solutions - kkuniyuk
QUIZ 3 (SECTIONS 16.3-16.9) SOLUTIONS MATH 252 – FALL 2008 – KUNIYUKI SCORED OUT OF 125 POINTS MULTIPLIED BY 0.84 105% POSSIBLE
r 1) Let f r, t = t 2 sin . Find f r r, t . (5 points) t
( )
( )
2 r f r r, t = Dr t sin "#" t r = t 2 Dr sin t
( )
r r = t 2 cos Dr t
t
r 1 2 = t cos Dr r t t
"#"
r 1 = t 2 cos t
t r = t cos t
(
)
(
)
(
)
2) Let f x, y, z = e xyz . Find f y x, y, z and use that to find f yz x, y, z . (8 points)
(
)
f y x, y, z = Dy e xyz = e xyz Dy xyz = e xyz Dy xz y "#" = e xyz xz
( )
= xze xyz
(
)
(
)
f yz x, y, z = Dz f y x, y, z x ze xyz = Dz "#" xyz = x Dz ze We will use a Product Rule for Differentiation. = x Dz z e xyz + z Dz e xyz = (1)
( )
()
( )
= x e xyz + z Dz e xyz = x e xyz + z e xyz Dz xy z "#" = x e xyz + z e xyz xy
(
= x e xyz + xyze xyz
(
)
)
= xe xyz 1 + xyz , or xe xyz + x 2 yze xyz
( )
3) Assume that f is a function of s and t. Write the limit definition of f s s, t using the notation from class. (4 points)
( )
f s s, t = lim
h0
(
)
( )
f s + h, t f s, t h
4) Find
z if z = f ( x, y) is a differentiable function described implicitly by the y
( )
equation tan y 3 z = x 2 yz . Use the Calculus III formula given in class. Simplify. (12 points)
( )
First, isolate 0 on one side: tan y 3 z x 2 + yz = 0 Let this be F ( x, y,z ) Find
z . When using the formula, treat x, y, and z as independent variables. y
( (
Fy x, y, z z = y Fz x, y, z
) )
"#" "#" 3 2 Dy tan y z x + y z =
Dz tan y 3 z x2 + y z "#" "#" "#" sec 2 y 3 z D y 3 z + z y =
sec 2 y 3 z D y 3 z + y z "#" sec 2 y 3 z z D y 3 + z y = 2 3 3 sec y z y + y 2 3 sec y z z 3y 2 + z = 2 3 3 sec y z y + y
( )
( )
( )
( )
( ) ( ) ( ) ( )
=
( ) y sec ( y z ) +
( )
3y 2 z sec 2 y 3 z + z 3
2
3
y
( ) ( )
z 3y 2 sec 2 y 3 z + 1 or y y 2 sec 2 y 3 z + 1
( )
5) Let f , g , and h be differentiable functions such that z = f u, v ,
(
)
(
)
u = g r, s, t , and v = h r, s, t . Use the Chain Rule to write an expression for
z . (5 points) t
z z u z v = + t u t v t
6) The temperature at any point ( x, y) in the xy-plane is given by
( )
f x, y = 5x 2 y + y 3 in degrees Fahrenheit. Assume x and y are measured in meters. Give appropriate units in your answers. (30 points total)
(
)
a) Find the maximum rate of change of temperature at the point 2, 3 . Approximate your final answer to four significant digits. (10 points)
( )
( ) ( )
f x, y = f x x, y , f y x, y
(
)
(
= Dx 5x 2 y + y 3 , Dy 5x 2 y + y 3
)
= 10xy, 5x 2 + 3y 2
(
)
( )( ) ( )
2
( )
f 2, 3 = 10 2 3 , 5 2 + 3 3
2
= 60, 47
(
)
The length of the gradient of f at 2, 3 gives the maximum rate of change of temperature at that point.
