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I am currently learning about differentiation equations and this is my first try at calculus. I have found it OK however I am struggling to figure out the mistake I'm making on this type of question:
Find the equations of the normals to the curves at given points:
1) $y = 2x^2 + 3$ at $x = -1 $
My workings: $$\frac{dy}{dx} = 4x \\ 4 \cdot -1 = -4 \\ -4 \longrightarrow \frac{1}{-4} \\ m = \frac{1}{-4} \\ 2 \cdot -1^2 +3 = 2 \cdot 1 + 3 = 2 + 3 = 5\\ P = (-1,5) \\ y - 5 = \frac{1}{-4}(x+1) \\ -4y + 20 = x + 1 \\ -x - 4y + 19 $$ However, this is wrong apparently but I do not know where I have gone wrong.

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negative reciprocal... –  David Mitra Feb 21 '12 at 22:52
    
$m$ should be equal to $\frac 1 4$. –  azarel Feb 21 '12 at 22:52
1  
Please explain your steps. More often than you'd think, it will solve your problem... –  Per Alexandersson Feb 21 '12 at 22:53

2 Answers 2

up vote 1 down vote accepted

You need to obtain the normal to $y=2x^2+3$ at $x=-1$. Since we know the equation to the tangent in $x=a$ is

$$y_T = f'(a)(x-a)+f(a)$$

we can also show the normal is given by

$$y_N = -\frac{1}{f'(a)}(x-a)+f(a)$$

Note the minus, which you missed.

We have that, as you correctly put,

$y' = 4x$

Thus $f'(-1) =-4$ and $f(-1) = 5$, so

$$y_N = \frac{1}{4}(x+1)+5 = \frac{x}{4}+\frac{21}{4}$$

or

$$4y +x+21=0$$

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I'm glad you're active on the site. :-) (+1) for working the solution to completion. –  user21436 Feb 21 '12 at 22:57
    
I knew about the negative reciprocal thing - I was reading my answer too hard to notice my mistake –  Kian Feb 21 '12 at 22:59
    
@KannappanSampath I really love this site. I've recently had a great reception here. Are you on the chat? –  Pedro Tamaroff Feb 21 '12 at 22:59
    
Yes. Sure I am. Will you be dropping by? –  user21436 Feb 21 '12 at 23:00
    
@KannappanSampath I'm there! –  Pedro Tamaroff Feb 21 '12 at 23:06

The product of the slope of perpendicular lines is $-1$ and not $1$ as indicated by $\rightarrow$ in your working.

Hence, $m=\dfrac{1}{4}$ and $m \neq \dfrac{-1}{4}$.

On a unrelated yet friendly note, please use \cdot to indicate product rather than that * you have used in your post.

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Thanks for the \cdot tip! I'll update now –  Kian Feb 21 '12 at 22:57

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