Respuesta :
Answer:
[tex]y = \frac{\sqrt{2}x}{3} - \frac{\sqrt{2}\pi}{3} + 1.5[/tex]
Step-by-step explanation:
The equation to the line normal to the curve has the following format:
[tex]y - y(x_{0}) = m(x - x_{0})[/tex]
In whicm m is the derivative of y at the point [tex]x_{0}[/tex]
In this problem, we have that:
[tex]x_{0} = \pi[/tex]
[tex]y(x) = 3\cos{\frac{x}{3}}[/tex]
[tex]y(\pi) = 3\cos{\frac{\pi}{3}} = \frac{3}{2}[/tex]
The derivative of [tex]\cos{ax}[/tex] is [tex]a\sin{ax}[/tex]
So
[tex]y(x) = 3\cos{\frac{x}{3}}[/tex]
[tex]y'(x) = 3*\frac{1}{3}\sin{\frac{x}{3}} = \sin{\frac{x}{3}}[/tex]
[tex]m = \sin{\frac{\pi}{3}} = \frac{\sqrt{2}}{3}[/tex]
The equation of the line normal to the curve of y=3cos1/3x is:
[tex]y - y(x_{0}) = m(x - x_{0})[/tex]
[tex]y - \frac{3}{2} = \frac{\sqrt{2}}{3}(x - \pi)[/tex]
[tex]y = \frac{\sqrt{2}}{3}(x - \pi) + \frac{3}{2}[/tex]
[tex]y = \frac{\sqrt{2}x}{3} - \frac{\sqrt{2}\pi}{3} + 1.5[/tex]
