MAT235CALCULUS II
SUMMER 2009
ASSIGNMENT #3, DUE ON JULY 9
SOLUTIONS
Problem 1
Gradient vector, and directional derivatives.
a) Suppose that the temperature at a point (
x, y
) in the plane is given by
T
(
x, y
) =
xye

x
2

y
2
.
Explain how the quantity
∇
T
(
x, y
) is related to the flow of heat in the plane. Evaluate
the integral
1
0
∇
T
(
x,
1)
·
<
0
,
1
> dx
and explain what it represents physically.
Solution: Recall that the vector
∇
T
(
x, y
) is called the gradient vector.
The direction of
this vector is the direction where the function
T
increases fastest. Also
∇
T
(
x, y
) is the
direction where
T
decreases faster.
This direction is naturally related to the flow of heat
because the heat flows from the warmer parts to the colder parts. To evaluate the integral
observe that
1
0
∇
T
(
x,
1)
·
<
0
,
1
> dx
=
∂T
∂y
(
x,
1)
dx.
(1)
On the other hand we have
∂T
∂y
(
x, y
) =
xe

x
2

y
2

2
xy
2
e

x
2

y
2
,
which gives that
∂T
∂y
(
x,
1) =

xe

x
2

1
,
(2)
which together with (1) gives that
1
0
∇
T
(
x,
1)
·
<
0
,
1
> dx
=
1
0

xe

x
2

1
dx.
(3)
Let
u
=

x
2

1. Since
du
= 2
xdx
, and the bound
x
= 0 gives
u
=

1, and the bound
x
= 1
gives
u
=

2 the integral becomes
1
0

xe

x
2

1
dx
=

2

1
e
u
2
du
=
1

e
2
e
2
.
Observe that
∇
T
(
x,
1)
·
<
0
,
1
>
gives us the directional derivative of
T
in the direction
<
0
,
1
>
. In other words, it gives us the rate at which energy flows to (
x,
1) from the upper
1
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2
half plane.
Integrating it over the line segment
{
(
x,
1) : 0
≤
x
≤
1
}
gives us the rate of
energy flow from the upper half plane to the lower half plane for 0
≤
x
≤
1.
Since the
integral is negative, energy flows from the lower half plane to the upper half plane. (Here
the upper half plane and the lower half plane are divided by the line
y
= 1.)
Another way to understand why the integral represents rate of flow of energy is to realize
that the dot product of
∇
T
(
x,
1)
·
<
0
,
1
>
tells us how much
∇
T
(
x,
1) points upwards
or downwards. Therefore it measures how fast the heat flows upwards or downwards. Since
1
0
∇
T
(
x,
1)
·
<
0
,
1
> dx >
0
,
it means that overall the
∇
T
(
x,
1) is pointing upwards, and therefore the heat flows up
wards.
b) Find all planes that are tangent to both
x
2
+
y
2
+
z
2
= 1 and 2
z
+
x
2
+
y
2
= 0.
Solution: The surfaces can be described by the equations
f
(
x, y, z
) = 1, and
g
(
x, y, z
) = 0,
where
f
(
x, y, z
) =
x
2
+
y
2
+
z
2
, and
g
(
x, y, z
) =
x
2
+
y
2
+2
z
. Suppose that the tangent plane
of the sphere
f
(
x, y, z
) at (
x
1
, y
1
, z
1
) is tangent to the elliptic paraboloid
g
(
x, y, z
) = 0 at the
point (
x
2
, y
2
, z
2
).
Since both
∇
f
(
x
1
, y
1
, z
1
)
,
and
∇
g
(
x
2
, y
2
, z
2
) are orthogonal to the plane, they will have
to be parallel to each other. This means that
∇
f
(
x
1
, y
1
, z
1
) =
λ
∇
g
(
x
2
, y
2
, z
2
)
.
or
< x
1
, y
1
, z
1
>
=
λ < x
2
, y
2
,
1
> .
(4)
On the other hand the equation of the tangent plane is going to be
∇
f
(
x
1
, y
1
, z
1
)
·
< x

x
1
, y

y
1
, z

z
1
>
= 0
.
and since
∇
f
(
x
1
, y
1
, z
1
) = 2
< x
1
, y
1
, z
1
>
this gives
xx
1
+
yy
1
+
zz
1
=
x
2
1
+
y
2
1
+
z
2
1
= 1
so the equation of the tangent plane to the sphere at (
x
1
, y
1
, z
1
) is
xx
1
+
yy
1
+
zz
1
= 1
(5)
Since (
x
2
, y
2
, z
2
) has to pass through this plane we obtain that
x
1
x
2
+
y
1
y
2
+
z
1
z
2
= 1
.
(6)
This together with (4) gives that
λ
(
x
2
2
+
y
2
2
+
z
2
) = 1
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 Summer '08
 Recio
 Derivative, Multivariable Calculus, Multiple integral

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