Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    1
    ReputationRep:
    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
    Offline

    15
    ReputationRep:
    (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?
    • Thread Starter
    Offline

    1
    ReputationRep:
    OK so now dz/dt = a(dz/du) + v(dz/dv), so how do you get x and y into the equation?
    • Wiki Support Team
    Offline

    14
    ReputationRep:
    Wiki Support Team
    (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.
    Offline

    15
    ReputationRep:
    (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.
    Online

    17
    ReputationRep:
    (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.
    • Wiki Support Team
    Offline

    14
    ReputationRep:
    Wiki Support Team
    (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.
    Offline

    15
    ReputationRep:
    (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.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What's your favourite Christmas sweets?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.