What is the slope of the line normal to the tangent line of #f(x) = sin(3x-pi) # at # x= pi/3 #?

Question

What is the slope of the line normal to the tangent line of #f(x) = sin(3x-pi)  # at # x= pi/3  #?

Evelyn Agard · Accepted Answer

#y = -1/3x + pi/9#

We start by finding the corresponding #y# coordinate.

#f(pi/3) = sin(3(pi/3) - pi)#

#f(pi/3) = sin(pi - pi)#

#f(pi/3) = sin(0)#

#f(pi/3) = 0#

We now differentiate. We let #y = sinu# and #u = 3x- pi#.

Then #dy/(du)= cosu# and #(du)/dx = 3#, since #pi# is but a constant.

By the chain rule,

#f'(x) = dy/(du) xx (du)/dx#

#f'(x) = cosu xx 3#

#f'(x) = 3cos(3x - pi)#

We now determine the slope of the tangent.

#f'(pi/3) = 3cos(3(pi/3) - pi)#

#f'(pi/3) = 3cos(0)#

#f'(pi/3) = 3#

The normal line is perpendicular to the tangent, so the slope will be the negative reciprocal.

#m_"normal" = -1/3#

We can now determine the equation.

#y - y_1 = m(x- x_1)#

#y - 0 = -1/3(x- pi/3)#

#y = -1/3x + pi/9#

Hopefully this helps!

Elijah Abram · Answer

undefined To find the slope of the line normal to the tangent line of $ f(x) = \sin(3x - \pi) $ at $ x = \frac{\pi}{3} $, we first need to find the slope of the tangent line at $ x = \frac{\pi}{3} $. The slope of the tangent line at a given point is equal to the derivative of the function evaluated at that point.

First, find the derivative of $ f(x) $:
$$ f'(x) = 3\cos(3x - \pi) $$

Evaluate $ f'(x) $ at $ x = \frac{\pi}{3} $:
$$ f'\left(\frac{\pi}{3}\right) = 3\cos\left(3\left(\frac{\pi}{3}\right) - \pi\right) = 3\cos(\pi - \pi) = 3\cos(0) = 3 $$

So, the slope of the tangent line at $ x = \frac{\pi}{3} $ is $ m_{\text{tangent}} = 3 $.

The slope of the line normal to the tangent line is the negative reciprocal of the tangent line's slope. Therefore, the slope of the line normal to the tangent line is:
$$ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{3} $$