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Math Help - bit rusty on partial derivatives

  1. #1
    Senior Member slevvio's Avatar
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    bit rusty on partial derivatives

    I have a function f(u,v) of two variables

    If I set w = u\sin \theta + v \cos \theta, how do I show then that

    \frac{ d f}{dw} = \sin \theta \frac{ \partial f}{ \partial u} + \cos \theta \frac{ \partial f}{ \partial v} ?

    My book states this but I was wondering what rule was used to get this

    Thanks very much
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  2. #2
    A Plied Mathematician
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    Well, the general rule would be

    \displaystyle{\frac{df}{dw}=\frac{\partial f}{\partial u}\,\frac{\partial u}{\partial w}+\frac{\partial f}{\partial v}\,\frac{\partial v}{\partial w}}.

    To me, off-hand, I'm a bit puzzled why the trig functions aren't in the denominators. Are you sure this is the correct expression?
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  3. #3
    Senior Member slevvio's Avatar
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    yeah its about a parabolic cylinder, whose bottom runs along the v direction. Theta is introduced as an angle between v and w to show that we can see what happens in all directions for all theta, except when theta = 0, i.e. what happens in the v-direction. So perhaps we are considering theta to be constant here. I don't know. thanks for reminding me of the chain rule
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  4. #4
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    I would definitely say that \theta is constant here. But what I can't get over is where the trig functions are in the expression you're trying to prove. You've got

    \displaystyle{\frac{\partial w}{\partial u}=\sin(\theta)}, so I would expect \displaystyle{\frac{\partial u}{\partial w}=\frac{1}{\sin(\theta)}.}

    A similar calculation would go for the other. I can't explain why this is not the case. Maybe there's something simple I'm missing. Maybe Danny could weigh in?
    Last edited by Ackbeet; August 18th 2010 at 11:17 AM. Reason: math tags around theta.
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  5. #5
    Senior Member slevvio's Avatar
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    I will write out a section: ''

    f has a maximum or a minimum (depending upon the sign) in the u-direction, but we do not yet know what happens in the v-direction. The surface z = f(x,y) is, to second order, a parabolic cylinder (Fig2.2)

    In fact we know what happens in every direction except the v-direction. For let w = u \sin \theta + v\cos \theta. Then at the origin

    \frac{d f}{d w} = \sin \theta \frac{ \partial f}{\partial u} + \cos \theta \frac{ \partial f}{\partial v} = 0

    and

    \frac{ d^2 f}{d w^2} = \sin^2 \theta \frac{\partial^2 f}{\partial u^2} + 2\sin \theta \cos \theta \frac{\partial^2 f}{\partial u \partial v} + \cos^2 \theta \frac{\partial^2 f}{\partial v^2} = \sin^2 \theta \frac{\partial^2 f}{\partial u^2}

    Hence f has the same sort of behaviour in the w-direction as in the u-direction, provided only that theta is not 0. IF theta is 0, i.e. in the v-direction, the Taylor series for f reduces to....

    "

    In case this is relevant this is an examination of what happens when the Hessian is 0 of a two variable function and not all the 2nd partial derivatives are zero
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  6. #6
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    Quote Originally Posted by slevvio View Post
    I have a function f(u,v) of two variables

    If I set w = u\sin \theta + v \cos \theta, how do I show then that

    \frac{ d f}{dw} = \sin \theta \frac{ \partial f}{ \partial u} + \cos \theta \frac{ \partial f}{ \partial v} ?

    My book states this but I was wondering what rule was used to get this

    Thanks very much
    I think what you're trying to do (please correct me if I'm wrong) is to establish the directional derivative.

    If we start at the point, say (a,b) and move in the direction of say {\bf w} = < \cos \theta, \sin \theta> then

    D_{\bf w}f = \displaystyle \lim_{h \to 0} \dfrac{f(a + h \cos \theta, b + h \sin \theta) - f(a,b)}{h}.

    If we define g(h) = f(a + h \cos \theta, b + h \sin \theta) then

    D_{\bf w}f = \displaystyle \lim_{h \to 0} \dfrac{g(h) - g(0)}{h}

    which, by definition is g'(0). Using the chain rule for functions of more than one variable

    g'(h) = f_x(a + h \cos \theta, b + h \sin \theta)\cos \theta + f_y(a + h \cos \theta, b + h \sin \theta) \sin \theta so

    D_{\bf w}f = g'(0) = \cos \theta f_x(a , b) + \sin \theta  f_y(a,b )

    noting that I've used x'\text{s} and y'\text{s} instead of u'\text{s} and v'\text{s}.

    I would, however, like to know your reference.
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  7. #7
    Senior Member slevvio's Avatar
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    the book is Saunders: an introduction to catastrophe theory.

    thanks fo the help, i think it is the directional derivative. you can work it too out by making a w' variable which is orthogonal to the w axis and then rearrange and doing partial derivatives
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