(
)
f 2, 3 = =
60, 47
( 60) + ( 47 )
= 5809 76.22 o
Answer:
About 76.22
F m
2
2
(
)
b) Find the rate of change of temperature at 2, 3 in the direction of
4i + j . Approximate your final answer to four significant digits. (10 points) Let a be the given direction vector 4i + j , or 4, 1 . Find the unit vector u in the direction of a. u=
a a 4, 1
= =
=
4, 1 4, 1
( 4) + (1) 2
4, 1
2
1
or
4, 1 or
17
17
(
17 4, 1 17
)
The directional derivative at 2, 3 in the direction of u is:
(
)
(
)
Du f 2, 3 = f 2, 3 • u 1 4, 1
= 60, 47 • 17 1 = 60, 47 • 4, 1 17 1 60 4 + 47 1 =
17 1 193 = 17
(
)
(
=
193
17 46.81
Answer:
About 46.81
o
F m
)( ) ( )( )
or
193 17 17
c) Use differentials to linearly approximate the change in temperature if ( x, y) changes from 2, 3 to 2.03, 3.01 . (7 points)
(
) (
)
(
)
From part a), we know that f 2, 3 = 60, 47 . In particular,
( ) ( 2, 3) = 47 .
f x 2, 3 = 60 , and fy
Find the changes in x and y: dy = new y old y
dx = new x old x = 2.03 2
( )
= 3.01 3
= 0.03
= 0.01
The approximate change in f is given by:
(
)
(
)
df = f x 2, 3 dx + f y 2, 3 dy = 60 0.03 + 47 0.01 = 2.27
(
)
(
)
Note: Actual change = f 2.03, 3.01 f 2, 3 2.29045 . 2.27 o F
Answer:
(
)
d) In what direction does the temperature decrease most rapidly at 2, 3 ? Give an appropriate non-0 direction vector. (3 points) We want a vector that points in the direction opposite from the gradient vector, f 2, 3 = 60, 47 .
(
Answer:
)
60 m , 47 m , or any positive scalar multiple of this.
7) Find an equation for the tangent plane to the graph of the equation z 3 = x xy at the point P 8, 26, 6 . (10 points)
(
)
(You may check that the coordinates of P satisfy the given equation, meaning that P lies on the graph of the equation.) Isolate 0 on one side of the given equation. z 3 x + xy = 0 = F ( x, y,z )
(
)
A normal vector for the desired tangent plane is given by F 8, 26, 6 .
F
P
= Fx
P
, Fy
P
, Fz
= 1 + y, x, 3z
(
)
P
2
or
(
)
y 1, x, 3z 2
()
F 8, 26, 6 = 1 + 26 , 8, 3 6
2
= 27, 8, 108 An equation for the tangent plane is given by:
( F )( x x ) + ( F )( y y ) + ( F )( z z ) = 0 ( 27 )( x 8) + (8)( y ( 26)) + (108)( z 6) = 0 x P
0
y P
0
z P
0
( ) ( ) ( ) 27 ( x 8) 8 ( y + 26 ) 108 ( z 6 ) = 0
27 x 8 + 8 y + 26 + 108 z 6 = 0 or
or 27x + 8y + 108z 224 = 0 or 27x 8y 108z + 224 = 0
Here’s a graph:
( )
8) Find all critical points of f x, y = 2x 4 + y 2 12xy , and classify each one as a local maximum, a local minimum, or a saddle point. Show all work, as we have done in class. You do not have to find the corresponding function values. (26 points) Step 1: Find any critical points (CPs).
( ) ( never DNE; is continuous on R ) ( x, y ) = 2 y 12x ( never DNE; is continuous on R )
f x x, y = 8x 3 12 y
2
fy
2
8x 3 12 y = 0 Solve the system: 2 y 12x = 0 (Optional.) If we divide both sides of the first equation by 4 and divide both sides of the second equation by 2, we obtain the simpler system:
2x 3 3y = 0 y 6x = 0 We can solve this using the Substitution Method. If we solve the second equation for y, we get:
y = 6x Incorporate this into the first equation.
2x 3 3y = 0, y = 6x
( )
2x 3 3 6x = 0 2x 3 18x = 0
(
)
2x x 2 9 = 0
(
)(
)
2x x + 3 x 3 = 0
x=0
or
x +3= 0 x = 3
or
x 3= 0 x =3
Find the corresponding x-values. Using y = 6x ,
x=0
y=0
x = 3
y = 18
x=3
y = 18
(0, 0) ( 3, 18) (3, 18)
Our critical points (CPs) are:
(0, 0) , ( 3, 18) , and (3, 18) . (Note that they are in the domain of f.)
