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    z = f(x+at,y+bt), show that  \frac{\partial{z}}{\partial{t}} = a\frac{\partial{z}}{\partial{x}} + b\frac{\partial{z}}{\partial{y}}

    By the chain rule  \frac{\partial{z}}{\partial{t}} = \frac{\partial{x}}{\partial{t}}\  frac{\partial{z}}{\partial{x}} + \frac{\partial{y}}{\partial{t}}\  frac{\partial{z}}{\partial{y}}

    But how do I show that a = dx/dt, b = dy/dt
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    (Original post by Kevin_B)
    z = f(x+at,y+bt), show that  \frac{\partial{z}}{\partial{t}} = a\frac{\partial{z}}{\partial{x}} + b\frac{\partial{z}}{\partial{y}}

    By the chain rule  \frac{\partial{z}}{\partial{t}} = \frac{\partial{x}}{\partial{t}}\  frac{\partial{z}}{\partial{x}} + \frac{\partial{y}}{\partial{t}}\  frac{\partial{z}}{\partial{y}}

    But how do I show that a = dx/dt, b = dy/dt
    Your choice of variable names is a little confusing. I think it might have been clearer if you'd written

    z = f(u,v) and u = x+at and v = y+bt. The chain rule would then read

     \frac{\partial{z}}{\partial{t}} = \frac{\partial{u}}{\partial{t}}\  frac{\partial{z}}{\partial{u}} + \frac{\partial{v}}{\partial{t}}\  frac{\partial{z}}{\partial{v}}

    Does that help?
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    OK so now dz/dt = a(dz/du) + v(dz/dv), so how do you get x and y into the equation?
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    (Original post by Kevin_B)
    OK so now dz/dt = a(dz/du) + v(dz/dv), so how do you get x and y into the equation?
    Well, you know u and v in terms of x and y. Think of a way of transforming dz/du and dz/dv into stuff involving dz/dx and dz/dy.
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    (Original post by Kevin_B)
    OK so now dz/dt = a(dz/du) + v(dz/dv), so how do you get x and y into the equation?
    What \frac{\partial{z}}{\partial{x}} is meant to be denoting in the question as written is partial differentiation of z with respect to its first variable - that's why I said the notation was somewhat confusing given x then appears in the substituted first variable x+at. So my \frac{\partial{z}}{\partial{u}} just denotes the same thing as the question was intending.
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    (Original post by RichE)
    What \frac{\partial{z}}{\partial{x}} is meant to be denoting in the question as written is partial differentiation of z with respect to its first variable
    I'm not entirely sure why you can say this is the case (it would seem a very odd way of writing the original question).

    It seems to me the point is that the chain rule gives us \frac{\partial}{\partial x} f(u, v) = \frac{\partial u}{\partial x} \frac{\partial}{\partial u} f(u, v) (assuming v doesn't depend on x), and then if u = x + at, then of course \frac{\partial u}{\partial x} = 1.

    In other words, partial differentiation of z w.r.t. its first variable is equal to partial differentiation of z w.r.t. x, because of the way x appears as the first variable of z in the simple form "x+at".

    But this was one of those topics I only every understood long enough to do the exam, and then largely forgot about, so I might be wrong.
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    (Original post by DFranklin)
    But this was one of those topics I only every understood long enough to do the exam, and then largely forgot about, so I might be wrong.
    Likewise, but I agree; this is what I was trying to get across in my previous post.
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    (Original post by DFranklin)
    I'm not entirely sure why you can say this is the case (it would seem a very odd way of writing the original question).

    It seems to me the point is that the chain rule gives us \frac{\partial}{\partial x} f(u, v) = \frac{\partial u}{\partial x} \frac{\partial}{\partial u} f(u, v) (assuming v doesn't depend on x), and then if u = x + at, then of course \frac{\partial u}{\partial x} = 1.

    In other words, partial differentiation of z w.r.t. its first variable is equal to partial differentiation of z w.r.t. x, because of the way x appears as the first variable of z in the simple form "x+at".

    But this was one of those topics I only every understood long enough to do the exam, and then largely forgot about, so I might be wrong.
    I think everything you say above is valid (and the way to go), and I should have looked at the question more thoroughly. I think my first post was just driven off-track a little by having to explain why dx/dt wasn't a, and got into my head that the notation of the given question was a bit dodgy.
 
 
 
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