Problem finding normal line equation using partial derivative and gradient vector

**x3bnm** · July 23rd 2010, 11:40 AM

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:
$ y - \sin{(x)} = 1$ at $P_0\left(\pi, 1) $

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

Then get the derivative $D_uf\left|_{(\pi,1)}$ which is(because we know that slope
of a function is it's derivative at that point):
$ 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:

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

Is this right or am i wrong? The answer back of my book is $y = x - \pi + 1$
Why book's answer is different? Any ideas?

**Ackbeet** · July 23rd 2010, 11:43 AM

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).

**Plato** · July 23rd 2010, 12:07 PM

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:
$y - \sin{(x)} = 1$ at $P_0\left(\pi, 1)$
The way i did was first find the gradient vector:
$\nabla f = \left(-\cos{(x)} . i\right) + 1.j = \left(-\cos{(x)} . i\right) + j$ .
The answer back of my book is $y = x - \pi + 1$
Why book's answer is different? Any ideas?

Here is another way to look at it.
The vector $\nabla f(\pi ,1)$ is the direction vector of the normal line to the curve at the point $(\pi,1)$ .
That is $\nabla f(\pi ,1)=i+j$ and the normal is $<\pi,1> +t<1,1>.$

**x3bnm** · July 23rd 2010, 12:12 PM

>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?

**Ackbeet** · July 23rd 2010, 12:19 PM

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.

**x3bnm** · July 23rd 2010, 12:34 PM

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.

**Ackbeet** · July 23rd 2010, 12:36 PM

No, we didn't post at the same time. I'm just fast that way, sometimes.

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