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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)
 one year ago
 one year ago
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)
 one year ago
 one year ago

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HoaBest ResponseYou've already chosen the best response.0
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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
where is v in your formula?
 one year ago

HoaBest ResponseYou've already chosen the best response.0
F(u(s,t),v(s,t)). where... yes. it is
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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)
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
looks like multivariable chain rule problem ...
 one year ago

HoaBest ResponseYou've already chosen the best response.0
@experimentX yes, it is. @TuringTest and then?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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.
 one year ago

HoaBest ResponseYou've already chosen the best response.0
you mean\[Ws = \frac{ \delta F }{ \delta u }\frac{ \delta u }{ \delta s }+\frac{ \delta F }{ \delta v }\frac{ \delta v }{ \delta s }\] and
 one year ago

HoaBest ResponseYou've already chosen the best response.0
\[\frac{ \delta F }{ \delta u }= F(2,3)\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
the second formula you wrote is not right, the first is
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
\[ F_u(u(1,0), v(1,0))\]
 one year ago

HoaBest ResponseYou've already chosen the best response.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
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
you can't simply equate the partial differential with the function itself.
 one year ago

HoaBest ResponseYou've already chosen the best response.0
yes, i don't understand the goal of this part. we take partial derivative for what? tangent line respect to coordinates?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
\[ \frac{ \partial F }{ \partial u }= F_u(2,3)\] 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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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'?
 one year ago

HoaBest ResponseYou've already chosen the best response.0
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
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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
 one year ago

HoaBest ResponseYou've already chosen the best response.0
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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
yeah ... it is much easier to solve it (for partial) ... we can solve for it without knowing how to evaluate that integral.
 one year ago

HoaBest ResponseYou've already chosen the best response.0
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
 one year ago

HoaBest ResponseYou've already chosen the best response.0
ok, sorry, I think tooooo far. waste your time
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
its okay ... the partial with respect to time will give you velocity along x and y axis
 one year ago

HoaBest ResponseYou've already chosen the best response.0
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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
i've been doing this for last 20 years
 one year ago

HoaBest ResponseYou've already chosen the best response.0
and never ask yourself the silly question like me?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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.
 one year ago

HoaBest ResponseYou've already chosen the best response.0
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
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
sure ... anytime!!
 one year ago
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