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

(b) At (0,0,0), ∇φ =0i+3j+0k =3j.
26.3 The gradient of a scalar field 853
(c) At(1,1,1),∇φ=(3×1 2 ×1+1)i+(1 3 +2×1×1+3)j+0k=4i+6j+0k
so that |∇φ| at (1,1,1) isequal to √ 4 2 +6 2 = √ 52.
So far we have been given φ and have calculated ∇φ. Sometimes we will be given ∇φ
and will needtofind φ. ConsiderExample 26.4.
Example26.4 IfF = ∇φ find φ whenF = (3x 2 +y 2 )i + (2xy +5)j.
Solution Note that in this example F has only two components. Consequently ∇φ will have two
components,thatisF = ∇φ = ∂φ
∂x i + ∂φ j. Therefore
∂y
(3x 2 +y 2 )i + (2xy +5)j = ∂φ
∂x i + ∂φ
∂y j
Equating theicomponents we have
∂φ
∂x =3x2 +y 2 (26.1)
Equating thejcomponents we have
∂φ
∂y
=2xy+5 (26.2)
Integrating Equation (26.1) w.r.t.xand treatingyas a constant we find
φ=x 3 +xy 2 +f(y) (26.3)
where f (y) is an arbitrary function of y which plays the same role as the constant of
integration does when there is only one independent variable. Note in particular that
∂
∂φ
f (y) = 0.Check by partialdifferentiation that
∂x ∂x =3x2 +y 2 .
Integrating Equation (26.2) w.r.t.yand treatingxas a constant we find
φ=xy 2 +5y+g(x) (26.4)
∂
whereg(x) is an arbitrary function ofx. Note that g(x) = 0. Check by partial differentiationthat
∂φ = 2xy+5.Comparingbothformsfor φ giveninEquations(26.3)and
∂y
∂y
(26.4) wesee thatby choosingg(x) =x 3 and f (y) = 5y wehave
φ=x 3 +xy 2 +5y
Check that F is indeed equal to ∇φ. Also check that by adding any constant to φ the
same property holds, thatisFisstillequal to ∇φ.