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

854 Chapter 26 Vector calculus
26.3.1 Physicalinterpretationof ∇φ
Suppose wethinkofthescalarfield φ(x,y,z)asdescribingthetemperaturethroughout
a region. This temperature will vary from point to point. At a particular point it can be
shown that ∇φ is a vector pointing in the direction in which the rate of temperature increaseisgreatest.
|∇φ|isthemagnitudeoftherateofincreaseinthatdirection.Similarly,
the rate of temperature decrease is greatest in the direction of −∇φ. Analogous interpretationsarepossibleforotherscalarfieldssuchaspressureandelectrostaticpotential.
Engineeringapplication26.1
Electrostaticpotential
Engineers working on the design of equipment such as cathode ray tubes and electricalvalves,whicharealsocommonlyknownbytheirgenerictermvacuumtubes,
needtocalculatetheelectrostaticpotentialthatresultsfromanaccumulationofstatic
chargesatvariouspointsinaregionofspace.Complicatedexamplesrequiretheuse
of a computer. Let us consider a simple example.
The electrostatic potential,V, inaregion isgiven by
y
V =
(x 2 +y 2 +z 2 ) 3/2
Suppose a unit charge is located in the region at the point with coordinates (2, 1, 0).
Find the direction atthis point,inwhich the rate ofdecrease inpotential isgreatest.
Solution
The rate of decrease isgreatest inthe direction of −∇V.
We first calculate the first partialderivatives ofV. Writing
we find
∂V
V =y(x 2 +y 2 +z 2 ) −3/2
∂x = (− 3 2
)
y(x 2 +y 2 +z 2 ) −5/2 (2x) = −3xy(x 2 +y 2 +z 2 ) −5/2
∂V
(−
∂y = 3 )
y(x 2 +y 2 +z 2 ) −5/2 (2y) + (x 2 +y 2 +z 2 ) −3/2
2
= −3y 2 (x 2 +y 2 +z 2 ) −5/2 + (x 2 +y 2 +z 2 ) −3/2
∂V
(−
∂z = 3 )
y(x 2 +y 2 +z 2 ) −5/2 (2z) = −3yz(x 2 +y 2 +z 2 ) −5/2
2
These partialderivatives can each be evaluated atthe point(2, 1,0). Thatis,
∂V
∂V
∂x ∣ = −0.107
∂V
(2,1,0)
∂y ∣ = 0.036
(2,1,0)
∂z ∣ = 0
(2,1,0)
and so
Finally,
∇V| (2,1,0)
= −0.107i +0.036j +0k
−∇V| (2,1,0)
= 0.107i −0.036j −0k