Multi Calculus- Partial Derivatives

[Roni1985]

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
e^yt+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 K_t(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 into H, no success.

Would appreciate any help.
Roni

 

[gabbagabbahey]

Start by applying the chain rule to $H$...

 

[Roni1985]

Start by applying the chain rule to $H$...

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

 

[Roni1985]

I tried the chain rule, but looks like I'm doing something wrong.

K_t=K_x*X_t+K_y*Y_t

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

 

[gabbagabbahey]

I tried the chain rule, but looks like I'm doing something wrong.

K_t=K_x*X_t+K_y*Y_t

I am assuming K(x,y)=H(x,y), a correct assumption ?

Well, sort of. $K(s,t)$ is explicitly a function of $s$ and $t$ only. You know that $x$ and $y$ are also some function of $s$ and $t$ 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]<br /> <br /> So, it is more correct to say <br /> <br /> [tex]K_t(s,t)=H_x x_t+H_y y_t[/itex]<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I am getting 0 and that's not the answer.<br /> K_x=2x-y<br /> K_y=2y-x </div> </div> </blockquote><br /> Well, $H_x=2x-y$ and $H_y=2y-x$...but what about $x_t$ and $y_t$...what are you getting for those?[/tex][/tex]

 

[Roni1985]

Well, sort of. $K(s,t)$ is explicitly a function of $s$ and $t$ only. You know that $x$ and $y$ are also some function of $s$ and $t$ 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]<br /> <br /> So, it is more correct to say <br /> <br /> [tex]K_t(s,t)=H_x x_t+H_y y_t[/itex]<br /> <br /> <br /> <br /> Well, $H_x=2x-y$ and $H_y=2y-x$...but what about $x_t$ and $y_t$...what are you getting for those?[/tex][/tex]

[tex]$$<br /> X_t=-x*cos(xt) <br /> Y_t=(-ye^yt)/(t*e^yt+s+1)<br /> <br /> But it doesn't matter what I'm getting for these because I multiply each one of them by a zero.$$[/tex]

 

[Roni1985]

ohh hold on, the s and t are zeros , so it doesn't mean x and y are zeros
wow

Edit:
got the answer,

thanks for the help

 

[gabbagabbahey]

X_t=-x*cos(xt)
Y_t=(-ye^yt)/(t*e^yt+s+1)

Your expression for $x_t$ doesn't look quite right.

But it doesn't 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. $H_x=2x-y\neq 0$ and $H_y=2y-x\neq 0$

 

[Roni1985]

Your expression for $x_t$ doesn't look quite right.

oh, it didn't influence my answer so I didn't see the mistake

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

Thank you.