Step 2: Find f xx and D. Remember,
( ) ( x, y ) = 2 y 12x
f x x, y = 8x 3 12 y fy
( ) ( x, y ) = 12 ( x, y ) = 2
f xx x, y = 24x 2 f xy f yy
D= =
(
( = f ( x, y ); note that " 12" is continuous.) yx
f xx
f xy
f yx
f yy
24x 2 12
12 2
)( ) ( )( )
= 24x 2 2 12 12 = 48x 2 144
Step 3: Classify the critical points (CPs). Critical Point
(0, 0)
D = 48x 2 144
()
irrelevant
Saddle Point
2
= 144
(" ")
( )
( )
f xx = 24 3
2
D = 48 3 144
>0
("+ ")
()
("+ ")
Local Minimum
Think: "Concave up" ()
()
f xx = 24 3
2
D = 48 3 144
2
= 216
= 288 >0
2
= 216
= 288 >0
(3, 18)
Classification
D = 48 0 144 0
("+ ")
("+ ")
Local Minimum
Think: "Concave up" ()
Note 1: The corresponding points of interest on the graph are: 0, 0, 0 , 3, 18, 162 , and 3, 18, 162 .
(
) (
)
(
)
( ) is positive everywhere in the xy-plane and in particular at ( 3, 18) and ( 3, 18) .
Note 2: Instead of using f xx = 24x 2 , it may be easier to use f yy x, y = 2 , which
Here are graphs. The second graph zooms in on the saddle point at the origin.
(
) (
) (
) ( ) squared distance of the point ( x, y, z ) from the point ( 3, 5, 7 ) in xyz-space. 2
2
2
9) The function f , where f x, y, z = x 3 + y 5 + z 7 , gives the Using the method of Lagrange multipliers, find the point on the unit sphere x 2 + y 2 + z 2 = 1 that is closest to the point 3, 5, 7 . The point will lie in
(
)
Octant I; you may use this fact without proof. You do not have to prove that your candidate point corresponds to an absolute minimum of f under the constraint. Give an exact, simplified answer, and rationalize denominators. (25 points) Take the constraint equation, and isolate 0 on one side: x2 + y2 + z2 = 1 x2 + y2 + z2 1 = 0 g ( x, y,z ) f ( x, y, z) = g ( x, y, z) Solve g ( x, y, z) = 0
(
)
( ) g ( x, y, z ) , g ( x, y, z ) , g ( x, y, z )
f x, y, z = g x, y, z
(
)
(
)
( ) 2 ( x 3) , 2 ( y 5) , 2 ( z 7 )
f x x, y, z , f y x, y, z , f z x, y, z
( ( (
) ) )
=
x
y
z
= 2x, 2 y, 2z
( ) ( Eq.1) ( ) ( Eq.2) ( ) ( Eq.3) ( Eq.C)
2 x 3 = 2x 2 y 5 = 2y Solve 2 z 7 = 2z 2 2 2 x + y + z 1 = 0
Solve the first three equations for . Observe that x = 0 , y = 0 , and z = 0 cannot solve the first three equations, so we may assume that x, y, and z are nonzero.
( Eq.1)
=
( Eq.2)
=
( Eq.3)
=
(
)
2 x3
2x 2 y5
(
(
)
2y
2 z7 2z
)
=
x3 x
=
y5 y
=
z7 z
Equate our expressions for .
=
x3 y5 z 7 = = x y z
We may then solve for y and z in terms of x. Remember that we may assume that x, y, and z are nonzero. x3 y5 = y x y 5 x 3 x y = x y x y
x3 z7 = z x x 3 z 7 x z =xz x z
y x3 = x y5
z x3 = x z7
xy 3y = xy 5x
xz 3z = xz 7x 3z = 7x 7 z= x 3
(
) (
(
)
3y = 5x 5 y= x 3
) (
)
5 7 Substitute y = x and z = x into (Eq.C). 3 3 5 7 y = x and z = x 3 3
x 2 + y 2 + z 2 1 = 0, 2
2
5 7 x + x + x 1 = 0 3 3 2
25 2 49 2 x + x 1= 0 9 9 9 2 25 2 49 2 x + x + x =1 9 9 9 83 2 x =1 9 9 x2 = 83
x2 +
x=± x=± x=± Take x =
9 83 3 83 3 83 83
3 83 . 83
( We restrict our attention to Octant I.)
