Question

Let R(s, t) = F(u(s, t), v(s, t)), and assume the following are true.

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 the values below.

Rs(1, 0) =
Rt(1, 0) =

Answer 1

To find Rs(1,0) and Rt(1,0), we need to use the chain rule of partial differentiation.

Rs(1,0) can be found by computing the partial derivative of R with respect to s, while holding t constant and then evaluating the result at (1,0). Similarly, Rt(1,0) can be found by computing the partial derivative of R with respect to t, while holding s constant and then evaluating the result at (1,0).

Using the chain rule, we have:

Rs = Fu * us + Fv * vs

Rt = Fu * ut + Fv * vt

We are given the values of u, v, us, ut, vs, vt, Fu, and Fv at the point (1,0) and the values of u and v at the point (2,3). We can use these values to compute Rs(1,0) and Rt(1,0) as follows:

Rs(1,0) = Fu(2,3) * us(1,0) + Fv(2,3) * vs(1,0)

= (-1) * (-2) + (10) * (5)

= 52

Rt(1,0) = Fu(2,3) * ut(1,0) + Fv(2,3) * vt(1,0)

= (-1) * (6) + (10) * (4)

= 34

Therefore, Rs(1,0) = 52 and Rt(1,0) = 34