Chapter (2)
Applications of Partial
Differentiation
2.1 TANGENT PLANES AND NORMAL VECTORS
Definition
Assume that 𝐹 (𝑥, 𝑦, 𝑧) has continuous first-order partial
derivatives and that 𝑃0 (𝑥0 , 𝑦0 , 𝑧0 ) is a point on the level
surface 𝑆: 𝐹(𝑥, 𝑦, 𝑧) = 𝑐. If ∇F(𝑥0 , 𝑦0 , 𝑧0 ) = 0, then 𝒏 =
𝛁𝐅(𝒙𝟎 , 𝒚𝟎 , 𝒛𝟎 ) is a normal vector to 𝑆 at 𝑃0 and the
tangent plane to S at 𝑃0 is the plane with equation
F𝑥 (𝑥0 , 𝑦0 , 𝑧0 )(𝑥 − 𝑥0 ) + F𝑦 (𝑥0 , 𝑦0 , 𝑧0 )(𝑦 − 𝑦0 )
+ F𝑧 (𝑥0 , 𝑦0 , 𝑧0 )(𝑧 − 𝑧0 ) = 0
Remark
n = ∇F = F𝑥 𝑖 + F𝑦 𝑗 + F𝑧 𝑘 =< F𝑥 , F𝑦 , F𝑧 >
where 𝑛 normal vector at point (𝑥0 , 𝑦0 , 𝑧0 )
Remark
The angle between two vectors 𝑛1 and 𝑛2 is
1
cos 𝜃 =
|𝑛1 ∙ 𝑛2 |
‖𝑛1 ‖‖𝑛2 ‖
Example 12 Consider the ellipsoid 𝑥 2 + 4𝑦 2 + 𝑧 2 = 18.
(a) Find an equation of the tangent plane to the ellipsoid at
the point (1,2,1).
(b) Find parametric equations of the line that is normal to the
ellipsoid at the point (1,2,1).
(c) Find the acute angle that the tangent plane at the
point(1,2,1) makes with the xy-plane.
Solution
Solution (a).
Since
∇F = F𝑥 𝑖 + F𝑦 𝑗 + F𝑧 𝑘 =< F𝑥 , F𝑦 , F𝑧 >
2
∇F = 2x𝑖 + 8y𝑗 + 2z𝑘 =< 2x, 8y, 2z >
we have
𝒏 = 𝜵𝑭(1,2,1) =< 2, 16,2 >
Hence, the equation of the tangent plane is
F𝑥 (𝑥0 , 𝑦0 , 𝑧0 )(𝑥 − 𝑥0 ) + F𝑦 (𝑥0 , 𝑦0 , 𝑧0 )(𝑦 − 𝑦0 )
+ F𝑧 (𝑥0 , 𝑦0 , 𝑧0 )(𝑧 − 𝑧0 ) = 0
By substituting, then
2(𝑥 − 1) + 16(𝑦 − 2) + 2(𝑧 − 1) = 0
Or
𝑥 + 8𝑦 + 𝑧 = 18
Solution (b).
Since 𝒏 = 𝜵𝑭(1,2,1) =< 2, 16,2 > at the point (1, 2, 1), it
follows from parametric equations for the normal line to the
ellipsoid at the point (𝑥0 , 𝑦0 , 𝑧0 ) = (1, 2, 1) are
𝑥 = 1 + 2𝑡, 𝑦 = 2 + 16𝑡, 𝑧 = 1 + 2𝑡
Solution (c).
To find the acute angle θ between the tangent plane and the
xy-plane (𝑧 = 0) , since
𝑛1 = 𝒏 =< 2, 16,2 > and 𝑛2 =< 0, 0,1 >.
This yields
3
cos 𝜃 =
|𝑛1 ∙ 𝑛2 |
|2(0) + 16(0) + 2(1)|
‖𝑛1 ‖‖𝑛2 ‖ √(22 + 162 + 22 )√(02 + 02 + 12 )
cos 𝜃 =
2
2√66
Thus,
𝜃 = cos−1 (
1
√66
) ≈ 83°
2.2 Linear Approximation.
Since the tangent plane to the surface 𝑧 = 𝑓(𝑥, 𝑦) at 𝑃 is
very close to the surface at least when it is near 𝑃, we may
use the function defining the tangent plane as a linear
approximation to 𝑓. Recall that the equation of the tangent
plane to the graph of 𝑓(𝑥, 𝑦) at 𝑃(𝑎, 𝑏, 𝑓(𝑎, 𝑏)) is
𝑧 = 𝑓(𝑎, 𝑏) + 𝑓𝑥 (𝑎, 𝑏)(𝑥 − 𝑎) + 𝑓𝑦 (𝑎, 𝑏)(𝑦 − 𝑏)
Definition.
