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#1 2006-01-11 13:49:07

RickyOswaldIOW
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Tangent to Normal.

I've been working out the gradient (and whole equation) of the tangent of certain points on a curve but this new set of questions wants the equation of the normal.  I figure I'd just work as I had been and get the equation of the tangent, then simply flip over the gradient and reverse the sign for the normal i.e.
tangent; y = 7x - 13
normal; y = -1/7x - 13

The full sum is as follows:

y = 2x^2 - 5x + 3    at (2, 1)

dy/dx = 6x - 5
            12 - 5 = 7

(y - 1)/(x - 2) = 7
y - 1 = 7(x - 2) = 7x - 14
y = 7x - 13
and thus the normal; y = -1/7x - 13.

The book claims x + 3y = 5


Aloha Nui means Goodbye.

#2 2006-01-11 14:25:04

Ricky
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Re: Tangent to Normal.

dy/dx = 6x - 5

dy/dx = 4x - 5, slope at (1, 2) is 3

Try it now.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

#3 2006-01-11 14:28:08

RickyOswaldIOW
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Re: Tangent to Normal.

roll


Aloha Nui means Goodbye.

#4 2006-01-11 14:32:54

Ricky
Moderator

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Re: Tangent to Normal.

Don't worry about it, on a math test, I wrote:

2x = 2
x = 2


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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