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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

If $ST$ and $SN$ are the lengths of the subtangent and the subnormal at the point $\theta = \frac{\pi}{2}$ on the curve $x = a(\theta + \sin \theta ), y = a(1 - \cos \theta ), a \neq 0$,then:

If $2y = 3x - 1$ is a tangent drawn to the curve $y^2 = ax^3 + b$ at $(1, 1)$,where $a$ and $b$ are constants,then $(a, b) = $

Find the points on the curve $y=x^{3}$ at which the slope of the tangent is equal to the $y$-coordinate of the point.

The normal to the curve $y(x - 2)(x - 3) = x + 6$ at the point where the curve intersects the $y$-axis passes through the point:

If the tangent to the curve $x^{2/3} + y^{2/3} = a^{2/3}$ meets the $X$-axis at $A$ and $Y$-axis at $B$,then $AB =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module