4
The linear function 𝐿 whose graph is this tangent plane is
given by
𝐿(𝑥, 𝑦) = 𝑓(𝑎, 𝑏) + 𝑓𝑥 (𝑎, 𝑏)(𝑥 − 𝑎) + 𝑓𝑦 (𝑎, 𝑏)(𝑦 − 𝑏)
L is called the linearization of 𝑓 at (𝑎, 𝑏). The approximation
𝑓(𝑥, 𝑦) ≈ 𝐿(𝑥, 𝑦)
= 𝑓 (𝑎, 𝑏) + 𝑓𝑥 (𝑎, 𝑏)(𝑥 − 𝑎) + 𝑓𝑦 (𝑎, 𝑏)(𝑦 − 𝑏)
is called the linear approximation or tangent plane
approximation of 𝑓 at (𝑎, 𝑏).
Example. 14
Let 𝑓(𝑥, 𝑦) = 𝑥𝑒 𝑥𝑦 . Find the linearization of 𝑓 at (1, 0).
Use it to approximate 𝑓(1.1, −0.1).
Solution
𝑓𝑥 = 𝑒 𝑥𝑦 + 𝑥𝑦𝑒 𝑥𝑦 , then 𝑓𝑥 (1,0) = 1
𝑓𝑦 = 𝑥 2 𝑒 𝑥𝑦 , then 𝑓𝑦 (1,0) = 1 and 𝑓(1, 0) = 1. Then
𝐿(1,0) = 𝑓(1, 0) + 𝑓𝑥 (1, 0)(𝑥 − 1) + 𝑓𝑦 (1, 0)(𝑦 − 0)
=𝑥+𝑦
The corresponding linear approximation is
𝑓(𝑥, 𝑦) = 𝑥𝑒 𝑥𝑦 ≈ 𝑥 + 𝑦
Then
5
𝑓(1.1, −0.1) = 1
The actual value of f(1.1, −0.1) is 0.98542 round up to 5
decimal places.
2.3. Maximum and Minimum
Definition
A point (a, b) is said to be relative maximum of a function
f(x, y), if (𝑎, 𝑏) in the domain of 𝑓 such that 𝑓(𝑎, 𝑏) ≥
𝑓(𝑥, 𝑦), for all (x, y) in a neighborhood of (a, b). and 𝑓 is
said to have an absolute maximum at (𝑎, 𝑏) if 𝑓 (𝑎, 𝑏) ≥
𝑓(𝑥, 𝑦)for all points (𝑥, 𝑦) in the domain of 𝑓.
Definition
A point (a, b) is said to be relative Minimum of a function
f(x, y), if (𝑎, 𝑏) in the domain of 𝑓 such that 𝑓(𝑎, 𝑏) ≤
𝑓(𝑥, 𝑦), for all (𝑥, 𝑦) in a neighborhood of (𝑎, 𝑏). and 𝑓 is
said to have an absolute Minimum at (𝑎, 𝑏) if 𝑓(𝑎, 𝑏) ≤
𝑓(𝑥, 𝑦), for all points (𝑥, 𝑦) in the domain of 𝑓.
6
If f has a relative maximum or a relative minimum at (𝑎, 𝑏),,
then we say that 𝑓 has a relative extremum at (𝑎, 𝑏),, and if
𝑓 has an absolute maximum or absolute minimum at (𝑎, 𝑏),
then we say that 𝑓 has an absolute extremum at (𝑎, 𝑏).
Necessary Condition
A necessary condition that 𝑓(𝑥, 𝑦) have a relative
extremum at (𝑎, 𝑏), an interior of the domain of 𝑓, is
𝑓𝑥 (𝑎, 𝑏) = 𝑓𝑦 (𝑎, 𝑏) = 0
Or one (or both) of partials does not exist
Definition (Critical point)
The point (𝑎, 𝑏) is called critical point of 𝑓 if
𝑓𝑥 (𝑎, 𝑏) = 𝑓𝑦 (𝑎, 𝑏) = 0
or if either partials does not exist at (𝑎, 𝑏).
