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anonymous
 3 years 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)
anonymous
 3 years 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)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it'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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1where is v in your formula?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0F(u(s,t),v(s,t)). where... yes. it is

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1W(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)

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1looks like multivariable chain rule problem ...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@experimentX yes, it is. @TuringTest and then?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1Fu(2,3) < this 2 and 3 are the values of u and v when (s,t) = (1,0) .. this is just change of variables.

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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1the second formula you wrote is not right, the first is

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1\[ F_u(u(1,0), v(1,0))\]

anonymous
 3 years ago
Best 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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1you can't simply equate the partial differential with the function itself.

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

experimentX
 3 years ago
Best 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.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1instead 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'?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hey. 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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1suppose ... 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok, 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.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0to 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok, sorry, I think tooooo far. waste your time

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1its okay ... the partial with respect to time will give you velocity along x and y axis

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah. 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.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1i've been doing this for last 20 years

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and never ask yourself the silly question like me?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1of 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks 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
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