let W(s,t) =F(u(s,t))where F,u,v are differentiable and u(1,0) =2; v(1.0)=3;us(1,0)=-2;vs(1,0)=5;ut(1,0)=6;vt(1,0)=4;Fu(2,3)=-1;Fv(2,3)=10. Find Ws(1,0) and Wt(1,0)

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let W(s,t) =F(u(s,t))where F,u,v are differentiable and u(1,0) =2; v(1.0)=3;us(1,0)=-2;vs(1,0)=5;ut(1,0)=6;vt(1,0)=4;Fu(2,3)=-1;Fv(2,3)=10. Find Ws(1,0) and Wt(1,0)

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it's quite easy to just put everything into the formula to get the answer is 34. my question is why dF/du at (1,0) can be calculate by Fu(2,3).the same problem with dF/dv
where is v in your formula?
F(u(s,t),v(s,t)). where... yes. it is

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W(1,0)=F(u(1,0),v(1,0))and we are told that u(1,0)=2 and v=(1,0)=3, so W(1,0)=F(2,3)
looks like multivariable chain rule problem ...
@experimentX yes, it is. @TuringTest and then?
Fu(2,3) <--- this 2 and 3 are the values of u and v when (s,t) = (1,0) .. this is just change of variables.
you mean\[Ws = \frac{ \delta F }{ \delta u }\frac{ \delta u }{ \delta s }+\frac{ \delta F }{ \delta v }\frac{ \delta v }{ \delta s }\] and
\[\frac{ \delta F }{ \delta u }= F(2,3)\]
is it right?
the second formula you wrote is not right, the first is
\[ F_u(u(1,0), v(1,0))\]
@TuringTest mine is wrong? why? @experimentX I think i got it. need time to make it clear in my mind. i will let you know when i get it perfectly. thanks anyway
you can't simply equate the partial differential with the function itself.
yes, i don't understand the goal of this part. we take partial derivative for what? tangent line respect to coordinates?
\[ \frac{ \partial F }{ \partial u }= F_u(2,3)\] is the correct expression ... this does't have any goal. this is simply like You know a function W of 's' and 't' ... but you do not know it as an explicit function of s and t.
instead you know it a function of 'u' and 'v' ... which in in turn function of 's' and 't' ... how do you find the tangent at given values of 's' and 't'?
Hey. think about t is time, s is the length, and v is velocity which function respect to s,t. and we need those part to get the function of v and some u(as a variable which respect to s, t , too. I don't think we study for nothing . if putting everything in big scenario, i think we get more effect than just formula, formula...and formula
suppose ... we have function like this, \[ \huge W(s,t) = \int_0^s \cosh (tx) dx + \sin(s)\cos(t) \] how do you calculate those partial values
ok, you are right, antiderivative for those parts? we must know the technique to solve that problem, but it must be used in somewhere, right? to me, the goal of partial derivative is find out the slope of tangent line in R3 respect to x,y,z.
yeah ... it is much easier to solve it (for partial) ... we can solve for it without knowing how to evaluate that integral.
to x axis, the tangent line has the slope totally different from it is to y axis and z axis as well. and those tangent line has the different equation, too. and then, to each dimension in R3, each of them respect to time, to velocity of something or at least to a stable origin as earth
ok, sorry, I think tooooo far. waste your time
its okay ... the partial with respect to time will give you velocity along x and y axis
yeah. when i were at kindergarden school, study addition and subtraction to count money or count my fingers. a little bit older, study multiplication and division to know how to count money and account. now, study partial derivative for ,,,,,don't know is really boring.
i've been doing this for last 20 years
and never ask yourself the silly question like me?
of course i've in the past .. i think you are just little confused with the notation. while doing these sorta question .. it does smoothly unless it's hard type.
Thanks for answering my question. I have many, many many homework from my classes. i wish i have 28 hours a day to solve all of them and to completely understand all of them, Thanks a lot. I have to go to school now. Hopefully I can get your help whenever I stuck at somewhere
sure ... anytime!!
Bye bye.

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