Definition (saddle point)
7
the function 𝑓 has neither a relative maximum nor a
relative minimum at critical point (a, b),the point (a, b) is
called a saddle point of 𝑓.
Example (14)
Let 𝑓(𝑥, 𝑦) = 4 − 𝑥 2 −𝑦 2 , find the extreme values of 𝑓
Solution
The partial exist every where, so the extreme values occur
only where
𝑓𝑥 (𝑥, 𝑦) = −2𝑥 = 0,
𝑓𝑦 (𝑥, 𝑦) = −2𝑦 = 0
Thus the only critical point is (0,0). The graph 𝑓 clearly
indicates that 𝑓(0,0) = 4 ≥ 𝑓(𝑥, 𝑦), is the absolute
maximum.
Second Partial Test
Suppose that 𝑓 has continuous second partials on a disc
about critical point (𝑎, 𝑏) and that
𝑓𝑥 (𝑎, 𝑏) = 𝑓𝑦 (𝑎, 𝑏) = 0
Define a new function
𝐹 (𝑥, 𝑦) = 𝑓𝑥𝑥 (𝑥, 𝑦)𝑓𝑦𝑦 (𝑥, 𝑦) − [𝑓𝑥𝑦 (𝑥, 𝑦)]
8
2
(1) If 𝐹 (𝑎, 𝑏) > 0, and 𝑓𝑥𝑥 (𝑎, 𝑏) < 0, then 𝑓(𝑎, 𝑏) is a
relative Maximum
(2) If 𝐹 (𝑎, 𝑏) > 0, and 𝑓𝑥𝑥 (𝑎, 𝑏) > 0, then 𝑓(𝑎, 𝑏) is a
relative Minimum
(3) If 𝐹 (𝑎, 𝑏) < 0, then 𝑓 has a saddle point at (𝑎, 𝑏).
(4) If 𝐹 (𝑎, 𝑏) = 0, then the test fails.
Remarks
(a) Since 𝑓 has continuous second partials, 𝑓𝑥𝑦 = 𝑓𝑦𝑥 so
either may be used for defining
If 𝐹 (𝑎, 𝑏) > 0, then 𝑓𝑥𝑥 and 𝑓𝑦𝑦 must have the
(b)
same sign.
Example (15)
Find the a relative maximum and the a relative minimum
of
𝑓(𝑥, 𝑦) = 𝑥 3 + 𝑦 3 − 3𝑥 − 12𝑦 + 20
Solution
𝑓𝑥 (𝑥, 𝑦) = 3𝑥 2 − 3,
𝑓𝑦 (𝑥, 𝑦) = 3𝑦 2 − 12
Therefore the only critical points are where, 3𝑥 2 − 3 = 0
and 3𝑦 2 − 12 = 0
Thus
𝑥 = ±1,
𝑦 = ±2
9
Thus the critical points are
(1,2), (1, −2), (−1,2) and (−1, −2)
𝑓𝑥𝑥 = 6𝑥,
𝑓𝑦𝑦 = 6𝑦,
𝑓𝑥𝑦 = 0
𝐹 (𝑥, 𝑦) = 𝑓𝑥𝑥 (𝑥, 𝑦)𝑓𝑦𝑦 (𝑥, 𝑦) − [𝑓𝑥𝑦 (𝑥, 𝑦)]
2
𝐹 (𝑥, 𝑦) = (6𝑥 )(6𝑦) − [0]2 = 36𝑥𝑦
Thus
𝐹 (1,2) = 72 > 0,
𝑓𝑥𝑥 (1,2) = 6 > 0
Thus (1,2) is relative Minimum
Thus
𝐹 (−1, −2) = 72 > 0,
𝑓𝑥𝑥 (−1, −2) = −6 < 0
Thus (−1, −2) relative maximum
𝐹 (−1,2) = −72 < 0, thus (−1,2) 𝑖𝑠 𝑎 𝑠𝑎𝑑𝑑𝑙𝑒 𝑝𝑜𝑖𝑛𝑡
𝐹 (1, −2) = −72 < 0, thus (1, −2) 𝑖𝑠 𝑎 𝑠𝑎𝑑𝑑𝑙𝑒 𝑝𝑜𝑖𝑛
FINDING ABSOLUTE EXTREMA ON CLOSED AND
BOUNDED SETS
How to Find the Absolute Extrema of a Continuous
Function f of two Variables on a Closed and Bounded Set
𝑅
10
Step 1. Find the critical points of 𝑓 that lie in the interior
of R.
