# Problem finding normal line equation using partial derivative and gradient vector

• Jul 23rd 2010, 11:40 AM
x3bnm
Problem finding normal line equation using partial derivative and gradient vector
I'm trying to solve a problem but the answer is different from the book.
So wondering what am i doing wrong. The problem is like this:

Find equations for the lines that are normal to function:
$\displaystyle y - \sin{(x)} = 1$ at $\displaystyle P_0\left(\pi, 1)$

The way i did was first find the gradient vector:
$\displaystyle \nabla f = \left(-\cos{(x)} . i\right) + 1.j = \left(-\cos{(x)} . i\right) + j$

Then get the derivative $\displaystyle D_uf\left|_{(\pi,1)}$ which is(because we know that slope
of a function is it's derivative at that point):
$\displaystyle D_uf\left|_{(\pi,1)} = \sqrt{\left(-\cos{(x)}\right)^2 + (1^2)} = \sqrt{1^2 + 1^2} = \sqrt{2}$

Finally the equation of normal line is:

$\displaystyle y - 1 = \sqrt{2}\left(x - \pi\right)$

Is this right or am i wrong? The answer back of my book is $\displaystyle y = x - \pi + 1$
Why book's answer is different? Any ideas?
• Jul 23rd 2010, 11:43 AM
Ackbeet
Well, one problem is that you can't use the slope of the tangent line as the slope of the normal line! They are related: once you find the slope of the tangent line, you can use a simple formula to find the slope of the normal line. Second, I'm not sure I agree with your method of computing the slope of the tangent line. Can't you just use the normal Calc I method? Your method got the wrong answer. (I agree with the book's answer completely, by the way).
• Jul 23rd 2010, 12:07 PM
Plato
Quote:

Originally Posted by x3bnm
I'm trying to solve a problem but the answer is different from the book. Find equations for the lines that are normal to function:
$\displaystyle y - \sin{(x)} = 1$ at $\displaystyle P_0\left(\pi, 1)$
The way i did was first find the gradient vector:
$\displaystyle \nabla f = \left(-\cos{(x)} . i\right) + 1.j = \left(-\cos{(x)} . i\right) + j$.
The answer back of my book is $\displaystyle y = x - \pi + 1$
Why book's answer is different? Any ideas?

Here is another way to look at it.
The vector $\displaystyle \nabla f(\pi ,1)$ is the direction vector of the normal line to the curve at the point $\displaystyle (\pi,1)$.
That is $\displaystyle \nabla f(\pi ,1)=i+j$ and the normal is $\displaystyle <\pi,1> +t<1,1>.$
• Jul 23rd 2010, 12:12 PM
x3bnm
>Well, one problem is that you can't use the slope of the tangent line as the slope of the normal line!

Okay.

>Can't you just use the normal Calc I method?

Can you kindly explain this method?
• Jul 23rd 2010, 12:19 PM
Ackbeet
Just solve your original equation for y = sin(x) + 1. Take its derivative: y' = cos(x). Plug in the value of x in question to get y'(pi) = -1. That's your slope of the tangent line. Then go from there.
• Jul 23rd 2010, 12:34 PM
x3bnm
Oops! looks like we posted at the same time. Sorry.
I thank you all for your help. Again those who are interested:

The slope of the normal line is the negative reciprocal of slope of the tangent line.
• Jul 23rd 2010, 12:36 PM
Ackbeet
No, we didn't post at the same time. I'm just fast that way, sometimes. (Wink)