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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
0
Given curve is $x=3 t^2+1$
Second curve is $y=t^3-1$
$\therefore \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=3 t^2 \times \frac{1}{6 t}=\frac{t}{2}$
But from (i) when $x=1$
we have $1=3 t^2+1 \Rightarrow 3 t^2=0 \Rightarrow t=0$
$\therefore$ When $\mathrm{x}=1$ then $\mathrm{t}=0 \quad \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\mathbf{0}$
Hence, slope of the tangent to the curve $=0$
Second curve is $y=t^3-1$
$\therefore \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=3 t^2 \times \frac{1}{6 t}=\frac{t}{2}$
But from (i) when $x=1$
we have $1=3 t^2+1 \Rightarrow 3 t^2=0 \Rightarrow t=0$
$\therefore$ When $\mathrm{x}=1$ then $\mathrm{t}=0 \quad \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\mathbf{0}$
Hence, slope of the tangent to the curve $=0$
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