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mixer [17]
4 years ago
5

Let w(s,t)=f(u(s,t),v(s,t)) where u(1,0)=−6,∂u∂s(1,0)=5,∂u∂1(1,0)=7 v(1,0)=−8,∂v∂s(1,0)=−8,∂v∂t(1,0)=6 ∂f∂u(−6,−8)=−1,∂f∂v(−6,−8

)=2
Mathematics
1 answer:
Blababa [14]4 years ago
6 0
w(s,t)=f(u(s,t),v(s,t))

From the given set of conditions, it's likely that you are asked to find the values of \dfrac{\partial w}{\partial s} and \dfrac{\partial w}{\partial t} at the point (s,t)=(1,0).

By the chain rule, the partial derivative with respect to s is

\dfrac{\partial w}{\partial s}=\dfrac{\partial f}{\partial u}\dfrac{\partial u}{\partial s}+\dfrac{\partial f}{\partial v}\dfrac{\partial v}{\partial s}

and so at the point (1,0), we have

\dfrac{\partial w}{\partial s}\bigg|_{(s,t)=(1,0)}=\dfrac{\partial f}{\partial 
u}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial u}{\partial s}\bigg|_{(s,t)=(1,0)}+\dfrac{\partial f}{\partial 
v}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial v}{\partial s}\bigg|_{(s,t)=(1,0)}
\dfrac{\partial w}{\partial s}\bigg|_{(s,t)=(1,0)}=(-1)(5)+(2)(-8)=-21

Similarly, the partial derivative with respect to t would be found via

\dfrac{\partial w}{\partial t}\bigg|_{(s,t)=(1,0)}=\dfrac{\partial f}{\partial 
u}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial u}{\partial t}\bigg|_{(s,t)=(1,0)}+\dfrac{\partial f}{\partial 
v}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial v}{\partial t}\bigg|_{(s,t)=(1,0)}
\dfrac{\partial w}{\partial t}\bigg|_{(s,t)=(1,0)}=(-1)(7)+(2)(6)=5
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