What is the slope of the tangent to the curve | vedclass

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(D) Given the parametric equations of the curve are $x = a \sin^3 t$ and $y = a \cos^3 t$.First,we find the derivatives with respect to $t$:$\frac{dx}

Vedclass · Accepted Answer

$-\frac{1}{\sqrt{3}}$

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(D) Given the parametric equations of the curve are $x = a \sin^3 t$ and $y = a \cos^3 t$.First,we find the derivatives with respect to $t$:$\frac{dx}{dt} = 3a \sin^2 t \cos t$$\frac{dy}{dt} = -3a \cos^2 t \sin t$Now,the slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.$\frac{dy}{dx} = \frac{-3a \cos^2 t \sin t}{3a \sin^2 t \cos t} = -\frac{\cos t}{\sin t} = -\cot t$.At $t = \frac{\pi}{3}$,the slope is:$\left( \frac{dy}{dx} \right)_{t = \pi/3} = -\cot\left( \frac{\pi}{3} \right) = -\frac{1}{\sqrt{3}}$.

What is the slope of the tangent to the curve $x = a \sin^3 t, y = a \cos^3 t$ at $t = \frac{\pi}{3}$?

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