# Thread: equation of the normal curve

1. ## equation of the normal curve

What is the equation of the normal line to the curve y = x^2 + 2x -1 at point (1,2)

attempt: dy/dx = 2x + 0
dy/dx = 2x,

im stuck here

2. ## Re: equation of the normal curve

Originally Posted by rcs
What is the equation of the normal line to the curve y = x^2 + 2x -1 at point (1,2)

attempt: dy/dx = 2x + 0
dy/dx = 2x,

im stuck here
Hint: The normal line is perpendicular to the tangent line. The gradients of perpendicular lines multiply to give -1.

3. ## Re: equation of the normal curve

Your derivative is not quite right:

$y=x^2+2x-1$

$\frac{dy}{dx}=2x+2$

The slope m of the normal line, however, is:

$m=-\frac{dx}{dy}=-\frac{1}{\frac{dy}{dx}}$

So, for $x=1$, what is the slope of the normal line?

4. ## Re: equation of the normal curve

Originally Posted by MarkFL2
Your derivative is not quite right:

$y=x^2+2x-1$

$\frac{dy}{dx}=2x+2$

The slope m of the normal line, however, is:

$m=-\frac{dx}{dy}=-\frac{1}{\frac{dy}{dx}}$

So, for $x=1$, what is the slope of the normal line?
m = 2(1) + 2 = 4 is the slope.

how is equation of the normal line ? how is that done?

thanks

5. ## Re: equation of the normal curve

You have found the slope of the tangent line. The normal line is perpendicular to the tangent line, so what must the slope of the normal line be?

6. ## Re: equation of the normal curve

hmmmm cant get it

7. ## Re: equation of the normal curve

Originally Posted by rcs
hmmmm cant get it
You can't get it or you haven't tried? I suggest you read my previous post.

8. ## Re: equation of the normal curve

really cant get it ... i tried but i really cant./. sorry

9. ## Re: equation of the normal curve

You found the slope of the tangent line $m_{\text{T}}$, but you need the slope of the normal line $m_{\text{N}}$. As Prove It pointed out, you require:

$m_{\text{N}}m_{\text{T}}=-1$

10. ## Re: equation of the normal curve

i dont have idea how it looks like... i cant visualize really

11. ## Re: equation of the normal curve

A normal line is perpendicular to the curve at the point of intersection, thus it is perpendicular to the tangent line at that point.

As far as the product of perpendicular slopes being -1, consider that the difference in their angles of inclination must be $\frac{\pi}{2}$:

$\tan^{-1}(m_1)-\tan^{-1}(m_2)=\frac{\pi}{2}$

Using the identity $\tan^{-1}(x)+\cot^{-1}(x)=\frac{\pi}{2}$ we may write:

$\tan^{-1}(m_1)-\tan^{-1}(m_2)=\tan^{-1}(m_1)+\cot^{-1}(m_1)$

$\tan^{-1}(-m_2)=\cot^{-1}(m_1)$

$\tan^{-1}(-m_2)=\tan^{-1}\left(\frac{1}{m_1} \right)$

$m_1m_2=-1$