Asked by jhon
find the approximate change in z when the point (x,y) changes from (1,1)to (1.01, 0.90)
f(x,y) = xe^(-y) + ye^(-x)
f(x,y) = xe^(-y) + ye^(-x)
Answers
Answered by
Steve
just use the total differential
dz = Zx dx + Zy dy
= (e^-y - ye^-x)dx + (-xe^-y + e^-x)dy
= (e^-1 - 1e^-1)(.01) + (-1e^-1 + e^-1)(-.1)
= 0
seems odd, but I don't see where I might have gone astray.
dz = Zx dx + Zy dy
= (e^-y - ye^-x)dx + (-xe^-y + e^-x)dy
= (e^-1 - 1e^-1)(.01) + (-1e^-1 + e^-1)(-.1)
= 0
seems odd, but I don't see where I might have gone astray.
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