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The slope of the tangent to the curve $x=3 t^2+1, y=t^3-1$ at $x=1$ is
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$0$
$\begin{aligned}& x=3 t^2+1, y=t^3-1 \\
& \frac{d x}{d t}=6 t, \quad \frac{d y}{d t}=3 t^2 \\
& \frac{d y}{d x}=\left(\frac{\frac{d y}{d t}}{\frac{d x}{dt}}\right) \left(\frac{3 t^2}{d t}\right)=\frac{t}{2 t}\end{aligned}$
For $x=1,3 t^2+1=1 \Rightarrow t=0 \Rightarrow$ Slope $=\frac{0}{2}=0$
& \frac{d x}{d t}=6 t, \quad \frac{d y}{d t}=3 t^2 \\
& \frac{d y}{d x}=\left(\frac{\frac{d y}{d t}}{\frac{d x}{dt}}\right) \left(\frac{3 t^2}{d t}\right)=\frac{t}{2 t}\end{aligned}$
For $x=1,3 t^2+1=1 \Rightarrow t=0 \Rightarrow$ Slope $=\frac{0}{2}=0$
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