The tangent to the curve y = x^{3} + 1 at (1, | vedclass

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(C) Let $\phi$ be the angle made by the tangent with the positive $x$-axis. The slope of the tangent $m$ is given by $	an \phi = \left. \frac{dy}{dx}

Vedclass · Accepted Answer

$-\frac{1}{3}$

Vedclass · Answer

(C) Let $\phi$ be the angle made by the tangent with the positive $x$-axis. The slope of the tangent $m$ is given by $	an \phi = \left. \frac{dy}{dx} ight|_{(1, 2)}$.Given $y = x^{3} + 1$,we have $\frac{dy}{dx} = 3x^{2}$.At $(1, 2)$,the slope $m = 	an \phi = 3(1)^{2} = 3$.From the geometry of the line and the $y$-axis,the angle $	heta$ made with the $y$-axis satisfies $	heta + \phi = 90^{\circ}$ or $	heta = 90^{\circ} - \phi$ (depending on the orientation). Based on the provided diagram,the angle $	heta$ is the obtuse angle between the tangent and the positive $y$-axis,such that $	heta = 90^{\circ} + \phi$.Then,$	an 	heta = 	an(90^{\circ} + \phi) = -\cot \phi$.Since $	an \phi = 3$,we have $\cot \phi = \frac{1}{3}$.Therefore,$	an 	heta = -\frac{1}{3}$.

The tangent to the curve $y = x^{3} + 1$ at $(1, 2)$ makes an angle $\theta$ with the $y$-axis. Then the value of $\tan \theta$ is

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