The equation of the normal drawn to the curve | vedclass

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Question

(B) Given curve is $y = \sin 3x$ ... $(i)$At $x = \frac{\pi}{4}$, $y = \sin\left(3 \cdot \frac{\pi}{4}\right) = \sin\left(\frac{3\pi}{4}\right) = \fra

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$y = \frac{\sqrt{2}}{3}\left(x + \frac{6-\pi}{4}\right)$

Vedclass · Answer

(B) Given curve is $y = \sin 3x$ ... $(i)$At $x = \frac{\pi}{4}$, $y = \sin\left(3 \cdot \frac{\pi}{4}\right) = \sin\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}}$.So, the point is $\left(\frac{\pi}{4}, \frac{1}{\sqrt{2}}\right)$.Differentiating equation $(i)$ with respect to $x$:$\frac{dy}{dx} = 3 \cos 3x$At $x = \frac{\pi}{4}$, the slope of the tangent is $m_T = \left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}} = 3 \cos\left(\frac{3\pi}{4}\right) = 3 \left(-\frac{1}{\sqrt{2}}\right) = -\frac{3}{\sqrt{2}}$.The slope of the normal $m_N$ is given by $m_N = -\frac{1}{m_T} = -\frac{1}{-3/\sqrt{2}} = \frac{\sqrt{2}}{3}$.The equation of the normal at $\left(\frac{\pi}{4}, \frac{1}{\sqrt{2}}\right)$ is:$y - \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{3} \left(x - \frac{\pi}{4}\right)$$y = \frac{\sqrt{2}}{3}x - \frac{\sqrt{2}\pi}{12} + \frac{1}{\sqrt{2}}$$y = \frac{\sqrt{2}}{3}x - \frac{\sqrt{2}\pi}{12} + \frac{6}{6\sqrt{2}}$$y = \frac{\sqrt{2}}{3}x - \frac{\sqrt{2}\pi}{12} + \frac{6\sqrt{2}}{12}$$y = \frac{\sqrt{2}}{3}x + \frac{\sqrt{2}(6-\pi)}{12}$$y = \frac{\sqrt{2}}{3} \left(x + \frac{6-\pi}{4}\right)$.

The equation of the normal drawn to the curve $y = \sin 3x$ at $x = \frac{\pi}{4}$ is

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