Multivariable calculus partial derivatives/chain rule

[kaitamasaki]

Homework Statement

[PLAIN]http://img195.imageshack.us/img195/8196/mathqshwk4.jpg
Find [URL]https://webwork.elearning.ubc.ca/webwork2_files/tmp/equations/f3/d5a53499bbf12cafa144440e3095781.png[/URL]
and [URL]https://webwork.elearning.ubc.ca/webwork2_files/tmp/equations/b1/ce11cb4b480b6159791b1605b6f2681.png[/URL]

Homework Equations

Chain rule from multivariable calculus.

The Attempt at a Solution

I have tried doing the obvious: multiplying the given numbers, but I am just confused by what F_u and F_v at 5,3 means.

Instinct tells me the question requires F_u at (1,0), multiplied by u_s at (1,0), but none of the answers I've tried have worked. I am more confused by the way this question is asked because I understand multivariable chain rule quite well already to do most other questions.

 

[HallsofIvy]

F_u(5,3) means means the partial derivative of F, with respect to u, evaluated at u= 5, v= 3. That's pretty much standard notation. When s= 1, t= 0, then u= 5, v= 3 and you evaluate F and its derivatives at u and v, not s and t. That's why they are labeling F(u(x,t), v(s,t)) as "W(s, t)" rather than "F(s,t)".

From the chain rule, directly,
W_s(1, 0)= F_u(5, 3)u_s(1, 0)+ F_v(5, 3)v_s(1, 0)
and
W_t(1, 0)= F_u(5, 3)u_t(1, 0)+ F_v(5, 3)v_t(1, 0)[/itex]<br /> <br /> and all of those numbers are given.