# intrinsic coordinated and radius of curvature?

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#1
can anyone help me?
exam on monday and i find thosequestions so hard
0
15 years ago
#2
I havent learnt them yet and I will have the exam also on Monday!
Can anyone suggest a good site for these or post up the notes here?
I cant find them in MathsNet and S-Cool.
0
15 years ago
#3
(Original post by kriztinae)
can anyone help me?
exam on monday and i find thosequestions so hard
This isn't Edexcel P2 is it? Cos i have an exam on Monday and i haven't done this??? 0
15 years ago
#4
That is P5... get your textbook, read it, and do all the questions. That is the best way learn it, except for being taught.
0
15 years ago
#5
(Original post by mik1a)
That is P5... get your textbook, read it, and do all the questions. That is the best way learn it, except for being taught.
I've got no textbooks for these.
Can you post up some notes?
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15 years ago
#6
(Original post by keisiuho)
I've got no textbooks for these.
Can you post up some notes?
These questions are easy - if its in cartesian coordinates just differentiate twice and plug in the values into the formula (in the formula book!) If its in parametric equations diff with respect to t twice and plug in formula (also in formula book!) if its in intrinsic coordinates use rho = ds/dpsi (might not be in the formula book but easty to remember) Thats all you ever have to do
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#7
(Original post by It'sPhil...)
These questions are easy - if its in cartesian coordinates just differentiate twice and plug in the values into the formula (in the formula book!) If its in parametric equations diff with respect to t twice and plug in formula (also in formula book!) if its in intrinsic coordinates use rho = ds/dpsi (might not be in the formula book but easty to remember) Thats all you ever have to do
believe it or notthat helped ! thanx!
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15 years ago
#8
Well, like (x, y) tell you the point of a curve, so do (s, ψ ). The (x, y) are cartesian coordinates, and (s, ψ ) are the intrinsic coordinates of the point on the curve, where s is the length of the arc from that pointnbeing described, and ψ is the gradient angle (angle that the tangent to the curve at that point makes with the x-axis).

Because the gradient of the curve at a point is dy/dx, and the gradient of the tangent is tan ψ , you can tell dy/dx = tan ψ (check if you want by drawing a triangle, Δy/Δx of this traingle should equal dy/dx of the curve, assuming the tangent is made from that point on the curve.

Now some copying:

dy/dx = tan ψ

At P,

(ds/dx)² = 1 + (dy/dx)²
(ds/dx)² = 1 + tan² ψ

But: sec² ψ = 1 + tan² ψ
So: ds/dx = sec ψ

And therefore, by reciprocating,

dx/ds = cos ψ

Similarly, you can show that:

dy/ds = sin ψ

By using these three bold relations, you can convert from cartesian coordinates into intrinsic coordinates (y = f(x) into s = g(ψ )), which can sometimes be useful.

The radius of curvature at a certain point is given by:

ρ (rho) = ds/dψ, or the rate of change of s w.r.t. ψ

(having a break)
0
#9
(Original post by mik1a)
Well, like (x, y) tell you the point of a curve, so do (s, ψ ). The (x, y) are cartesian coordinates, and (s, ψ ) are the intrinsic coordinates of the point on the curve, where s is the length of the arc from that pointnbeing described, and ψ is the gradient angle (angle that the tangent to the curve at that point makes with the x-axis).

Because the gradient of the curve at a point is dy/dx, and the gradient of the tangent is tan ψ , you can tell dy/dx = tan ψ (check if you want by drawing a triangle, Δy/Δx of this traingle should equal dy/dx of the curve, assuming the tangent is made from that point on the curve.

Now some copying:

dy/dx = tan ψ

At P,

(ds/dx)² = 1 + (dy/dx)²
(ds/dx)² = 1 + tan² ψ

But: sec² ψ = 1 + tan² ψ
So: ds/dx = sec ψ

And therefore, by reciprocating,

dx/ds = cos ψ

Similarly, you can show that:

dy/ds = sin ψ

By using these three bold relations, you can convert from cartesian coordinates into intrinsic coordinates (y = f(x) into s = g(ψ )), which can sometimes be useful.

