# Tangent and Normal Lines

• Nov 25th 2009, 10:49 PM
mj.alawami
Tangent and Normal Lines
Question:
Find the equation of the normal lines to the curve $\displaystyle y=x^4$ which is parallel to the line $\displaystyle 2x+y=3$

Thank you
• Nov 25th 2009, 10:58 PM
Arturo_026
First you need to find the derivative:

$\displaystyle y' = 4x^3$

Then, from the equation of the line you are given, you see that the slopes of the normal line is -2 ; and since you want the slope of the tangets, you get the opposite reciprocal of -2, which is m=1/2.

Now you have y', just plug it in the derivative function and find your x.
Then with that, go back to the original equation and find y.
And you have all the ingredients necessary to set a line equation of the normals.
• Nov 26th 2009, 12:53 AM
mj.alawami
Quote:

Originally Posted by Arturo_026
First you need to find the derivative:

$\displaystyle y' = 4x^3$

Then, from the equation of the line you are given, you see that the slopes of the normal line is -2 ; and since you want the slope of the tangets, you get the opposite reciprocal of -2, which is m=1/2.

Now you have y', just plug it in the derivative function and find your x.
Then with that, go back to the original equation and find y.
And you have all the ingredients necessary to set a line equation of the normals.

So From what i understood?

$\displaystyle 4x^3=\frac{1}{2}$
$\displaystyle x=\frac{1}{2}$

$\displaystyle y-\frac{1}{16}=\frac{1}{2}(x-\frac{1}{2}$
$\displaystyle x-2y+\frac{3}{8}$

Is my answer correct??

• Nov 26th 2009, 02:11 AM
Quote:

Originally Posted by mj.alawami
So From what i understood?

$\displaystyle 4x^3=\frac{1}{2}$
$\displaystyle x=\frac{1}{2}$

$\displaystyle y-\frac{1}{16}=\frac{1}{2}(x-\frac{1}{2}$
$\displaystyle x-2y+\frac{3}{8}$

Is my answer correct??

HI

your values of x and y are both correct . The gradient of normal is -2 because the normal line is parallel to the line y=-2x+3
• Nov 26th 2009, 02:38 AM
Hello mj.alawami
Quote:

Originally Posted by mj.alawami
So From what i understood?

$\displaystyle 4x^3=\frac{1}{2}$
$\displaystyle x=\frac{1}{2}$

$\displaystyle y-\frac{1}{16}=\frac{1}{2}(x-\frac{1}{2}$
$\displaystyle x-2y+\frac{3}{8}$

Is my answer correct??

Your answer is partly correct. You have the right values of $\displaystyle x$ and $\displaystyle y$, but the wrong gradient. The gradient of the normal is $\displaystyle -2$, so the equation is:
$\displaystyle y-\frac{1}{16}=-2\left(x - \frac12\right)$
i.e. $\displaystyle 16y-1=-32x + 16$
i.e. $\displaystyle 32x+16y-17 =0$