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Question

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


A
12
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B
0
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C
2
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D
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Solution

The correct option is D $$0$$
Given curves are,
$$x=3t^2+1$$           ---- ( 1 )
$$y=t^3-1$$             ---- ( 2 )
Substituting $$x=1$$ in ( 1 ) we get,
$$\Rightarrow$$  $$3t^2+1=1$$
$$\Rightarrow$$  $$3t^2=0$$
$$\Rightarrow$$  $$t=0$$
Differentiate ( 1 ) w.r.t. $$t,$$ we get
$$\Rightarrow$$  $$\dfrac{dx}{dt}=6t$$
Differentiate ( 2 ) w.r.t. $$t,$ we get
$$\Rightarrow$$  $$\dfrac{dy}{dt}=3t^2$$

$$\Rightarrow$$  $$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{3t^2}{6t}$$

$$\therefore$$  $$\dfrac{dy}{dx}=\dfrac{t}{2}$$

Slope of the tangent $$=\left(\dfrac{dy}{dx}\right)_{t=0}=\dfrac{0}{2}=0$$


Mathematics

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