5 7 Find our candidate point. Remember: y = x and z = x . 3 3 x=
3 83 83
y=
5 3 83 5 83 = 83 3 83
and
z=
7 3 83 7 83 . = 83 3 83
3 83 5 83 7 83 Candidate point: , , . 83 83 83
Our candidate is in the domain of f , which is R 3 . If we had not restricted our attention to Octant I, we would have obtained two 3 83 5 83 7 83 3 83 5 83 7 83 candidate points: , , , , and . 83 83 83 83 83 83 We know that our candidates for local extrema are absolute extrema, because there are only two, the graph of the constraint is a closed surface (in this case, a sphere), and f is continuous on it. We may identify which is the absolute maximum and which is the absolute minimum by evaluating and comparing values of f at these candidate points. It turns out that: 3 83 5 83 7 83 f , , = 84 2 83 65.8 , while 83 83 83 3 83 5 83 7 83 f , , = 84 + 2 83 102.2 . 83 83 83 3 83 5 83 7 83 Therefore, , , is the absolute minimum of f on the given 83 83 83 unit sphere. In other words, it is the point on the unit sphere that is closest to the 3 83 5 83 7 83 point 3, 5, 7 ; the point , , is the point on the sphere that has 83 83 83 the absolute minimum squared distance (and therefore the absolute minimum distance) from the point 3, 5, 7 .
(
)
(
Answer:
)
3 83 5 83 7 83 , , 83 83 83
(
)
Note: Observe that the origin, this point on the sphere, and the point 3, 5, 7 all lie on a straight line. Our visual intuition leads us to believe that we have, in fact, found the correct point.
r 1) Let f r, t = t 2 sin . Find f r r, t . (5 points) t
( )
( )
2 r f r r, t = Dr t sin "#" t r = t 2 Dr sin t
( )
r r = t 2 cos Dr t
t
r 1 2 = t cos Dr r t t
"#"
r 1 = t 2 cos t
t r = t cos t
(
)
(
)
(
)
2) Let f x, y, z = e xyz . Find f y x, y, z and use that to find f yz x, y, z . (8 points)
(
)
f y x, y, z = Dy e xyz = e xyz Dy xyz = e xyz Dy xz y "#" = e xyz xz
( )
= xze xyz
(
)
(
)
f yz x, y, z = Dz f y x, y, z x ze xyz = Dz "#" xyz = x Dz ze We will use a Product Rule for Differentiation. = x Dz z e xyz + z Dz e xyz = (1)
( )
()
( )
= x e xyz + z Dz e xyz = x e xyz + z e xyz Dz xy z "#" = x e xyz + z e xyz xy
(
= x e xyz + xyze xyz
(
)
)
= xe xyz 1 + xyz , or xe xyz + x 2 yze xyz
( )
3) Assume that f is a function of s and t. Write the limit definition of f s s, t using the notation from class. (4 points)
( )
f s s, t = lim
h0
(
)
( )
f s + h, t f s, t h
4) Find
z if z = f ( x, y) is a differentiable function described implicitly by the y
( )
equation tan y 3 z = x 2 yz . Use the Calculus III formula given in class. Simplify. (12 points)
( )
First, isolate 0 on one side: tan y 3 z x 2 + yz = 0 Let this be F ( x, y,z ) Find
z . When using the formula, treat x, y, and z as independent variables. y
( (
Fy x, y, z z = y Fz x, y, z
) )
"#" "#" 3 2 Dy tan y z x + y z =
Dz tan y 3 z x2 + y z "#" "#" "#" sec 2 y 3 z D y 3 z + z y =
sec 2 y 3 z D y 3 z + y z "#" sec 2 y 3 z z D y 3 + z y = 2 3 3 sec y z y + y 2 3 sec y z z 3y 2 + z = 2 3 3 sec y z y + y
( )
( )
( )
( )
( ) ( ) ( ) ( )
=
( ) y sec ( y z ) +
( )
3y 2 z sec 2 y 3 z + z 3
2
3
y
( ) ( )
z 3y 2 sec 2 y 3 z + 1 or y y 2 sec 2 y 3 z + 1
( )
5) Let f , g , and h be differentiable functions such that z = f u, v ,
(
)
(
)
u = g r, s, t , and v = h r, s, t . Use the Chain Rule to write an expression for
z . (5 points) t
z z u z v = + t u t v t
6) The temperature at any point ( x, y) in the xy-plane is given by
( )
f x, y = 5x 2 y + y 3 in degrees Fahrenheit. Assume x and y are measured in meters. Give appropriate units in your answers. (30 points total)
(
)
a) Find the maximum rate of change of temperature at the point 2, 3 . Approximate your final answer to four significant digits. (10 points)
( )
( ) ( )
f x, y = f x x, y , f y x, y
(
)
(
= Dx 5x 2 y + y 3 , Dy 5x 2 y + y 3
)
= 10xy, 5x 2 + 3y 2
(
)
( )( ) ( )
2
( )
f 2, 3 = 10 2 3 , 5 2 + 3 3
2
= 60, 47
(
)
The length of the gradient of f at 2, 3 gives the maximum rate of change of temperature at that point.
