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Multi Calculus- Partial Derivatives

  • Thread starter Roni1985
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  • #1
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Homework Statement



I am translating the question from another language so it might not be word to word with the original question.

assume x(s,t) and y(s,t) determined by these two functions:

sin(xt) +x+s=1
eyt+y(s+1)=1

The following function is defined H(x,y)=x2+y2-xy
such that
K(s,t)=H(x(s,t),y(s,t))

find the partial derivative Kt(0,0) (the derivative of K in terms of t)

Homework Equations





The Attempt at a Solution



first I wanted to solve the first two equation system in terms of x and y, but it didn't work.
tried also plugging them in to H, no success.

Would appreciate any help.
Roni
 

Answers and Replies

  • #2
gabbagabbahey
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Start by applying the chain rule to [itex]H[/itex]...
 
  • #3
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Start by applying the chain rule to [itex]H[/itex]...
what do you mean by the chain rule ?

should I first replace the x,y with s,t ?
I don't really see the lead ....
would appreciate if you could give any more hints but I'm going to try it also
 
  • #4
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I tried the chain rule, but looks like I'm doing something wrong.

Kt=Kx*Xt+Ky*Yt

I am assuming K(x,y)=H(x,y), a correct assumption ?
I am getting 0 and that's not the answer.
Kx=2x-y
Ky=2y-x

no matter what I multiply these by, I must get a zero but the answer is not zero.
So I guess I am missing something
 
  • #5
gabbagabbahey
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I tried the chain rule, but looks like I'm doing something wrong.

Kt=Kx*Xt+Ky*Yt

I am assuming K(x,y)=H(x,y), a correct assumption ?
Well, sort of. [itex]K(s,t)[/itex] is explicitly a function of [itex]s[/itex] and [itex]t[/itex] only. You know that [itex]x[/itex] and [itex] y[/itex] are also some function of [itex]s[/itex] and [itex]t[/itex] and that

[tex]K(s,t)=H(x(s,t),y(s,t))=[x(s,t)]^2+[y(s,t)]^2+x(s,t)y(s,t)[/itex]

So, it is more correct to say

[tex]K_t(s,t)=H_x x_t+H_y y_t[/itex]

I am getting 0 and that's not the answer.
Kx=2x-y
Ky=2y-x
Well, [itex]H_x=2x-y[/itex] and [itex]H_y=2y-x[/itex]....but what about [itex]x_t[/itex] and [itex]y_t[/itex]....what are you getting for those?
 
  • #6
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Well, sort of. [itex]K(s,t)[/itex] is explicitly a function of [itex]s[/itex] and [itex]t[/itex] only. You know that [itex]x[/itex] and [itex] y[/itex] are also some function of [itex]s[/itex] and [itex]t[/itex] and that

[tex]K(s,t)=H(x(s,t),y(s,t))=[x(s,t)]^2+[y(s,t)]^2+x(s,t)y(s,t)[/itex]

So, it is more correct to say

[tex]K_t(s,t)=H_x x_t+H_y y_t[/itex]



Well, [itex]H_x=2x-y[/itex] and [itex]H_y=2y-x[/itex]....but what about [itex]x_t[/itex] and [itex]y_t[/itex]....what are you getting for those?
Xt=-x*cos(xt)
Yt=(-yeyt)/(t*eyt+s+1)

But it doesnt matter what I'm getting for these because I multiply each one of them by a zero.
 
  • #7
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ohh hold on, the s and t are zeros , so it doesnt mean x and y are zeros
wow

Edit:
got the answer,

thanks for the help
 
  • #8
gabbagabbahey
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Xt=-x*cos(xt)
Yt=(-yeyt)/(t*eyt+s+1)
Your expression for [itex]x_t[/itex] doesn't look quite right.

But it doesnt matter what I'm getting for these because I multiply each one of them by a zero.
I'm not sure why you think this. [itex]H_x=2x-y\neq 0[/itex] and [itex]H_y=2y-x\neq 0[/itex]
 
  • #9
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Your expression for [itex]x_t[/itex] doesn't look quite right.
oh, it didn't influence my answer so I didn't see the mistake

I think it's gotta be
-x*cos(xt)/(t*cos(xt)+1)

Thank you.
 

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