Step 2. Find all boundary points at which the absolute
extrema can occur
Step 3. Evaluate 𝑓(𝑥, 𝑦) at the points obtained in the
preceding steps. The largest of these values is the
absolute maximum and the smallest the absolute
minimum
Example
Find the absolute maximum and minimum values of
𝑓 (𝑥, 𝑦) = 3𝑥𝑦 − 6𝑥 − 3𝑦 + 7
(1)
on the closed triangular region 𝑅 with vertices
(0, 0), (3, 0), and (0, 5).
Solution. The region R is shown in Figure
11
We have
𝑓𝑥 = 3𝑦 − 6 = 0 𝑎𝑛𝑑 𝑓𝑦 = 3𝑥 − 3 = 0
Solving these equations yields 𝑥 = 1 and 𝑦 = 2, so
(1, 2) is the only critical point.
As shown in Figure, this critical point is in the interior of R.
Next we want to determine the locations of the points on
the boundary of R at which the absolute extrema might
occur.
1. The line segment between (0, 0) and (3, 0):
On this line segment we have 𝑦 = 0, so (1)
simplifies to a function of the single variable 𝑥,
𝑢(𝑥) = 𝑓(𝑥, 0) = −6𝑥 + 7, 0 ≤ 𝑥 ≤ 3
This function has no critical points because 𝑢′ (𝑥) = −6
is nonzero for all 𝑥. Thus the extreme values of 𝑢(𝑥)
occur at the endpoints 𝑥 = 0 and 𝑥 = 3, which
correspond to the points (0, 0) and (3, 0) of R.
2. The line segment between (0, 0) and (0, 5):
On this line segment we have 𝑥 = 0, so (1) simplifies
to a function of the single variable 𝑦,
𝑣(𝑦) = 𝑓(0, 𝑦) = −3𝑦 + 7, 0 ≤ 𝑦 ≤ 5. This function
has no critical points because 𝑣 ′ (𝑥) = −3 is nonzero
for all 𝑦. Thus, the extreme values of 𝑣(𝑦) occur at the
12
endpoints 𝑦 = 0 and 𝑦 = 5, which correspond to the
points (0, 0) and (0, 5) of 𝑅 .
3. The line segment between (3, 0) 𝑎𝑛𝑑 (0, 5):
In the 𝑥𝑦-plane, an equation for this line segment is
𝑦−5 0−5
=
𝑥−0 3−0
or
−5
𝒚 =
𝒙 + 5, 0 ≤ 𝒙 ≤ 3
(2)
3
so (*) simplifies to a function of the single variable x,
−5
𝑤 (𝑥 ) = 𝒇 (𝒙,
𝒙 + 5)
3
−5
= 3𝒙 (
𝒙 + 5) − 6𝒙
3
−5
− 3(
𝒙 + 5) + 7
3
= −5𝒙𝟐 + 14𝒙 − 8, 0 ≤ 𝒙 ≤ 3
Since 𝑤 ′ (𝑥) = −10𝑥 + 14, the equation 𝑤 ′ (𝑥 ) = 0
yields 𝑥 =
7
5
as the only critical point of 𝑤. Thus, the
extreme values of w occur either at the critical point
𝑥=
7
5
or at the endpoints x = 0 and x = 3. The
endpoints correspond to the points (0, 5) and (3, 0) of
R, and from (2) the critical point corresponds
to(
7
5
8
, ) .
3
Finally, Table lists the values of 𝑓(𝑥, 𝑦) at the interior
critical point and at the points on the boundary where
an absolute extremum can occur. From the table we
13
conclude that the absolute maximum value of f
is 𝑓(0, 0) = 7 and the absolute minimum value is
𝑓(3, 0) = −11.
(x, y) (0, 0) (3, 0) (0, 5) (
f(x, y)
7
-11
-8
7 8
, )
5 3
9
5
(1, 2)
1
3.4 Total Differential.
Let 𝑧 = 𝑓(𝑥; 𝑦). As in the case of functions of one
variable, we take the differentials 𝑑𝑥 and 𝑑𝑦 to be
independent variables.
