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    [Edexcel- Heinemann P5]

    Since this text book seems woefully short of examples to practise converting Intrinsic to Cartesian coordinates and vice versa, I thought I would make my own questions by doing the given examples backwards. So I decided to start with the first given example [p104 example 10] which takes y=cosh x and derives s=tanψ from it.

    Derive the Cartesian equation of s=tanψ

    [Had to type psi in a couple of places. The editor seems to be playing up]

    ds/dψ = sec²ψ
    ds= sec²ψ.dψ

    dy/ds = sinψ
    dy = sinψ.ds

    So dy = sinψ.sec²ψ.dψ
    y=∫sinψ.sec²ψ.dψ

    Parts: v=sinψ =>dv/dψ=cosψ
    du/dψ = sec²ψ =>u=tanψ

    y=sinψ.tanψ -∫tanψ.cosψ dψ
    y=sinψ.tanψ -∫sinψ dψ
    y=sinψ.tanψ +cosψ
    y=sin²ψ/cosψ + cos²ψ/cosψ
    y=secψ

    [I've just noticed that sinψ.sec²ψ = secψ.tanψ, so I've gone a long way round here and I could have just quoted the standard ∫secψ.tanψ dψ = secψ, rather than showing it! Never mind. Good exercise...]


    s=tanψ = dy/dx
    1/s = dx/dy
    dx = (1/s)dy
    ∫1.dx =∫(1/s)dy

    and dy/ds = sinψ
    dy = sinψ ds

    So ∫1.dx =∫(1/s).sinψ ds =∫(1/tan psi).sinψ ds
    ds=sec²ψ.dψ
    so x=∫(1/tan psi).sinψ.sec²ψ.dψ = ∫secψ dψ
    x= ln|secψ + tan ψ|

    tanψ = √(sec²ψ -1)
    and since y=secψ
    x= ln|y + √(y² -1)| [logarithmic form of arcosh]
    so x= arcsosh y
    y = coshx QED

    I'm sure a lot of this goes the long way round, but it's interesting.
    I might not do a lot of these...

    Anyone got any other Intrinsic-Cartesian workouts that are not in the Heinemann text?

    Aitch
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    I agree that the heinemann book isn't exactly great at explaining this topic. It's important to have the right method for the possible situations:

    1: Converting intrinsic form [s=f(w)] to cartesian form [y=g(x)].
    Differentiate to give (ds/dw)=f'(w) so ds=f'(w) dw

    (dy/ds)=sinw
    ∫1 dy = ∫sinw ds
    y=∫sinw.f'(w) dw
    Integrate to give a relationship between y and w. Remember not to neglect the arbitrary constant.

    (dx/ds)=cosw
    ∫1 dx = ∫cosw ds
    x=∫cosw.f'(w) dw
    Integrate to give a relationship between x and w. Remember not to neglect the arbitrary constant.

    Eliminate the parameter 'w' to give the cartesian relationship between y and x. Typically you'll use a trigonometric identity or make w the subject in the two equations to eliminate w.

    2: Converting cartesian form [y=f(x)] to intrinsic form [s=g(w)]
    y=f(x) so (dy/dx)=f'(x)=tanw
    s=∫[1+(dy/dx)^2]^0.5 dx with lower limit a(where arc length is measured from) and upper limit x.
    s=∫[1+(f(x))^2]^0.5 dx with lower limit a and upper limit x.
    Perform the integration to find the relation s=h(x) and eliminate x using (dy/dx)=tanw, to form a relation involving only s and w.
    Tou can convert from x=f(y) or parametric equations x=f(t), y=g(t) in a similar manner using the appropriate arc length formula.

    I'm not sure whether anyone will find that useful but it's already typed now, it helps my own revision anyway at least :rolleyes:
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    (Original post by Gaz031)
    I agree that the heinemann book isn't exactly great at explaining this topic. It's important to have the right method for the possible situations:

    1: Converting intrinsic form [s=f(w)] to cartesian form [y=g(x)].
    Differentiate to give (ds/dw)=f'(w) so ds=f'(w) dw

    (dy/ds)=sinw
    ∫1 dy = ∫sinw ds
    y=∫sinw.f'(w) dw
    Integrate to give a relationship between y and w.

    (dx/ds)=cosw
    ∫1 dx = ∫cosw ds
    x=∫cosw.f'(w) dw
    Integrate to give a relationship between x and w.

    Eliminate the parameter 'w' to give the cartesian relationship between y and x. Typically you'll use a trigonometric identity or make w the subject in the two equations to eliminate w.
    I have to acknowledge that one of my most useful pieces of revision material is a printout of one of your earlier posts on this topic!

    Thanks again!

    Aitch
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    Thanks, guys.

    Muchly useful.
 
 
 
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