Mathematical Tools for Physics

8—Multivariable Calculus 215

For the case of three independent variables, I’ll leave the sketch to you.

Examples
The temperature on the surface of a heated disk is given to beT(r,θ) =T 0 +T 1

(

1 −r^2 /a^2

)

,

whereais the radius of the disk andT 0 andT 1 are constants. If you start at positionx=c < a,
y= 0and move parallel to they-axis at speedv 0 what is the rate of change of temperature
that you feel?
Use Eq. ( 7 ), and the relationr=

√

x^2 +y^2.

dT

dt

=

(

∂T

∂r

)

θ

dr

dt

+

(

∂T

∂θ

)

r

dθ

dt

=

(

∂T

∂r

)

θ

[(

∂r

∂x

)

y

dx

dt

+

(

∂r

∂y

)

x

dy

dt

]

=

(

− 2 T 1

r

a^2

)

[

y

√

x^2 +y^2

v 0

]

=− 2 T 1

√

c^2 +v 02 t^2

a^2

. v

2

√^0 t

c^2 +v 02 t^2

=− 2 T 1

v^20 t

a^2

As a check, the dimensions are correct (are they?). At time zero, this vanishes, and that’s what I expect
because at the beginning of the motion you’re starting to move in the directionperpendicularto the direction in
which the temperature is changing. The farther you go, the more nearly parallel to the direction of the radius
you’re moving. If you are moving exactly parallel to the radius, this time-derivative is easier to calculate; it’s then
almost a problem in a single variable.

dT

dt

≈

dT

dr

dr

dt

≈− 2 T 1

r

a^2

v 0 ≈− 2 T 1

V 0 t

a^2

v 0

So the approximate and the exact calculation agree. In fact they agree so well that you should try to find out
if this is a lucky coincidence or if there some special aspect of the problem that you might have seen from the
beginning and that would have made the whole thing much simpler.