Find the equation of normal line to the graph $y=2(x-1)^3$
Find the equation of normal line to the graph $y=2(x-1)^3$ at the point where $x=\frac12$.
So far, I found the derivative:
$$\frac{dy}{dx}= 6(x-1)^2 $$
What to do next?
$\endgroup$ 31 Answer
$\begingroup$The normal line is perpendicular to the tangent line. Therefore, its slope is the negative reciprocal of the tangent line.
To find the slope of the tangent line at the point where $x = 1/2$, you evaluate the derivative of the equation $y = 2(x - 1)^3$ when $x = 1/2$.
As you found, the derivative is
$$y' = 6(x - 1)^2$$
Hence, when $x = 1/2$, the tangent line has slope
$$y' = 6\left(\frac{1}{2} - 1\right)^2 = 6\left(-\frac{1}{2}\right)^2 = 6\left(\frac{1}{4}\right) = \frac{3}{2}$$
Since the slope of the normal line is the negative reciprocal of the slope of the tangent line, the normal line has slope $-2/3$.
When $x = 1/2$,
$$y = 2(x - 1)^3 = 2\left(\frac{1}{2} - 1\right)^3 = 2\left(-\frac{1}{2}\right)^3 = 2\left(-\frac{1}{8}\right) = -\frac{1}{4}$$
so the tangent line and normal line both pass through the point $(1/2, -1/4)$.
If you use the point-slope form of the equation of a line
$$y - y_0 = m(x - x_0)$$
you obtain
$$y + \frac{1}{4} = -\frac{2}{3}\left(x - \frac{1}{2}\right)$$
for the equation of the normal line.
$\endgroup$