Definition.
The
differential
𝑑𝑧,
or
the
total
differential, is defined to be
𝑑𝑧 = 𝑓𝑥 (𝑥; 𝑦)𝑑𝑥 + 𝑓𝑦 (𝑥; 𝑦)𝑑𝑦
Consider the differential of 𝑓 at the point (𝑎; 𝑏). The
tangent
plane
approximation
of
𝑓
at
(𝑎; 𝑏) implies that for a small change 𝑑𝑥 of 𝑎 and a
small change 𝑑𝑦 of 𝑏, the actual change ∆𝑧
of 𝑧 is approximately equal to 𝑑𝑧. In other words,
∆𝑧 ≈ 𝑑𝑧 = 𝑓𝑥 (𝑎; 𝑏)𝑑𝑥 + 𝑓𝑦 (𝑎; 𝑏)𝑑𝑦
Similarly, for a function w = f(x, y, z) of three variables
we have the total differential of w at (𝑎; 𝑏; 𝑐)
14
∆𝑤 ≈ 𝑑𝑤 = 𝑓𝑥 (𝑎; 𝑏, 𝒄)𝑑𝑥 + 𝑓𝑦 (𝑎; 𝑏; 𝑐)𝑑𝑦 + 𝑓𝑧 (𝑎; 𝑏; 𝑐)𝑑𝑧
which is also referred to as the total differential of 𝑓 at
(𝑎, 𝑏, 𝑐).
Example
Let
𝑓(𝑥; 𝑦) = 𝑥 2 + 3𝑥𝑦 − 𝑦 2 .
𝑑𝑧.
Find
If
𝑥 changes from 2 to 2.05 and 𝑦 changes from 3 to 2.96,
compare the values of ∆𝑧 and 𝑑𝑧.
Solution
𝑑𝑧 =
𝜕𝑧
𝜕𝑧
𝑑𝑥 +
𝑑𝑦 = (2𝑥 + 3𝑦)𝑑𝑥 + (3𝑥 − 2𝑦)𝑑𝑦
𝜕𝑥
𝜕𝑦
At the point (2; 3),
𝑑𝑧 = ((2)(2) + 3(3))𝑑𝑥 + ((3)(2) − (2)(3))𝑑𝑦
That is 𝑑𝑧 = 13𝑑𝑥.
Now we take
𝑑𝑥 = 2.05 − 2 = 0.05 and
𝑑𝑦 = 2.96 − 3 = −0.04.
Thus, 𝑑𝑧 = 13(0.05) = 0.65.
For
the
actual
change
in
z,
we
have
∆𝑧 = 𝑓(2.05; 2.96) − 𝑓(2; 3) = 0.6449:
Remark
if ∆𝑥 and .∆𝑦 are close to 0, then the magnitude of the
error |∆𝑧 − 𝑑𝑧| will be much smaller than the distance
√∆𝑥 2 + ∆𝑦 2 between (𝑎, 𝑏) and (𝑎 + ∆𝑥, 𝑏 + ∆𝑦)
Example
For the function 𝑧 = 𝑥𝑦 2
15
1. Find the total differential for 𝑧 = 𝑥𝑦 2
2. Use ∆𝑧 ≈ 𝑑𝑧 to approximate the change in 𝑧 from
its value at (0.5, 1.0) to its value at (0.503, 1.004).
3. Compute the error in approximating ∆𝑧 by 𝑑𝑧
4. Compare the magnitude of the error in this
approximation with the distance between the points
(0.5, 1.0) and (0.503, 1.004).
