Question

x=a cos^(3)t,y=a sin^(3)t

dy/dx = ?

Answer 1

Answer:

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Answer 2

$\underline{\large{\sf Answer:}}$

$\sf x = a cos^3t$,$\sf y = a sin^3t$

here we have given, parameter (t) so first differentiate the functions,

$\sf x = a cos^3t$and $\sf y = a sin^3t$ with respect to t

therefore,

x = a cos³t , Differentiate wrt t

$\implies \sf \frac{dx}{dt}=\frac{d}{dt}(a cos^2t )$

$\implies \sf a 3cos^2t (\frac{d}{dt}(cost))$

$\implies \sf (3acos^2t )(-sint)$

$\implies \sf -(3a cos^2t)(sin t)$

Now ,

y = a sin³t , Differentiate wrt t

$\implies \sf \frac{dy}{dt}=\frac{d}{dt}(a sin^3t)$

$\implies \sf 3a sin^2t (\frac{d}{dt}(sint))$

$\implies \sf 3a sin^2t (cost)$

{we know, $\sf \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ }

Therefore,

$\implies \sf \frac{dy}{dx}=\frac{3a sin^2t (cost)}{-(3a cos^2t)(sin t)}$

$\implies -(\sf tan^2t \times cot t)$

$\implies \sf -tan^2t (\frac{1}{tan t})$

$\implies {\boxed{\large{\sf -tant}}}$