Question

let ( r(s,t)=g(u(s,t),v(s,t)) ), where ( g ), ( u ), and ( v ) are differentiable, and the following applies.

( u(-2,-1)=8 ) ( v(-2,-1)=-6 )
( u_{s}(-2,-1)=-3 ) ( v_{s}(-2,-1)=9 )
( u_{t}(-2,-1)=4 ) ( v_{t}(-2,-1)=1 )
( g_{u}(8,-6)=7 ) ( g_{v}(8,-6)=3 )

find ( r_{s}(-2,-1) ) and ( r_{t}(-2,-1) ).
( r_{s}(-2,-1)= )
( r_{t}(-2,-1)= )

Explanation:

Step1: Use the chain rule for partial derivatives

The chain rule for \(R_s\) is \(R_s = G_u\cdot u_s+G_v\cdot v_s\).
Substitute the given values: \(G_u = 7\), \(u_s=-3\), \(G_v = 3\), \(v_s = 9\).

$$R_s=7\times(-3)+3\times9$$

Step2: Calculate \(R_s\)

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Step3: Use the chain rule for partial derivatives

The chain rule for \(R_t\) is \(R_t=G_u\cdot u_t+G_v\cdot v_t\).
Substitute the given values: \(G_u = 7\), \(u_t = 4\), \(G_v=3\), \(v_t = 1\).

$$R_t=7\times4+3\times1$$

Step4: Calculate \(R_t\)

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Answer:

\(R_s(-2,-1)=6\)
\(R_t(-2,-1)=31\)