Solution
1. For 𝑧 = 𝑥𝑦 2 , we have 𝑑𝑧 = 𝑦 2 𝑑𝑥 + 2𝑦𝑥𝑑𝑦.
2. Evaluating this differential at (𝑥, 𝑦) = (0.5,1.0),
𝑑𝑥 = ∆𝑥 = 0.503 − 0.5 = 0.003, and
𝑑𝑦 = 𝑦 = 1.004 − 1.0 = 0.004
yields
𝑑𝑧 = 1.02 (0.003) + 2(0.5)(1.0)(0.004)=0.007
𝑧 = 0.5 at
Since
(𝑥, 𝑦) = (0.5, 1.0) and
𝑧 = 0.507032048 at (𝑥, 𝑦) = (0.503, 1.004),
we have
∆𝑧 = 0.507032048 − 0.5 = 0.007032048
3. The error in approximating ∆𝑧 by 𝑑𝑧 has
magnitude
|𝑑𝑧 − ∆𝑧| = |0.007 − 0.007032048|
= 0.000032048
4. Since the distance between (0.5,1.0) and
(0.503,1.004) = (0.5 + ∆𝑥, 1.0 + ∆𝑦) is
16
√(∆𝑥)2 + (∆𝑦)2 = √(0.003)2 + (0.004)2
= √0.000025 = 0.005
We have
|𝑑𝑧 − ∆𝑧| = 0.000032048 < √(∆𝑥)2 + (∆𝑦)2 == 0.005
Lagrange Multipliers
In this section we consider the problem of maximizing or
minimizing a function 𝑓(𝑥; 𝑦) subject to a constraint
𝑔(𝑥; 𝑦) = 0.
1. Find all 𝑥; 𝑦; 𝑧 and 𝜆 such that
𝑓𝑥 = 𝜆𝑔𝑥 ; 𝑓𝑦 = 𝜆𝑔𝑦 ; 𝑓𝑧 = 𝜆𝑔𝑧
The number 𝜆 is called a Lagrange Multiplier.
2. Evaluate 𝑓 at all points (𝑥; 𝑦; 𝑧) obtained in (1). The
largest of these values is the absolute maximum of 𝑓;
the smallest is the absolute minimum of 𝑓.
Example. Find the extreme values of 𝑓(𝑥; 𝑦) = 𝑥 2 + 2𝑦 2
on the circle 𝑥 2 + 𝑦 2 = 1. See figure
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Solution
Let 𝑓 (𝑥; 𝑦) = 𝑥 2 + 2𝑦 2 𝑎𝑛𝑑 𝑔(𝑥, 𝑦) = 𝑥 2 + 𝑦 2 − 1
𝑓𝑥 = 𝜆𝑔𝑥 ;
𝑓𝑦 = 𝜆𝑔𝑦 ;
equivalent to
2𝑥 = 2𝜆 𝑥;
4𝑦 = 2𝜆𝑦
we need to solve the following system of equations in
𝑥; 𝑦 and 𝜆
2𝑥 (𝜆 − 1) = 0
2𝑦(𝜆 − 2) = 0
𝑥 2 + 𝑦 2 = 1 (constraint equation)
The first equation gives either 𝑥 = 0 𝑜𝑟 𝜆 = 1.
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If 𝑥 = 0, then the constraint equation gives 𝑦 = ±1.
Thus we have the solutions (0; −1); (0; 1).
If 𝜆 = 1, then the second equation gives 𝑦 = 0. Thus,
by the constraint equation, we have the solutions
(−1; 0); (1; 0).
Consequently, we have four solutions
(0; 1); (0; −1); (−1; 0); (1; 0)
Now
𝑓 (0; 1) = 2; 𝑓 (0; −1) = 2;
𝑓(−1; 0) = 1; 𝑓(1; 0) = 1.
Therefore, the absolute maximum value is 2 at
(0; 1), (0; −1) and the absolute minimum value is 1 at
(−1; 0), (1; 0)
Example
Find the point on the sphere
𝑥 2 + 𝑦2 + 𝑧2 = 4
that are closest to and farthest from the point (3; 1; −1).
