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

25.3 Partial derivatives 827
Engineeringapplication25.3
Eddycurrentlosses
Eddycurrentsarecirculatingcurrentsthatariseinironcoresofelectricalequipment
as a resultof ana.c. magnetic field. They lead toenergy lossesgiven by
where
P e
=k e
f 2 B 2 max
P e
= eddy currentlosses (Wper unit mass),
B max
= maximum value of the magnetic field wave (T),
f = frequency of the magnetic field wave (Hz),
k e
= a constant thatdepends upon factors such as the lamination thickness of
the ironcore.
Calculate ∂P e
∂f , and ∂P e
∂B max
.
Solution
∂P e
∂f
=2k e
fB 2 max
and
∂P e
∂B max
=2k e
f 2 B max
Example25.3 IfV(x,y) = sin(xy),find ∂V
∂x
and
∂V
∂y .
Solution To find ∂V
∂x we treat y as a constant. Recalling that d
(sinkx) = kcoskx we find
dx
∂V
∂V
∂V
=ycos(xy).Tofind wetreatxas a constant. Thus
∂x ∂y ∂y =xcos(xy).
Example25.4 Findthe first partialderivatives ofzwhere
(a) z(x,y) =yxe x
(b) z(x,y) =x 2 sin(xy)
Solution (a) To find ∂z we must treatyas a constant. However, the differentiation of the factor
∂x
xe x will require use of the product rule.We find
∂z
∂x =y ∂ ∂x (xex )
=y((1)e x +xe x )
=ye x (x+1)