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Homework Help: Chain rule and partial derivatives

  1. Mar 31, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose the differentiable function f(x,y,z) has the partial derivatives fx(1,0,1) = 4, fy(1,0,1) = 1 and fz(1,0,1) = 0. Find g'(0) if g(t) = f(t2 + 1, t2-t, t+1).

    2. Relevant equations

    3. The attempt at a solution
    Ok I'm given the solution for this and I'm trying to work through it but I'm confused.

    I understand we have g(t) = f(x(t),y(t),z(t)) where x = t2+1; y = t2-t; z = t+1.

    So i thought I would just use the chain rule to find g'(t) and plug in 0 for t. But I check the solutions sheet and he uses a very different (easier) method that I don't understand. ( I just realized that we aren't given a function f in terms of x y or z so thats why this wouldn't work and we need another method. Still I don't understand this other method.)

    First he says when t=0 (x,y,z)|t=0 = (1,0,0) (I think he may have made a mistake, shouldn't it be (1,0,1) ?) And then he gets a point P = (1,-1,0) No idea were this point comes from...

    Next he sets up the chain rule equation: [tex]\frac{dg}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial z} \frac{dz}{dt}[/tex]

    Now I think he somehow uses that point and the partial derivatives given above to solve the equation and he gets: = 1*0 + 1*(-1) + (0*1) = -1 (depending on whether P was correct above may change the answer here I think)

    So could someone please explain the method used here?
  2. jcsd
  3. Mar 31, 2010 #2


    Staff: Mentor

    Another way to write this is g'(t) = fx * x'(t) + fy * y'(t) + fz * z'(t), and what you want is g'(0). When t = 0, what are x(0), y(0), and z(0)? You need to evaluate all three partial deriviatives at the point (x, y, z) for which t = 0, and you need to evaluate all three ordinary derivatives at t = 0.
  4. Mar 31, 2010 #3
    Ah, so then x'(0) = 2t = 2(0) = 0
    y'(0) = 2t -1 = -1
    z'(0) = 1

    Now I understand where those come from but I'm still not sure what you mean by "You need to evaluate all three partial derivatives at the point (x, y, z) for which t = 0".
  5. Mar 31, 2010 #4


    Staff: Mentor

    Not quite. You are confusing the derivatives of x(t), y(t), and z(t) with the value of each derivative at a specific value of t. IOW, x'(t) = 2t, but x'(0) = 0, and similarly for y'(t) vs. y'(0) and z'(t) vs. z'(0). In each case, the deriviative is a function, while the derivative evaluated at a particular value of t is just a number. x'(t) is different from x'(0), y'(t) is different from y'(0), and z'(t) is different from z'(0).

    The same difference exists between g'(t) and g'(0). Using the chain rule you can write g'(t) = fx(x,y,z) * x'(t) + fy(x,y,z) * y'(t) + fz(x,y,z) * z'(t). Since you aren't given any details about f(x, y, z), it's not possible to calculate the functions fx(x, y, z), fy(x, y, z), and fz(x, y, z). You are, however given the values of these partials at a particular point (1, 0, 1) that corresponds to t = 0.

    So g'(0) = fx(1,0,1) * x'(0) + fy(1,0,1) * y'(0) + fz(1,0,1) * z'(0). The only reason I have used the subscript notation (as opposed to the Leibniz notation you used) for the partial derivatives is that it's a little easier to explicitly show that they are to be evaluated at a particular point.
  6. Mar 31, 2010 #5
    Isn't this what I did? I think I may have messed up my notation a bit but I have x'(t) = 2t then x'(0) = 0; y'(t) = 2t-1 then y'(0) = -1; z'(t) = 1 then z'(0) = 1;

    So here since the point (1,0,1) corresponds to g(t) when t is 0 we can use the value of the partial at (1,0,1) as the value of fx(x,y,z)? Which would mean at (1,0,1) fx = 4; fy = 1; fz = 0.

    So if everything above is correct that gives me: 4*0 + 1*(-1) + 0*1 = -1.

    Also, does that mean (if this is all correct) this method can only be used if the point the partials are evaluated at and the value of t are directly related as in this example, correct?
  7. Mar 31, 2010 #6


    Staff: Mentor

    It's not really "messing up the notation a bit." You wrote some things that just plain aren't true. Here's what you wrote that I commented on.
    x'(0) [itex]\neq[/itex] 2t [itex]\neq[/itex] 0 and
    y'(0) [itex]\neq[/itex] 2t - 1 [itex]\neq[/itex] -1

    Since you apparently didn't understand the difference between a function of t and the value of a function at a specific number, I assumed that was the reason you were having such difficulties understanding that you needed to evaluated the partial derivatives at the point corresponding to t = 0, namely (1, 0, 1).
    A less roundabout way to say "g(t) when t is 0" is g(0).
    Here are the relationships.
    x(t) = t2 + 1
    y(t) = t2 - t
    z(t) = t + 1

    x(0) = 1
    y(0) = 0
    z(0) = 1

    We don't know the formulas of any of the three partials, so for example, we can't evaluate fx(x, y, z) at an arbitrary value of t - which determines the values of x, y, and z, but we are given fx(1, 0, 1) as being 4.
    Similarly, we don't know fy(x, y, z) or fz(x, y, z), but we're given the values of these partials when t = 0, which means we know the value of each partial evaluated at (1, 0, 1).
    Yes, but you should write this in context. This is g'(0) = -1.
    Correct. If we didn't have the information about the values of the partials for t = 0 (i.e., at (1, 0, 1)), we wouldn't be able to answer the question.
  8. Mar 31, 2010 #7
    Ok. My professor made several errors in his work on the solutions sheet which made this even more confusing for me. Thanks a lot for your help.
  9. Apr 1, 2010 #8


    Staff: Mentor

    Sure, you're welcome. I'm hopeful that this is less confusing for you now.
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