If x=t^2+t+1 and y=sin left(frac{t pi}{2}righ | vedclass

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(D) Given $x = t^2 + t + 1$ and $y = \sin \left(\frac{t \pi}{2}\right) + \cos \left(\frac{t \pi}{2}\right)$.First,differentiate $x$ with respect to $t

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$\frac{-\pi}{6}$

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(D) Given $x = t^2 + t + 1$ and $y = \sin \left(\frac{t \pi}{2}\right) + \cos \left(\frac{t \pi}{2}\right)$.First,differentiate $x$ with respect to $t$:$\frac{dx}{dt} = \frac{d}{dt}(t^2 + t + 1) = 2t + 1$.Next,differentiate $y$ with respect to $t$:$\frac{dy}{dt} = \frac{d}{dt}\left(\sin \left(\frac{t \pi}{2}\right) + \cos \left(\frac{t \pi}{2}\right)\right) = \frac{\pi}{2} \cos \left(\frac{t \pi}{2}\right) - \frac{\pi}{2} \sin \left(\frac{t \pi}{2}\right)$.Now,find $\frac{dy}{dx}$ using the chain rule:$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{\pi}{2} \left(\cos \left(\frac{t \pi}{2}\right) - \sin \left(\frac{t \pi}{2}\right)\right)}{2t + 1}$.At $t=1$:$\frac{dy}{dx} = \frac{\frac{\pi}{2} \left(\cos \left(\frac{\pi}{2}\right) - \sin \left(\frac{\pi}{2}\right)\right)}{2(1) + 1} = \frac{\frac{\pi}{2} (0 - 1)}{3} = \frac{-\pi/2}{3} = -\frac{\pi}{6}$.Thus,the correct option is $D$.

If $x=t^2+t+1$ and $y=\sin \left(\frac{t \pi}{2}\right)+\cos \left(\frac{t \pi}{2}\right)$,then find the value of $\frac{dy}{dx}$ at $t=1$.

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