⋅cost+sin3t⋅sin2t⋅tcos2t⋅cos2t−−−−−√

dxdt=cos⁡2t⋅3sin2⁡t⋅cos⁡t+sin3⁡t⋅sin⁡2t⋅tcos⁡2t⋅cos⁡2t

Similarly Given

y=cos3tcos2t−−−−−√.

y=cos3⁡tcos⁡2t.

So

dydt=−cos2t−−−−−√⋅3cos2t⋅sint−cos3t⋅12cos2t√⋅−sin2t⋅2tcos2t

dydt=−cos⁡2t⋅3cos2⁡t⋅sin⁡t−cos3⁡t⋅12cos⁡2t⋅−sin⁡2t⋅2tcos⁡2t

So we get

dydt=−cos2t⋅3cos2t⋅sint+cos3t⋅sin2t⋅tcos2t⋅cos2t−−−−−√

dydt=−cos⁡2t⋅3cos2⁡t⋅sin⁡t+cos3⁡t⋅sin⁡2t⋅tcos⁡2t⋅cos⁡2t

So

dydx=−3cos2t⋅sint⋅cos2t+cos3t⋅sin2t⋅tcos2t⋅3sin2t⋅cost+sin3t⋅sin2t⋅t

dydx=−3cos2⁡t⋅sin⁡t⋅cos⁡2t+cos3⁡t⋅sin⁡2t⋅tcos⁡2t⋅3sin2⁡t⋅cos⁡t+sin3⁡t⋅sin⁡2t⋅t