Question

Find $\frac{dy}{dx}$, if x = a cos³ t, y = sin³ t, at t = π/3

Answer 1

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Answer 2

Answer:

At t = $\dfrac \pi 3$,

$\rm\dfrac{dy}{dx}=-\dfrac{1.732}{a}.$

Step-by-step explanation:

Given:

• x = a cos³ t.

• y = sin³ t.

We know,  

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

Now,  

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

Therefore,

$\rm\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{ 3\sin^2t\cos t}{a\ 3\cos^2t\ (-\sin t)}=\dfrac{\sin t}{-a\cos t}=-\dfrac 1a \tan t.$

At t = $\dfrac \pi 3$,

$\rm\dfrac{dy}{dx}=-\dfrac 1a \tan \left (\dfrac \pi 3 \right )=-\dfrac{1.732}{a}.$