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# Chain Rule watch

1. z = f(x+at,y+bt), show that

By the chain rule

But how do I show that a = dx/dt, b = dy/dt
2. (Original post by Kevin_B)
z = f(x+at,y+bt), show that

By the chain rule

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

Does that help?
3. OK so now dz/dt = a(dz/du) + v(dz/dv), so how do you get x and y into the equation?
4. (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.
5. (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 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 just denotes the same thing as the question was intending.
6. (Original post by RichE)
What 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 (assuming v doesn't depend on x), and then if u = x + at, then of course .

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.
7. (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.
8. (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 (assuming v doesn't depend on x), and then if u = x + at, then of course .

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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