Find the equation of normal line to the graph $y=2(x-1)^3$

20,938

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.

Share:
20,938

Related videos on Youtube

Tayyab Gulsher Vohra
Author by

Tayyab Gulsher Vohra

I am a Expert Seo And A Biginner Developer Of Php

Updated on October 21, 2020

Comments

  • Tayyab Gulsher Vohra
    Tayyab Gulsher Vohra about 2 years

    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?

    • Jack
      Jack about 8 years
      If the gradient of the tangent was $a$, what would the gradient of the normal be?
    • N. F. Taussig
      N. F. Taussig about 8 years
      The normal line is perpendicular to the tangent line. What does that tell you about its slope?
    • Tayyab Gulsher Vohra
      Tayyab Gulsher Vohra about 8 years
      how to solve this further @N.F.Taussig