082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

838 Chapter 25 Functions of several variables
We note the following properties of the second-order Taylor polynomial:
(1) The values of the Taylor polynomial and the function areidentical at (a,b).
(2) The values of the first partial derivatives of the Taylor polynomial and the function
areidentical at (a,b).
(3) The values of the second partial derivatives of the Taylor polynomial and the function
areidentical at (a,b).
(4) Thesecond-orderTaylorpolynomialcontainsquadraticterms,thatistermsinvolvingx
2 ,y 2 andxy.
Example25.12 A function, f, and its first and second partial derivatives are evaluated at (2,−1). The
values are
f = 3
∂f
∂x = 4
∂f
∂y = −3
∂ 2 f
∂x 2 = 1
∂ 2 f
∂x∂y = 2
(a) State the second-order Taylor polynomial generated by f about (2,−1).
(b) Estimate f (1.8,−0.9).
(c) Verifythatthe Taylor polynomial and f have identical values at (2,−1).
∂ 2 f
∂y 2 = −1
(d) Verify that the first partial derivatives of the Taylor polynomial and f are identical
at (2,−1).
(e) VerifythatthesecondpartialderivativesoftheTaylorpolynomialand f areidentical
at (2,−1).
Solution (a) We puta = 2 andb=−1 inthe formulafor p 2
(x,y):
p 2
(x,y)=f +(x−2) ∂f
∂x +(y+1)∂f ∂y
+ 1 (
)
(x−2) 2∂2 f
2 ∂x +2(x −2)(y +1) ∂2 f f
2 ∂x∂y +(y+1)2∂2 ∂y 2
where f and its derivatives areevaluated at (2,−1). So
p 2
(x,y) =3+(x−2)4+(y+1)(−3)+ 1 2 {(x −2)2 +2(x −2)(y +1)2
+(y +1) 2 (−1)} = x2
2 − y2
2 +2xy+4x−8y−21 2
(b) The value of p 2
(1.8,−0.9) isan estimateof f (1.8,−0.9):
p 2
(1.8,−0.9) = (1.8)2
2
= 1.875
− (−0.9)2 +2(1.8)(−0.9)+4(1.8) −8(−0.9)− 21 2
2
(c) p 2
(2,−1) = 22
2 − (−1)2 +2(2)(−1) +4(2) −8(−1) − 21 2
2 = 3
f(2,−1)=3
Hence p 2
(2,−1) = f (2,−1); that is, the Taylor polynomial and the function have
identical values at (2,−1).
(d) The first partialderivatives of p 2
are
∂p 2
∂x =x+2y+4 ∂p 2
∂y =−y+2x−8