What is the slope of the line normal to the tangent line of $f (x) = sin (3 x - π)$ $f(x) = sin(3x-pi)$ at $x = \frac{π}{3}$ $x= pi/3$ ?

Question

3037 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-28T02:36:07

$y = - \frac{1}{3} x + \frac{π}{9}$ $y = -1/3x + pi/9$

We start by finding the corresponding $y$ $y$ coordinate.

$f (\frac{π}{3}) = sin (3 (\frac{π}{3}) - π)$ $f(pi/3) = sin(3(pi/3) - pi)$

$f (\frac{π}{3}) = sin (π - π)$ $f(pi/3) = sin(pi - pi)$

$f (\frac{π}{3}) = sin (0)$ $f(pi/3) = sin(0)$

$f (\frac{π}{3}) = 0$ $f(pi/3) = 0$

We now differentiate. We let $y = sin u$ $y = sinu$ and $u = 3 x - π$ $u = 3x- pi$ .

Then $\frac{d y}{d u} = cos u$ $dy/(du)= cosu$ and $\frac{d u}{d x} = 3$ $(du)/dx = 3$ , since $π$ $pi$ is but a constant.

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

$f'(x) = cosu xx 3$

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

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!

What is the slope of the line normal to the tangent line of f(x) = sin(3x-pi) f(x)=sin(3x−π)f(x) = sin(3x-pi) at x= pi/3 x=π3 x= pi/3 ?