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(A) Given equations are $x = \sin t$ and $y = \cos 2t$. Differentiating $x$ with respect to $t$: $\frac{dx}{dt} = \frac{d}{dt}(\sin t) = \cos t$. Diff

Vedclass · Accepted Answer

$-4 \sin t$

Vedclass · Answer

(A) Given equations are $x = \sin t$ and $y = \cos 2t$. Differentiating $x$ with respect to $t$: $\frac{dx}{dt} = \frac{d}{dt}(\sin t) = \cos t$. Differentiating $y$ with respect to $t$: $\frac{dy}{dt} = \frac{d}{dt}(\cos 2t) = -\sin 2t \cdot \frac{d}{dt}(2t) = -2 \sin 2t$. Using the chain rule for parametric differentiation: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-2 \sin 2t}{\cos t}$. Using the trigonometric identity $\sin 2t = 2 \sin t \cos t$: $\frac{dy}{dx} = \frac{-2(2 \sin t \cos t)}{\cos t} = -4 \sin t$.

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x = \sin t, y = \cos 2t$.

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