Question

If x = a cos³ t, y = a sin³ t, show that $\frac{dy}{dx}= -(\frac{y}{x})^{\frac{1}{3} }$

Answer 1

Answer:

Given:

• x = a cos³ t.
• y = a sin³ t.

We know,

$\rm \dfrac {dy}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$.

Now,

$\rm \dfrac{dy}{dt} = \dfrac{d}{dt}(a\sin^3t)\\= a\dfrac{d}{dt}(\sin^3t)\\= a\ 3\sin^2t\dfrac{d}{dt}(\sin t)\\= a\ 3\sin^2t\cos t.\\\\\dfrac{dx}{dt} = \dfrac{d}{dt}(a\cos^3t)\\= a\dfrac{d}{dt}(\cos^3t)\\= a\ 3\cos^2t\dfrac{d}{dt}(\cos t)\\= a\ 3\cos^2t\ (-\sin t).$

We have to show that:

$\rm \dfrac{dy}{dx}=-\left (\dfrac{y}{x} \right )^{\frac 13}.$

LHS of the equation:

$\rm\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{a\ 3\sin^2t\cos t}{a\ 3\cos^2t\ (-\sin t)}=\dfrac{-\sin t}{\cos t}=-\tan t.$

RHS of the equation:

$\rm -\left (\dfrac{y}{x} \right )^{\frac 13}=-\left (\dfrac{a\sin^3 t}{a\ cos^3 t} \right )^{\frac 13}\\=-\left (\tan^3 t \right )^{\frac 13}\\=-\tan t.$

Clearly, LHS = RHS.