a curve in the plane is defined parametrically by the equations x = t^3 - 6t and y = sqrt(2 - t^2). find…

Answer:

A. $-\frac{t}{(3t^{2}-6)(\sqrt{2 - t^{2}})}$

Explanation:

Step1: Differentiate $x$ with respect to $t$

$\frac{dx}{dt}=\frac{d}{dt}(t^{3}-6t)=3t^{2}-6$

Step2: Differentiate $y$ with respect to $t$

Using the chain - rule, if $y = \sqrt{2 - t^{2}}=(2 - t^{2})^{\frac{1}{2}}$, then $\frac{dy}{dt}=\frac{1}{2}(2 - t^{2})^{-\frac{1}{2}}\cdot(-2t)=-\frac{t}{\sqrt{2 - t^{2}}}$

Step3: Use the formula $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$\frac{dy}{dx}=\frac{-\frac{t}{\sqrt{2 - t^{2}}}}{3t^{2}-6}=-\frac{t}{(3t^{2}-6)\sqrt{2 - t^{2}}}$