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Homework Help: Changing the variable in a differential system

  1. Apr 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Hello everyone.

    I have differential system consists of:

    I want to change variable using:
    where z is a unit vector
    and rewrite the differential equations

    2. Relevant equations

    3. The attempt at a solution

    For the first equation:
    [itex]a_1\frac{dz_1}{dt}=a_1 z_1(1-2a_1 z_1+a_2 z_2)[/itex]
    [itex]a_2\frac{dz_2}{dt}=3*a_2 z_2(a_1 z_1-a_2 z_2)[/itex]

    Is this correct?
    I didn't use the part where it say: z is unit vector.

    Thanks a lot.
  2. jcsd
  3. Apr 22, 2012 #2


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    welcome to pf!

    hi fred_91! welcome to pf! :smile:
    yes, that looks ok :smile:

    (you can fix the "unit vector" part later :wink:)

    EDIT: oh, i've misunderstood the question :rolleyes:

    it meant ∑ aixi

    ignore the above, and follow sharks' :smile: advice ​
    Last edited: Apr 22, 2012
  4. Apr 22, 2012 #3


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    You should first find the derivatives of [itex]x_1[/itex] and [itex]x_2[/itex] with respect to [itex]z_i[/itex]:
    Then, apply the chain rule for both.
    [tex]\frac{dx_1}{dz_1}=\frac{dx_1}{dt} \times \frac{dt}{dz_1}
    \\\frac{dx_2}{dz_2}=\frac{dx_2}{dt} \times \frac{dt}{dz_2}[/tex]
    Last edited: Apr 22, 2012
  5. Apr 22, 2012 #4
    Re: welcome to pf!

    Thank you very much :)

    So, how would I use the information that z is a unit vector?

    sharks: thank you, do I have to use the chain rule?
  6. Apr 22, 2012 #5


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    Re: welcome to pf!

    Normally, you use the chain rule in this type of problem, to make it easier to see what derivative is required and what derivatives you already have. How did you do it?

    If z is a unit vector, then what is the result of the product of two unit vectors?
  7. Apr 22, 2012 #6
    Re: welcome to pf!

    I think I'm a bit confused now.
    The question says: I have to find the differential equations for A and Z
    x=Az and A>0 and z is a unit vector

    I simply used: [itex]z_i=\frac{x_i}{a_i}[/itex]
    and substituted it into the original differential equations for each i
    but now i'm starting to think that i made mistake.

    So, using the chain rule, I have:
    similarly for the second:

    Is this correct?
    But, I think my differential equations should be in terms of A and Z.

    For the unit vector part:
    dot product of 2 unit vectors:

    is this what you mean?
  8. Apr 22, 2012 #7


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    You haven't specified the exact or entire question for that matter. If the question requires "differential equations for A and Z", then you do need to give 2 differential equations, in which case your work is correct. If it had required you to give the answer is terms of z only, then you would need to work out another chain rule and you would then end up with a single differential equation, involving only [itex]a_1[/itex], [itex]a_2[/itex], [itex]z_1[/itex] and [itex]z_2[/itex], but no [itex]t[/itex].

    No, a unit vector is of the form (1,0) or (0,1) for a 2D coordinate system.
    Last edited: Apr 22, 2012
  9. Apr 22, 2012 #8
    Thank you.
    And sorry for not specifying clearly.
    The question is:
    x=Az where A>0 and z unit vector. What are the differential equations for A and z?

    So, does this mean my equations in the original attempt are correct?

    The product of 2 unit vectors is then: (1,0).(0,1)=0
    should this be included somehow?

    thanks again.
  10. Apr 22, 2012 #9


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    Your equations are correct, but you need to express them in terms of [itex]\frac{dz_1}{dt}[/itex] and [itex]\frac{dz_2}{dt}[/itex] only (everything else should be moved to the R.H.S.).

    [itex]z_1[/itex] and [itex]z_2[/itex] are unit vectors and they both have only one form (since they are both from the same given general equation, [itex]x_i=a_iz_i[/itex]) either (1,0) or (0,1).
    So, assuming [itex]z_1=(1,0)[/itex], then [itex]z_2=(1,0)[/itex].
    Now, [itex]z_1 \times z_2[/itex] gives...
    Last edited: Apr 22, 2012
  11. Apr 22, 2012 #10
    this gives:
    the cross product of a 2 by 2 is simply 1*0-1*0=0
  12. Apr 22, 2012 #11


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    That's incorrect. The scalar product of two vectors: [itex](A_x,B_x).(A_y,B_y) = A_x.A_y+B_x.B_y[/itex]
  13. Apr 22, 2012 #12
    Oh right, the scalar product. this is equal 2
    (sorry i thought you meant the cross product)

    so, should this be included in the differential equations?
  14. Apr 22, 2012 #13
    Sorry, i meant 0 (not 2) :)
  15. Apr 22, 2012 #14


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    That's incorrect. Write the two vectors (1,0) and (1,0) in their proper column forms on paper (so you can better understand), and then follow the formula that i've given above for the scalar product.
  16. Apr 22, 2012 #15
    Oh i see...sorry about that
    I thought they have to be 2 different unit vectors: (1,0) and (0,1).

    But, if they are the same: (1,0),and (1,0)
    then the cross product is = 1
  17. Apr 22, 2012 #16


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    Correct. Now, you can proceed to simplify your two differential equations.
  18. Apr 22, 2012 #17
    [itex]\frac{dz_1}{dt}=z_1−2a_1z_1z_1 +a_2 z_1z_2 [/itex]
    [itex]\frac{dz_2}{dt}=3a_1z_2z_1-3a_2z_2z_2 [/itex]
    [itex]\frac{dz_1}{dt}=z_1−2a_1 +a_2 [/itex]
    [itex]\frac{dz_2}{dt}=3a_1-3a_2 [/itex]

    Is this correct?
  19. Apr 22, 2012 #18


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    Correct. :smile:
  20. Apr 22, 2012 #19
    How can you solve this question x+y+z=1 x^2+y^2+z^2=35 x^3+y^3+z^3=97
  21. Apr 22, 2012 #20


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    Post your question in a new topic. Read the forum rules before you post or you could get banned.
  22. Apr 22, 2012 #21
    Thank you very much sharks.

    i'm not quite sure. is it relevant?
  23. Apr 24, 2012 #22
    I have another query about this problem:
    If z is a unit vector, then how can we find the derivative with respect to t?
    (isnt it just 0?)

    thank you
    Last edited: Apr 24, 2012
  24. Apr 24, 2012 #23


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    If [itex]\vec z[/itex] is (1,0), then [itex]\frac{d\vec z}{dx}[/itex] works just like divergence of [itex]\vec z[/itex].
    Another way to rewrite [itex]\vec z[/itex] would be: [itex]\vec i+0\vec j[/itex]
    [itex]div\, \vec z=\frac{\partial 1}{\partial x}+\frac{\partial 0}{\partial y}=0[/itex]
    However, you are not differentiating w.r.t.x but w.r.t.t which invalidates the concept. If you want to simplify your answer, you could integrate both sides w.r.t.t but then it wouldn't help unless you have some values to plug into those equations involving a1, a2, z1, z2 and now t.
    You could eliminate t by dividing those two resulting equations or simply by using chain rule on your 2 answers, to obtain a single equation involving only a1, a2, z1 and z2. But i don't think you're being asked to simplify any further in this problem.
    So, to answer your question - no, you cannot differentiate with respect to t.
    Last edited: Apr 24, 2012
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