The equation of the normal to the curve sin y | vedclass

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

Question

(A) Given the equation of the curve is $\sin y = \sqrt{3} x \sin \left(\frac{\pi}{6} + y\right) \quad (i)$.At $x = 0$, $\sin y = 0$, which implies $y

Vedclass · Accepted Answer

$2x + \sqrt{3}y = 0$

Vedclass · Answer

(A) Given the equation of the curve is $\sin y = \sqrt{3} x \sin \left(\frac{\pi}{6} + y\right) \quad (i)$.At $x = 0$, $\sin y = 0$, which implies $y = 0$.Now, differentiating equation $(i)$ with respect to $x$ using the product rule:$\cos y \frac{dy}{dx} = \sqrt{3} \sin \left(\frac{\pi}{6} + y\right) + \sqrt{3} x \cos \left(\frac{\pi}{6} + y\right) \frac{dy}{dx}$.At the point $(0, 0)$, substituting $x = 0$ and $y = 0$:$\cos(0) \frac{dy}{dx} = \sqrt{3} \sin \left(\frac{\pi}{6} + 0\right) + \sqrt{3}(0) \cos \left(\frac{\pi}{6} + 0\right) \frac{dy}{dx}$.$1 \cdot \frac{dy}{dx} = \sqrt{3} \sin \left(\frac{\pi}{6}\right) = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}$.Thus, the slope of the tangent $m_t = \frac{\sqrt{3}}{2}$.The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}}$.The equation of the normal at $(0, 0)$ is given by $y - y_1 = m_n(x - x_1)$:$y - 0 = -\frac{2}{\sqrt{3}}(x - 0)$.$\sqrt{3}y = -2x$, which simplifies to $2x + \sqrt{3}y = 0$.

The equation of the normal to the curve $\sin y = \sqrt{3} x \sin \left(\frac{\pi}{6} + y\right)$ at $x = 0$ is:

Similar Questions

The tangent to the curve $y=e^x$ drawn at the point $(c, e^c)$ intersects the line joining the points $(c-1, e^{c-1})$ and $(c+1, e^{c+1})$:

Find the equation of the tangent to the curve $y = be^{-x/a}$ at the point where it crosses the $y$-axis.

Let $y=e^{x^{2}}$ and $y=e^{x^{2}} \sin x$ be two given curves. Then, the angle between the tangents to the curves at any point of their intersection is:

Find the equation of the normal to the curve $2y = 3 - x^2$ at the point $(1, 1)$.

Find the slope of the tangent to the curve $y = x^{3} - 3x + 2$ at the point whose $x$-coordinate is $3$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module