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2009-09-27 16:50:22
CanadianLoser
Meow :3 :3
+1,148|6509
QUESTION:
considering the curve y=3x^2 - x.  Using the equation for the normal line to a curve, find the coordinates of the closest point on the curve to the point (1,1).

Hint: The line passing through (1,1) and the closest point (x,y) to it on the curve is the normal line through curve at (x,y).  Using the equation for the normal line, and the fact that it passes through (1,1), solve for (x,y)

HELPFUL EQUATIONS:
normal line to a curve: mN = - 1/m   where mN is the slope of the normal and m is the slope of the curve at the intersection point.

f`(x) = 6x - 1 = m




not sure where to go with this?

p.s. this may not be my last question as I'm doing a 6 question assignment and this is question 2
2009-09-27 16:51:06
Gooners
Wiki Contributor
+2,700|6633

42
2009-09-27 16:54:58
mcminty
Moderating your content for the Australian Govt.
+879|6722|Sydney, Australia

CanadianLoser wrote:

QUESTION:
considering the curve y=3x^2 - x.  Using the equation for the normal line to a curve, find the coordinates of the closest point on the curve to the point (1,1).

Hint: The line passing through (1,1) and the closest point (x,y) to it on the curve is the normal line through curve at (x,y).  Using the equation for the normal line, and the fact that it passes through (1,1), solve for (x,y)

HELPFUL EQUATIONS:
normal line to a curve: mN = - 1/m   where mN is the slope of the normal and m is the slope of the curve at the intersection point.

f`(x) = 6x - 1 = m




not sure where to go with this?

p.s. this may not be my last question as I'm doing a 6 question assignment and this is question 2
I'll help you out in about 40 minutes.

I'm currently at uni, in a tutorial. But shhh..
2009-09-27 16:56:14
Red Forman
Banned
+402|5401

Gooners wrote:

42
you forgot to carry the five
2009-09-27 16:57:02
CanadianLoser
Meow :3 :3
+1,148|6509
sweet man, I'll still be here so if we haven't figured it out by then I'd love your help
2009-09-27 17:49:23
Vub
The Power of Two
+188|6495|Sydney, Australia

mcminty wrote:

CanadianLoser wrote:

QUESTION:
considering the curve y=3x^2 - x.  Using the equation for the normal line to a curve, find the coordinates of the closest point on the curve to the point (1,1).

Hint: The line passing through (1,1) and the closest point (x,y) to it on the curve is the normal line through curve at (x,y).  Using the equation for the normal line, and the fact that it passes through (1,1), solve for (x,y)

HELPFUL EQUATIONS:
normal line to a curve: mN = - 1/m   where mN is the slope of the normal and m is the slope of the curve at the intersection point.

f`(x) = 6x - 1 = m




not sure where to go with this?

p.s. this may not be my last question as I'm doing a 6 question assignment and this is question 2
I'll help you out in about 40 minutes.

I'm currently at uni, in a tutorial. But shhh..
Hahaha, it's our non teaching week
2009-09-27 17:50:42
mcminty
Moderating your content for the Australian Govt.
+879|6722|Sydney, Australia
Wow, I've gotta say.. I've forgotten this shit. It might take a little longer than I thought. I've got a lecture in 15 minutes, so you might have to wait a little.

But forgotten, I mean the basic parabola stuff. Cause like, you could always use the distance formula between (1,1) and (x, 3x2 - x), then solve it for a minimum distance or something, right?
2009-09-27 18:17:54
Vub
The Power of Two
+188|6495|Sydney, Australia
OK, been awhile, but here's my attempt:

y' = 6x - 1
therefore mN = 1/(1-6x)
Let (a,b) be the point nearest to (1,1), hence using the point gradient form:

y - b = (x - a)/(1 - 6a)

Sub in (1,1) and solve for a.

1 - b = (1 - a)/(1 - 6a)

but b = 3a^2 - a

Solve the cubic and you have 3 solutions:

a = 0, (9 + sqrt(369))/36 or (9 - sqrt(369))/36.

To determine which is the closest, I guess you have to sketch it. You'll find that a = (9 + sqrt(369))/36 is the closest. Use b = 3a^2 - a to find b.

Last edited by Vub (2009-09-27 18:18:50)

2009-09-27 18:24:53
WldctARCHe
Member
+9|5359|Kansas
Okay, i'm not 100% sure on this but I think the correct way to solve this is to determine the normal line (y=6x-1) and the line that is perpendicular to the normal line and passes through the point (1,1) which would be y= (-1/6)x + (7/6).  Then your closet point on your original line would be where the perpendicular line and your original line (3x^2-x) intersect, so you just set the two equations equal to each other and solve (i.e.  3x^2-x= (-1/6)x+(7/6))
x= (7/9)
y= (28/27)

Hope this helps
2009-09-27 18:31:51
Vub
The Power of Two
+188|6495|Sydney, Australia

WldctARCHe wrote:

Okay, i'm not 100% sure on this but I think the correct way to solve this is to determine the normal line (y=6x-1) and the line that is perpendicular to the normal line and passes through the point (1,1) which would be y= (-1/6)x + (7/6).  Then your closet point on your original line would be where the perpendicular line and your original line (3x^2-x) intersect, so you just set the two equations equal to each other and solve (i.e.  3x^2-x= (-1/6)x+(7/6))
x= (7/9)
y= (28/27)

Hope this helps
But the gradient of a parabola is a function of x, not a constant number, and (1,1) does not lie on the parabola itself.

Last edited by Vub (2009-09-27 18:32:34)

2009-09-27 18:39:40
Bevo
Nah
+718|6522|Austin, Texas
Use the slope and the point 1,1 to make an equation. Use polynomial equation thingies (add/subtract them) to find common x/y.

ed: hrm, you don't have a slope and you can't just pick and nilly willy point for a derivative.

Last edited by Bevo (2009-09-27 18:40:43)

 

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