Question

Find dy/dx , if x = acos¢ , y = asin¢ ​

Answer 1

Hi !!

Solution :- Given that ,

x = acos¢ , y = asin¢

therefore dx/d¢ = - asin¢ , dy/d¢ = acos¢

Hence, dy/dx = dy/d¢ /dx/d$ = acos¢/-asin¢ = - cot¢ Answer

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Hope it helps you !!!

@Raj ❤

Answer 2

$\bold{\huge{\underline{Answer-}}}$

-cot

$\bold{\huge{\underline{\underline{Used\:Formulas}}}}$

$\boxed{1. \tt \: \frac{d \: sin \theta}{d \theta} = cos \theta } \\ \boxed{2. \tt \frac{d \: cos \theta}{d \theta} = - sin \theta}$

$\bold{\huge{\underline{Solution-}}}$

Given - x = acos$\theta$ , y = asin$\theta$

Now we need to find dy/dx but our Differentiability is based on $\theta$ so we need to divide dy/dx by d$\theta$ let's see

$\implies \sf \frac{dy}{dx} = \frac{ \frac{dy}{d \theta} }{ \frac{dx}{d \theta} } \\ \\ \implies \sf \: \frac{ \frac{d \: asin \theta}{d \theta} }{ \frac{d \: acos \theta}{d \theta} } \\ \\ \implies \sf \: \frac{a \frac{ \: dsin \theta}{d \theta} }{a \frac{dcos \theta}{d \theta} } \\ \\ \implies \sf \: \frac{cos \theta}{ - sin \theta} \\ \\ \implies \sf \: - cot \theta$