Question

Let w(s,t)=F(u(s,t),v(s,t)),where F,u and v are differentiable,u(1,0)=2

us=(1,0)=-2,ut=(1,0)=6,v(1,0)=3,vs(1,0)=5,vt(1,0)=4,Fu(2,3)=-1, Fv(2,3)=10

Find ws(1,0) AND wt(1,0)

Answer 1

Using the chain rule for partial derivatives, we found that ws(1,0) is 52 and wt(1,0) is 34.

Step-by-step explanation:

To find ws(1,0) and wt(1,0), we need to use the chain rule for partial derivatives because w is defined as a composition of other functions: w(s,t) = F(u(s,t), v(s,t)). The chain rule in this context will help us find how the function w changes with respect to s and t.

Firstly, using the information for partial derivatives of u and v at the point (1,0) and the values of F with respect to u and v when u=2 and v=3, we find ws(1,0) as follows:

ws(1,0) = Fu(u(1,0),v(1,0))*us(1,0) + Fv(u(1,0),v(1,0))*vs(1,0)
= (-1)*(-2) + (10)*(5) = 2 + 50 = 52.

Similarly, to find wt(1,0), we apply the chain rule:

wt(1,0) = Fu(u(1,0),v(1,0))*ut(1,0) + Fv(u(1,0),v(1,0))*vt(1,0)
= (-1)*(6) + (10)*(4) = -6 + 40 = 34.

Answer 2

`ws(1,0)=52`

`wt(1,0)=34`

Step-by-step explanation:

u v

`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`

To find
`w_(s)` :

we have to use the above equation, which is subscription notation

w(s, t) = F(u(s, t), v(s, t))

(1, 0) = ((u(1, 0), v(1, 0)), (u(1, 0), v(1, 0))) · ((1, 0), (1, 0))

= ((2, 3), (2, 3)) · (−2, 5)

=
`(-1, 10) * (-2, 5)` expand

=
`(2)+(50)`

=
`52`

To find
`w_(t)` :

wt(1, 0) = ((u(1, 0), v(1, 0)), (u(1, 0), v(1, 0))) · ((1, 0), (1, 0))

= ((2, 3), (2, 3)) · (6, 4)

=
`(-1, 10) * (6, 4)` expand

=
`(-6)+(40)`

=
`34`