Question

find dy/dx
when
$x = a {cos}^{3} t \: \: and \: \: y = a {sin}^{3} t$
​

Answer 1

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Answer 2

Answer:

= - tan t

Step-by-step explanation:

We have,

$(x = a \: { \cos }^{3} t \: )(y = a \: {sin}^{3} t)$

$\frac{dx}{dt} = 3a \: {cos}^{2} t \: \frac{d}{dt} cos(t) = - 3a {cos}^{2} t \: \sin(t)$

$= \frac{dy}{dt} = 3a {sin}^{2} t \: \frac{d}{dt} (sin \: t) = 3a \: {sin}^{2} t \: cos \: t$

$= \frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } = \frac{3a \: {sin}^{2}t \: cos \: t }{ - 3a \: {cos}^{2}t \: sin \: t \: }$

= - tan t

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