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# Normal of a curve Help Please watch

1. Got a question that I'm stuck on :

Find the equation of the normal to the curve y=2x^2 + 3x +4 which is perpendicular to the line y=7x-5

========

What I do is first get the gradient of the perpendicular, which is 7 . The normal will then be -1/7 right ?

Then differentiate the curve for gradient of the curve = 4x+3 .

I make -1/7 = 4x+3 , to find x, but I'm not sure how to do it, I bring the +3 to the left, to give -22/7 = 4x . But after finding x , it appears wrong, as its a decimal
2. You were on the right lines. You want the normal which is perpendicular to the line y=7x-5, so yes, the normal will have gradient, -1/7. However, when the normal has gradient -1/7, what is the gradient of the tangent?

This gradient (of the tangent) is what you need to set 4x+3 equal to.
3. Oh Okay, why do I set it for the tangent though ?

So I done : 31 (tangent gradient is 7*4+3 right , because normal is -1/7 ? ) = 4x+3 . Bring 3 to left, 28=4x , x = 7 . y = 31 ?

I'm way off the mark though in terms of the answer, I end up with:
7y= -x +216 ?

EDIT: All wrong
4. (Original post by Gelato)
Got a question that I'm stuck on :

Find the equation of the normal to the curve y=2x^2 + 3x +4 which is perpendicular to the line y=7x-5

========

What I do is first get the gradient of the perpendicular, which is 7 . The normal will then be -1/7 right ?

Then differentiate the curve for gradient of the curve = 4x+3 .

I make -1/7 = 4x+3 , to find x, but I'm not sure how to do it, I bring the +3 to the left, to give -22/7 = 4x . But after finding x , it appears wrong, as its a decimal

first, if slope here is 7,
4x + 3 = 7
x = 1
so, y = 9

now, normal =
5. I tried that method and got it right at the end.

Could you explain the steps to me ?

I always use the tangent to find x and Y right ?

Then I use the gradient from the normal to find the equation ?

Thanks
6. (Original post by Gelato)
I always use the tangent to find x and Y right ?

Then I use the gradient from the normal to find the equation ?
That's correct, yes.
7. Thanks for the help =)

I'm not entirely sure why I need to make the DY/DX = 7 though , which was why I was confused at the beginning , as I've only been working with points so far, till this one.
8. (Original post by Gelato)
I tried that method and got it right at the end.

Could you explain the steps to me ?

I always use the tangent to find x and Y right ?

Then I use the gradient from the normal to find the equation ?

Thanks
Find the equation of the normal to the curve y=2x^2 + 3x +4 which is perpendicular to the line y=7x-5
1. "perpendicular to the line y=7x-5" so the gradient of the tangent would be 7
2. find x and y using the fact that gradient of the tangent is 7
3. So the gradient of the normal would be -1/7
4. Find the equation using -1/7 and (x,y)

hope this helps
9. (Original post by Gelato)
Thanks for the help =)

I'm not entirely sure why I need to make the DY/DX = 7 though , which was why I was confused at the beginning , as I've only been working with points so far, till this one.
The gradient of the normal is so the normal is -1/7 when, and ONLY at the point when the gradient () is 7.
10. (Original post by Gelato)
Thanks for the help =)

I'm not entirely sure why I need to make the DY/DX = 7 though , which was why I was confused at the beginning , as I've only been working with points so far, till this one.
look at this picture (sorry I'm terrible at drawing.. on computers especially)
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Updated: September 26, 2010
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