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

826 Chapter 25 Functions of several variables
Example25.1 Givenz(x,y) =x 2 y +sinx +xcosyfind ∂z ∂z
and
∂x ∂y .
Solution To find ∂z we differentiate z w.r.t. x, treating y as a constant. Note that since y is a
∂x
constant then soiscosy.
∂z
∂x =2xy+cosx+cosy
Infinding ∂z
∂y ,xand hencex2 and sinx areheld fixed, thus
∂z
∂y =x2 −xsiny
Example25.2 Givenz(x,y) = 3e x −2e y +x 2 y 3
(a) findz(1,1)
(b) find ∂z ∂z
and
∂x ∂y whenx=y=1.
Solution (a) z(1,1) = 3e 1 −2e 1 +1 = 3.718
∂z
(b)
∂x =3ex +2xy 3
∂z
∂y =−2ey +3x 2 y 2
Whenx =y=1,then
∂z
∂x =3e+2=10.155
∂z
∂y =−2e+3=−2.437
Atthepoint(1,1,3.718)onthesurfacedefinedbyz(x,y),theheightofthesurface
abovethex--yplaneisincreasinginthexdirection,anddecreasingintheydirection.
Note thatwecould alsowrite
∂z
∂z
∂x∣ = 10.155 and
(1,1)
∂y∣ = −2.437
(1,1)
or
∂z
∂x (1,1) = 10.155 and ∂z
(1,1) = −2.437
∂y
Both notations areincommon use.