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.
Asked
Active
Viewed 2,464 times
0
-
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
-
1Please explain your steps. More often than you'd think, it will solve your problem... – Per Alexandersson Feb 21 '12 at 22:53
2 Answers
1
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.
1
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
$$4y +x+21=0$$
Pedro
- 122,002
-
I'm glad you're active on the site. :-) (+1) for working the solution to completion. – 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 Feb 21 '12 at 22:59
-
-