Solution
19
we will find points on the sphere that minimize and
maximize the square of the distance to (3, 1, −1). Thus,
we want to find the relative extrema of
𝑓 (𝑥, 𝑦, 𝑧) = (𝑥 − 3)2 + (𝑦 − 1)2 + (𝑧 + 1)2
subject to the constraint
𝑔(𝑥, 𝑦, 𝑧) = 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝟒
𝑓𝑥 = 𝜆𝑔𝑥 ;
𝑓𝑦 = 𝜆𝑔𝑦
𝑓𝑧 = 𝜆𝑔𝑧 ;
which leads to the equations
2(𝑥 − 3) = 2𝑥𝜆,
2(𝑦 − 1) = 2𝑦𝜆,
2(𝑧 + 1) = 2𝑧𝜆
Or
(𝑥 − 3)
(𝑦 − 1)
(𝑧 + 1)
= 𝜆;
= 𝜆;
=𝜆
𝑥
𝑦
𝑧
The first two equations imply that
(𝑥 − 3) (𝑦 − 1)
=
𝑥
𝑦
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from which it follows that
3𝑦 = 𝑥
Similarly, the first and third equations imply that
(𝑦 − 1) (𝑧 + 1)
=
𝑦
𝑧
from which it follows that
𝑦 = −𝑧
Substituting 3𝑦 = 𝑥 and 𝑦 = −𝑧 in the constraint equation,
we obtain
9𝑦 2 + 𝑦 2 + 𝑦 2 = 4
11𝑦 2 = 4 𝑜𝑟 𝑦 = ±
2
√11
Substituting these values in 𝑥 = 3𝑦and 𝑧 = −𝑦 yields two
points
6
2 −2
−6 −2 2
(
,
,
),(
,
,
)
√11 √11 √11
√11 √11 √11
Since
𝑓(
6
,
2
,
−2
√11 √11 √11
) = √11 – 2,𝑓 (
it follows that (
6
,
2
,
−2
−6
,
−2
,
2
√11 √11 √11
) = √11 + 2
) is the point on the
√11 √11 √11
sphere closest to (3, 1, -1), and (
−6
,
−2
,
2
√11 √11 √11
that is farthest
Exercises
21
) is the point
1. Consider the surface 𝑥𝑧 − 𝑦𝑧 3 + 𝑦𝑧 2 = 2.
(a) Find an equation of the tangent plane to the
surface at the point (2, −1, 1).
(b) Find parametric equations of the line that is
normal to the surface at the point (2, −1,1).
(c) Find the acute angle that the tangent plane at the
point (2, −1,1) makes with the 𝑥𝑦 −plane.
2. Locate all relative maxima, relative minima, and
saddle points, if any
(i) 𝑓 (x, y) = y 2 + xy + 3y + 2x + 3
(ii) 𝑓(𝑥, 𝑦) = 𝑥 2 + 𝑥𝑦 − 2𝑦 − 2𝑥 + 1
(iii) 𝑓(𝑥, 𝑦) = 𝑦 sin 𝑦
(iv) 𝑓(𝑥, 𝑦) = 𝑒 𝑥 sin 𝑦
(v) 𝑓(𝑥, 𝑦) = 𝑥𝑒 𝑦
(vi) 𝑓(𝑥, 𝑦) = 𝑥 2 + 𝑦 − 𝑒 𝑦
3. Find the absolute extrema of the given function on
the indicated closed and bounded set R.
i. f(x, y) = xy − x − 3y; R is the triangular region
with vertices (0, 0), (0, 4), and (5, 0).
ii. 𝑓(𝑥, 𝑦) = 𝑥𝑦 − 2𝑥; 𝑅 is the triangular region with
vertices (0, 0), (0, 4), and (4, 0)
iii. f(x, y) = x 2 − 3y 2 − 2x + 6y; R
is the region
bounded by the square with vertices (0, 0), (0,
2), (2, 2), and (2, 0).
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iv. 𝑓 (𝑥, 𝑦) = 𝑥 2 + 2𝑦 2 − 𝑥; R is the disk 𝑥 2 + 𝑦 2 ≤ 4
v. 𝑓 (𝑥, 𝑦) = 𝑥𝑦 2 ; 𝑅 is the region that satisfies the
inequalities 𝑥 ≥ 0, 𝑦 ≥ 0, and 𝑥 2 + 𝑦 2 ≤ 1
4. Use a total differential to approximate the change in
the values of 𝑓 from 𝑃 to 𝑄. Compare your estimate
with the actual change in 𝑓 .
i.
𝑓 (𝑥, 𝑦) = 𝑥 2 + 2𝑥𝑦 − 4𝑥; 𝑃 (1,2), 𝑄(1.01,2.04)
ii.
𝑓 (𝑥, 𝑦, 𝑧) =
2𝑥𝑦 2 𝑧 3 ; 𝑃(1, −1, 2), 𝑄(0.99, −1.02,2.02)
5. At what point or points on the circle 𝑥 2 + 𝑦 2 = 1
does 𝑓(𝑥, 𝑦) = 𝑥𝑦 have an absolute maximum, and
what is that maximum
6. Find the points on the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 36 that
are closest to and farthest from the point (1, 2, 2).
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