(
)
f 2, 3 = =
60, 47
( 60) + ( 47 )
= 5809 76.22 o
Answer:
About 76.22
F m
2
2
(
)
b) Find the rate of change of temperature at 2, 3 in the direction of
4i + j . Approximate your final answer to four significant digits. (10 points) Let a be the given direction vector 4i + j , or 4, 1 . Find the unit vector u in the direction of a. u=
a a 4, 1
= =
=
4, 1 4, 1
( 4) + (1) 2
4, 1
2
1
or
4, 1 or
17
17
(
17 4, 1 17
)
The directional derivative at 2, 3 in the direction of u is:
(
)
(
)
Du f 2, 3 = f 2, 3 • u 1 4, 1
= 60, 47 • 17 1 = 60, 47 • 4, 1 17 1 60 4 + 47 1 =
17 1 193 = 17
(
)
(
=
193
17 46.81
Answer:
About 46.81
o
F m
)( ) ( )( )
or
193 17 17
c) Use differentials to linearly approximate the change in temperature if ( x, y) changes from 2, 3 to 2.03, 3.01 . (7 points)
(
) (
)
(
)
From part a), we know that f 2, 3 = 60, 47 . In particular,
( ) ( 2, 3) = 47 .
f x 2, 3 = 60 , and fy
Find the changes in x and y: dy = new y old y
dx = new x old x = 2.03 2
( )
= 3.01 3
= 0.03
= 0.01
The approximate change in f is given by:
(
)
(
)
df = f x 2, 3 dx + f y 2, 3 dy = 60 0.03 + 47 0.01 = 2.27
(
)
(
)
Note: Actual change = f 2.03, 3.01 f 2, 3 2.29045 . 2.27 o F
Answer:
(
)
d) In what direction does the temperature decrease most rapidly at 2, 3 ? Give an appropriate non-0 direction vector. (3 points) We want a vector that points in the direction opposite from the gradient vector, f 2, 3 = 60, 47 .
(
Answer:
)
60 m , 47 m , or any positive scalar multiple of this.
7) Find an equation for the tangent plane to the graph of the equation z 3 = x xy at the point P 8, 26, 6 . (10 points)
(
)
(You may check that the coordinates of P satisfy the given equation, meaning that P lies on the graph of the equation.) Isolate 0 on one side of the given equation. z 3 x + xy = 0 = F ( x, y,z )
(
)
A normal vector for the desired tangent plane is given by F 8, 26, 6 .
F
P
= Fx
P
, Fy
P
, Fz
= 1 + y, x, 3z
(
)
P
2
or
(
)
y 1, x, 3z 2
()
F 8, 26, 6 = 1 + 26 , 8, 3 6
2
= 27, 8, 108 An equation for the tangent plane is given by:
( F )( x x ) + ( F )( y y ) + ( F )( z z ) = 0 ( 27 )( x 8) + (8)( y ( 26)) + (108)( z 6) = 0 x P
0
y P
0
z P
0
( ) ( ) ( ) 27 ( x 8) 8 ( y + 26 ) 108 ( z 6 ) = 0
27 x 8 + 8 y + 26 + 108 z 6 = 0 or
or 27x + 8y + 108z 224 = 0 or 27x 8y 108z + 224 = 0
Here’s a graph:
( )
8) Find all critical points of f x, y = 2x 4 + y 2 12xy , and classify each one as a local maximum, a local minimum, or a saddle point. Show all work, as we have done in class. You do not have to find the corresponding function values. (26 points) Step 1: Find any critical points (CPs).
