anonymous
  • anonymous
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))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
pretty complicated chain rule problem, i'd really appreciate some help on this one
anonymous
  • anonymous
what do you mean (du/dx)^2 is that the second derivative ?
anonymous
  • anonymous
no, its the square of du/dx

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anonymous
  • anonymous
no, its the square of du/dx
anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
you mean have x+y be the function? f(x,y)=x+y?
anonymous
  • anonymous
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?
anonymous
  • anonymous
yes, i copied it exactly.
anonymous
  • anonymous
what does it mean to find du/dx?
anonymous
  • anonymous
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
anonymous
  • anonymous
ok well thanks anyway for your help
anonymous
  • anonymous
Still there?
anonymous
  • anonymous
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anonymous
  • anonymous
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.
anonymous
  • anonymous
when you say u sub x times c, what is c?
anonymous
  • anonymous
Oh, sorry, I use "c" and "s" for cos(t), sin(t) here.
anonymous
  • anonymous
Where I write "NTS", that means, "need to show".
anonymous
  • anonymous
It's messy.
anonymous
  • anonymous
yes, but you have no idea how messy mine is after working on it for over an hour now. thank you so much!
anonymous
  • anonymous
du/ds and du/dt...that comes from a result in calculus (chain rule for functions of many variables).
anonymous
  • anonymous
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.
anonymous
  • anonymous
Is that okay?
anonymous
  • anonymous
yes, thank you.
anonymous
  • anonymous
You're welcome :)

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