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SOVA2 [1]
3 years ago
14

X= a ( cost +logtan t/2 y = asin t find dy / dx​

History
1 answer:
Wewaii [24]3 years ago
5 0

Answer: \dfrac{dy}{dx} = \tan t

Explanation:

x = a(\cos t + \log \tan\dfrac{t}2)

Differentiate with respect to t,

\dfrac{dx}{dt} =a \dfrac{d{(\cos t + \log\tan \dfrac{t}2)}}{dt}  

= a[\dfrac{d(\cos t)}{dt} + \dfrac{d(\log\tan \dfrac{t}2)}{dt}]

= a[ -\sin t + \dfrac{1}{\tan  \dfrac{t}2 }\times sec^2\dfrac{t}2\times \dfrac12]

= a [ -\sin t + \dfrac12 \dfrac{1}{\sin \dfrac{t}2\cdot\cos \dfrac{t}2 }]

Since 2\sin A\cdot\cos A = \sin2 A

= a [ -\sin t + \dfrac1{\sin t}]

= a \dfrac{ 1 - \sin^2 t}{\sin t}

= a\dfrac{\cos^2t}{\sin t}

= a\cot t\cdot \cos t .......(1)

again, y = a\sin t

\dfrac{dy}{dt} = a\cos t .......(2)

now, \dfrac{dy}{dx} =\dfrac {\dfrac{dy}{dt}}{\dfrac{dx}{dt}} [from (1) and (2)]  

=\dfrac {a\cos t}{a\cot t.\cos t}  

= \dfrac{1}{\cot t }

Hence, \dfrac{dy}{dx} = \tan t

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