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anonymous

  • 5 years ago

If u=f(x,y) where x=(e^s)(cos(t)) and y=(e^s)(sin(t)), show that (du/dx)^2 + (du/dy)^2 = (e^(-2s))((du/ds)^2+(du/dt)^2))

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  1. anonymous
    • 5 years ago
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    pretty complicated chain rule problem, i'd really appreciate some help on this one

  2. anonymous
    • 5 years ago
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    what do you mean (du/dx)^2 is that the second derivative ?

  3. anonymous
    • 5 years ago
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    no, its the square of du/dx

  4. anonymous
    • 5 years ago
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    no, its the square of du/dx

  5. anonymous
    • 5 years ago
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    Its kinda confusing since usually f(x,y) is something like X^2 + y^3 something like that. Another confusing thing about the problem is that you have the derivative of u in respect to x and (du/dx) and derivative of u in respect to y (du/dy) which doesnt make sense.

  6. anonymous
    • 5 years ago
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    i mean you can say u= x+ y then the question would make more sense but in that case you have to find (dx/dt) or (dx/ds) or vice versa with y

  7. anonymous
    • 5 years ago
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    you mean have x+y be the function? f(x,y)=x+y?

  8. anonymous
    • 5 years ago
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    no usually the problem is set up this way f(x,y)= z if z= w/e ( it has an x and a y) where y= (something that has s and t) and x= ( something that has s and t) but your question does not make sense to me. Are you sure you wrote it right?

  9. anonymous
    • 5 years ago
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    yes, i copied it exactly.

  10. anonymous
    • 5 years ago
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    what does it mean to find du/dx?

  11. anonymous
    • 5 years ago
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    To find the derivative of the function u in respect to x for example lets say u= x^2 if you want to find du/dx = 2* x

  12. anonymous
    • 5 years ago
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    ok well thanks anyway for your help

  13. anonymous
    • 5 years ago
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    Still there?

  14. anonymous
    • 5 years ago
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  15. anonymous
    • 5 years ago
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    YES I AM. you really are a hero! the only thing i dont understand is what du/ds and du/dt are equal to, and how you got that.

  16. anonymous
    • 5 years ago
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    when you say u sub x times c, what is c?

  17. anonymous
    • 5 years ago
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    Oh, sorry, I use "c" and "s" for cos(t), sin(t) here.

  18. anonymous
    • 5 years ago
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    Where I write "NTS", that means, "need to show".

  19. anonymous
    • 5 years ago
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    It's messy.

  20. anonymous
    • 5 years ago
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    yes, but you have no idea how messy mine is after working on it for over an hour now. thank you so much!

  21. anonymous
    • 5 years ago
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    du/ds and du/dt...that comes from a result in calculus (chain rule for functions of many variables).

  22. anonymous
    • 5 years ago
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    You have u = f(x(s,t), y(s,t)). The parameters that do all the 'controlling' in the function are s and t, so you take the partial derivative with respect to them.

  23. anonymous
    • 5 years ago
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    Is that okay?

  24. anonymous
    • 5 years ago
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    yes, thank you.

  25. anonymous
    • 5 years ago
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    You're welcome :)

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