# intrinsic coordinated and radius of curvature?

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#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.

Can anyone suggest a good site for these or post up the notes here?

I cant find them in MathsNet and S-Cool.

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#3

(Original post by

can anyone help me?

exam on monday and i find thosequestions so hard

**kriztinae**)can anyone help me?

exam on monday and i find thosequestions so hard

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#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.

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#5

(Original post by

That is P5... get your textbook, read it, and do all the questions. That is the best way learn it, except for being taught.

**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.

Can you post up some notes?

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#6

(Original post by

I've got no textbooks for these.

Can you post up some notes?

**keisiuho**)I've got no textbooks for these.

Can you post up some notes?

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(Original post by

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

**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

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#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:

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,

Similarly, you can show that:

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)

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)

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(Original post by

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:

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,

Similarly, you can show that:

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)

**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)

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#10

(Original post by

believe it or notthat helped ! thanx!

**kriztinae**)believe it or notthat helped ! thanx!

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!)

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#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)...

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)...

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(Original post by

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)...

**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)...

wait do u have the formula booklet?

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#13

**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 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

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(Original post by

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!)

**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!)

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#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.

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.

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#17

(Original post by

How to change it to intrinsic coordinates:

2x^2 + y^2 = 4 ?

**keisiuho**)How to change it to intrinsic coordinates:

2x^2 + y^2 = 4 ?

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

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#18

(Original post by

How to change it to intrinsic coordinates:

2x^2 + y^2 = 4 ?

**keisiuho**)How to change it to intrinsic coordinates:

2x^2 + y^2 = 4 ?

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#19

(Original post by

Where is this from? Either I am missing somthing obvious or this is a very difficult question.

**mikesgt2**)Where is this from? Either I am missing somthing obvious or this is a very difficult question.

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#20

**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

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