The slope of the tangent to the curve x = 3t^ | vedclass

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(A) Given the parametric equations of the curve are $x = 3t^2 + 1$ and $y = t^3 - 1$.First,we find the derivatives with respect to $t$:$\frac{dx}{dt}

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$0$

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(A) Given the parametric equations of the curve are $x = 3t^2 + 1$ and $y = t^3 - 1$.First,we find the derivatives with respect to $t$:$\frac{dx}{dt} = 6t$$\frac{dy}{dt} = 3t^2$Now,the slope of the tangent $\frac{dy}{dx}$ is given by:$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2}{6t} = \frac{t}{2}$ (for $t 
eq 0$).At $x = 1$,we have $3t^2 + 1 = 1$,which implies $3t^2 = 0$,so $t = 0$.As $t 	o 0$,the slope $\frac{dy}{dx} = \frac{t}{2} 	o 0$.Thus,the slope of the tangent at $x = 1$ is $0$.

The slope of the tangent to the curve $x = 3t^2 + 1, y = t^3 - 1$ at $x = 1$ is

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