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Thread: 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 $\displaystyle w = u\sin \theta + v \cos \theta$, how do I show then that

    $\displaystyle \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

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

    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 $\displaystyle \theta$ 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

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

    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; Aug 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 $\displaystyle w = u \sin \theta + v\cos \theta$. Then at the origin

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

    and

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

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

    $\displaystyle \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 $\displaystyle (a,b)$ and move in the direction of say $\displaystyle {\bf w} = < \cos \theta, \sin \theta>$ then

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

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

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

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

    $\displaystyle 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

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

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

    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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