The radius of curvature at a certain point is given by:

ρ (rho) = ds/dψ, or the rate of change of s w.r.t. ψ

(having a break)
thanx im printing that out if you dont mind!
0
15 years ago
#10
(Original post by kriztinae)
believe it or notthat helped ! thanx!
Ok cool. Also when converting cartesian to intrinsic I always start off by drawing a right angled triangle, with angle psi. The length is dx the height is dy and the hyptonuse is ds. From this you can get the equations dy/dx = tan(psi) , dx/ds = cos(psi) , dy/ds = sin(psi). To convert intrinsic to cartesian use...

dx/ds = (dx/ds)(dp/dp) p = psi

so dx/ds = dx/dp . dp/ds

so dx/dp = cosp ds/dp you can find ds/dp from youre intrinsic equation

so x = INT (cosp ds/dp) dp If you do the same for y you will get parametric equations for x and y ie

x = f(p) y = g(p) . Then its a case of arranging these to eliminate p. Usually you have to use some trig manipulation. Hope that helps (writing it certainly helped me to remember it!)
0
15 years ago
#11
I don't... maybe Heinemann will though lol
Can you tell me what s means? Where is it measured from? I know its the length of the arc, and that there's some complicated integral to find it (I think)... 0
#12
(Original post by mik1a)
I don't... maybe Heinemann will though lol
Can you tell me what s means? Where is it measured from? I know its the length of the arc, and that there's some complicated integral to find it (I think)... hmm dont think iget what you mean
wait do u have the formula booklet?
0
15 years ago
#13
(Original post by mik1a)
I don't... maybe Heinemann will though lol
Can you tell me what s means? Where is it measured from? I know its the length of the arc, and that there's some complicated integral to find it (I think)... The idea behind intrinsic coordinates is that they describe the intrinsic nature of curves, ie how bendy they are (curvature), the angle the tangent makes to the horizontal (psi) and the length of the curve from a chosen point. This means they do not need axes as all the information is given, so s can be taken from an arbritrarily chosen point.

The integral comes in from (roughly) a right triangle with sides dx, dy and hypotenuse ds. By pythag (ds)² = (dx)² + (dy)². dividing by (dx)² gives (ds/dx)² = 1 + (dy/dx)² so ds/dx = (1 + (dy/dx)²)^(1/2). so s is the integral wrt x of this. This is not a very rigorous explaination but its the rough idea
0
#14
(Original post by It'sPhil...)
Ok cool. Also when converting cartesian to intrinsic I always start off by drawing a right angled triangle, with angle psi. The length is dx the height is dy and the hyptonuse is ds. From this you can get the equations dy/dx = tan(psi) , dx/ds = cos(psi) , dy/ds = sin(psi). To convert intrinsic to cartesian use...

dx/ds = (dx/ds)(dp/dp) p = psi

so dx/ds = dx/dp . dp/ds

so dx/dp = cosp ds/dp you can find ds/dp from youre intrinsic equation

so x = INT (cosp ds/dp) dp If you do the same for y you will get parametric equations for x and y ie

x = f(p) y = g(p) . Then its a case of arranging these to eliminate p. Usually you have to use some trig manipulation. Hope that helps (writing it certainly helped me to remember it!)
thanx!! im printing this out too! 0
15 years ago
#15
dy/dx = tan ψ
dy/ds = sin ψ
dx/ds = cos ψ

If anyone needs an easy way to remember these, draw a right angled triangle with an angle ψ where the opposite side is dy, the adjacent is dx and the hypotenuse is ds. Then all the equations follow from considering the tan, sin and cos of the angle ψ.
0
15 years ago
#16
How to change it to intrinsic coordinates:
2x^2 + y^2 = 4 ?
0
15 years ago
#17
(Original post by keisiuho)
How to change it to intrinsic coordinates:
2x^2 + y^2 = 4 ?
find the arclength of the equation from 0 ( im assuming it is ... )

see wat u get after integrating ..

ull end up with S = f(x)

use the fact tht dy/dx is always tan psi .... and try to find a simple relation between tan psi and ur original (dy/dx)

subsititute tht for psi in the S = f(x) ...

hopefully ull end up with S= G(psi) .. which is an intrinsic equation
0
15 years ago
#18
(Original post by keisiuho)
How to change it to intrinsic coordinates:
2x^2 + y^2 = 4 ?
Where is this from? Either I am missing somthing obvious or this is a very difficult question.
0
15 years ago
#19
(Original post by mikesgt2)
Where is this from? Either I am missing somthing obvious or this is a very difficult question.
Im left with INT (2 + 2cos^2t)^(1/2) dt which I cant do
0
15 years ago
#20
(Original post by mikesgt2)
Where is this from? Either I am missing somthing obvious or this is a very difficult question.
Is it OK to use the forumula right away?

I wonder if this type of question will come up in real exam papers
0
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