Question

Find the equation of the line normal to the curve of y=3cos1/3x, Where x=\pi

Answer 1

Answer:

$y = \frac{\sqrt{2}x}{3} - \frac{\sqrt{2}\pi}{3} + 1.5$

Step-by-step explanation:

The equation to the line normal to the curve has the following format:

$y - y(x_{0}) = m(x - x_{0})$

In whicm m is the derivative of y at the point $x_{0}$

In this problem, we have that:

$x_{0} = \pi$

$y(x) = 3\cos{\frac{x}{3}}$

$y(\pi) = 3\cos{\frac{\pi}{3}} = \frac{3}{2}$

The derivative of $\cos{ax}$ is $a\sin{ax}$

So

$y(x) = 3\cos{\frac{x}{3}}$

$y'(x) = 3*\frac{1}{3}\sin{\frac{x}{3}} = \sin{\frac{x}{3}}$

$m = \sin{\frac{\pi}{3}} = \frac{\sqrt{2}}{3}$

The equation of the line normal to the curve of y=3cos1/3x is:

$y - y(x_{0}) = m(x - x_{0})$

$y - \frac{3}{2} = \frac{\sqrt{2}}{3}(x - \pi)$

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

$y = \frac{\sqrt{2}x}{3} - \frac{\sqrt{2}\pi}{3} + 1.5$