Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Multivariable Chain Rule & Variable-Dependence

  1. Nov 1, 2007 #1
    I am having a terrible hard time with the multivariable chain rule and its related stuff (I read my textbook many times, but it doesn't help that much because the explanations are very limited). I hope that someone can help me to withdraw from this darkness of confusion.

    1) (Differentiation along curves) Find the rate of change of f(x,y) = 1 - x^2 - y^2 along the ellipse g(t) = (2cos(t),sin(t) ) at the point where t=pi/4.

    Using chain rule, we can get that the answer is d(f(g(t))/dt [evaluated at t=pi/4] =3.
    But what does this represent geometrically? I have no idea what is going on geometrically about differentiation along curves.
    Is the ellipse even lying on the surface f(x,y) = 1 - x^2 - y^2?
    If not, then why does it even make sense to talk about "the rate of change of f(x,y) = 1 - x^2 - y^2 along the ellipse g(t) = (2cos(t),sin(t) ) "?



    2) From my textbook:
    "Suppose w= f(x,y,t,s) and that x and y are themselves functions of the independent variables t and s.
    [note: Let D=curly d, representing partial derivatives]
    If we write Dw/Dt = (Dw/Dx)(Dx/Dt) + (Dw/Dy)(Dy/Dt) + (Dw/Dt) using the chain rule, this is nonsense because the Dw/Dt on the left and right denote different things. We definitely cannot cancel out the two Dw/Dt."

    Now this is driving me crazy, when I use the chain rule, the above expression is what I get. How can this be wrong?
    I absolutely don't understand why the Dw/Dt would be denoting different things.
    And also, how can we cope with this problem? (i.e. what would be the correct way to write the expression?)



    3) [note: Let D=curly d, representing partial derivatives]
    If, for example, w=f(x,y,z), are Dw/Dx and Df/Dx always always equal? My textbook seems to use Dw/Dx and Df/Dx quite interchangably in most cases, but I am not sure whether they are ALWAYS equal. The variable-dependence thing is just driving me crazy...


    Any help is greatly appreciated!:smile:
     
    Last edited: Nov 1, 2007
  2. jcsd
  3. Nov 1, 2007 #2
    Think of f as a scalar field (e.g. temperature), and you want to know the rate of scalar changes along the ellipse g.
     
    Last edited: Nov 1, 2007
  4. Nov 1, 2007 #3

    Avodyne

    User Avatar
    Science Advisor

    1) andytoh has the right idea. f(x,y) does not define a surface, it gives you the value of something (like temperature) at point in the x-y plane. The partial derivative with respect to x would tell you how fast f changes as you change x, while holding y fixed. This can be thought of as how fast f changes as you move along a particular curve in the x-y plane, the "curve" in this case being a straight line parallel to the x-axis. Now, generalize this to motion along a more general curve. That's the idea.

    2) Most books would write this as dw/dt = (Dw/Dx)(Dx/Dt) + (Dw/Dy)(Dy/Dt) + (Dw/Dt), where dw/dt denotes the total derivative of w with respect to t, as distinct from the partial derivative Dw/Dt, where you differentiate only with respect to the explicit dependence on t. The problem with calling this a total derivative is that there is still dependence on s. So, if we wrote out x(t,s) and y(t,s) explicitly, so that w was just a function of t and s, then what I have called the total derivative would normally be called the partial derivative with respect to t, with s held fixed. But if we call both these derivatives partial, then we can get confused, because what do we mean by Dw/Dt? This appears to be what your book is trying to explain.

    3) Once we have agreed on what we mean by the partial derivative, Dw/Dt and Df/Dt are always equal.
     
    Last edited: Nov 1, 2007
  5. Nov 3, 2007 #4
    Thanks a lot, but I have one more question.

    4) [Let D=curly d representing partial derivatives.]
    Q:Suppose u = f(x-ct) + g(x+ct), where c is a constant. Show that D^2u/Dx^2 = (1/c^2)(D^2u/Dt^2)

    The hint I get from the solutions manual is that
    http://www.geocities.com/asdfasdf23135/advcal6.JPG

    Firstly, I don't understand why it would even make sense to talk about f ''(x-ct) and g ''(x+ct) in this situation. t is not a constant, it's a variable, and the prime notation is used for functions of one variable only, but here f depends on 2 variables, x and t, why can we still use the prime notation? Shouldn't we be using the "curly d" notation here? I am absolutely lost here...

    Secondly, how can I actually find Du/Dx here? Can someone please show me the steps in detail? (I would really appreciate it) I know that I have to use the chain rule for sure, but I am EXTREMELY confused about keeping track of which variables depend on which variables, and I end up spending an hour looking at the question and fooling around, but still can't understand anything...

    Chain rule is just driving me crazy...

    I hope that someone can help me out! Thanks a million!
     
    Last edited: Nov 3, 2007
  6. Nov 3, 2007 #5

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    f is a function of the single variable y=x-ct, that is and g is function of the single variable z=x+ct.

    Thus, the expression f'' means [tex]\frac{d^{2}f}{dy^{2}}[/tex]
     
  7. Nov 3, 2007 #6

    Avodyne

    User Avatar
    Science Advisor

    Following arildno's notation, u(x,t) = f(y) + g(z). So Du/Dx = (df/dy)(Dy/Dx)+(dg/dz)(Dz/Dx)=f'(y)(1)+g'(z)(1)=f'(x-ct)+g'(z+ct), where the prime denotes a derivative with respect to the argument.
     
  8. Nov 3, 2007 #7

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    To remove ambiguities in notation:
    We have y=Y(x,t)=x-ct, z=Z(x,t)=x+ct, and u(x,t)=f(Y(x,t))+g(Z(x,t))[/tex]
    Thus, for example,
    [tex]\frac{\partial{u}}{\partial{x}}=\frac{df}{dy}\mid_{y=Y(x,t)}\frac{\partial{Y}}{\partial{x}}+\frac{dg}{dz}\mid_{z=Z(x,t)}\frac{\partial{Z}}{\partial{x}}[/tex]
     
  9. Nov 3, 2007 #8
    Why can x-ct be treated as ONE variable?

    Also, what is the difference between your notation and Avodyne's notation? How does it remove ambiguitites? Sorry, I don't understand this part.
     
  10. Nov 4, 2007 #9

    HallsofIvy

    User Avatar
    Science Advisor

    ANYTHING can be treated as one variable! Just make a substitution. For example, if
    F(x,t)= cos(x- ct) then you can let u= x- ct so that F(x,t)= F(u)= cos(u) and apply the chain rule:
    [tex]\frac{\partial F}{\partial x}= \frac{dF}{du}\frac{\partial u}{\partial x}[/tex]
    [tex]\frac{\partial F}{\partial x}= -sin(u)(1)= -sin(x- ct)[/tex]
    and
    [tex]\frac{\partial F}{\partial t}= \frac{dF}{du}\frac{\partial u}{\partial t}[/tex]
    [tex]\frac{\partial F}{\partial t}= -sin(u)(-c)= c sin(x- ct)[/tex]

    The difference between Avodyne's and Arildno's notations is that Arildno actually wrote down the substitutions: y=x-ct, z= x+ ct instead of just indicating them.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook