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kriztinae

can anyone help me?

exam on monday and i find thosequestions so hard

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???

keisiuho

I've got no textbooks for these.

Can you post up some notes?

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

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!

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)

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)

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

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!

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!)

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)...

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

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!)

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!

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.

keisiuho

How to change it to intrinsic coordinates:

2x^2 + y^2 = 4 ?

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

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

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