( ) ( never DNE; is continuous on R ) ( x, y ) = 2 y 12x ( never DNE; is continuous on R )
f x x, y = 8x 3 12 y
2
fy
2
8x 3 12 y = 0 Solve the system: 2 y 12x = 0 (Optional.) If we divide both sides of the first equation by 4 and divide both sides of the second equation by 2, we obtain the simpler system:
2x 3 3y = 0 y 6x = 0 We can solve this using the Substitution Method. If we solve the second equation for y, we get:
y = 6x Incorporate this into the first equation.
2x 3 3y = 0, y = 6x
( )
2x 3 3 6x = 0 2x 3 18x = 0
(
)
2x x 2 9 = 0
(
)(
)
2x x + 3 x 3 = 0
x=0
or
x +3= 0 x = 3
or
x 3= 0 x =3
Find the corresponding x-values. Using y = 6x ,
x=0
y=0
x = 3
y = 18
x=3
y = 18
(0, 0) ( 3, 18) (3, 18)
Our critical points (CPs) are:
(0, 0) , ( 3, 18) , and (3, 18) . (Note that they are in the domain of f.)
Step 2: Find f xx and D. Remember,
( ) ( x, y ) = 2 y 12x
f x x, y = 8x 3 12 y fy
( ) ( x, y ) = 12 ( x, y ) = 2
f xx x, y = 24x 2 f xy f yy
D= =
(
( = f ( x, y ); note that " 12" is continuous.) yx
f xx
f xy
f yx
f yy
24x 2 12
12 2
)( ) ( )( )
= 24x 2 2 12 12 = 48x 2 144
Step 3: Classify the critical points (CPs). Critical Point
(0, 0)
D = 48x 2 144
()
irrelevant
Saddle Point
2
= 144
(" ")
( )
( )
f xx = 24 3
2
D = 48 3 144
>0
("+ ")
()
("+ ")
Local Minimum
Think: "Concave up" ()
()
f xx = 24 3
2
D = 48 3 144
2
= 216
= 288 >0
2
= 216
= 288 >0
(3, 18)
Classification
D = 48 0 144 0
("+ ")
("+ ")
Local Minimum
Think: "Concave up" ()
Note 1: The corresponding points of interest on the graph are: 0, 0, 0 , 3, 18, 162 , and 3, 18, 162 .
(
) (
)
(
)
( ) is positive everywhere in the xy-plane and in particular at ( 3, 18) and ( 3, 18) .
Note 2: Instead of using f xx = 24x 2 , it may be easier to use f yy x, y = 2 , which
Here are graphs. The second graph zooms in on the saddle point at the origin.
(
) (
) (
) ( ) squared distance of the point ( x, y, z ) from the point ( 3, 5, 7 ) in xyz-space. 2
2
2
9) The function f , where f x, y, z = x 3 + y 5 + z 7 , gives the Using the method of Lagrange multipliers, find the point on the unit sphere x 2 + y 2 + z 2 = 1 that is closest to the point 3, 5, 7 . The point will lie in
(
)
Octant I; you may use this fact without proof. You do not have to prove that your candidate point corresponds to an absolute minimum of f under the constraint. Give an exact, simplified answer, and rationalize denominators. (25 points) Take the constraint equation, and isolate 0 on one side: x2 + y2 + z2 = 1 x2 + y2 + z2 1 = 0 g ( x, y,z ) f ( x, y, z) = g ( x, y, z) Solve g ( x, y, z) = 0
(
)
( ) g ( x, y, z ) , g ( x, y, z ) , g ( x, y, z )
f x, y, z = g x, y, z
(
)
(
)
( ) 2 ( x 3) , 2 ( y 5) , 2 ( z 7 )
f x x, y, z , f y x, y, z , f z x, y, z
( ( (
) ) )
=
x
y
z
= 2x, 2 y, 2z
( ) ( Eq.1) ( ) ( Eq.2) ( ) ( Eq.3) ( Eq.C)
2 x 3 = 2x 2 y 5 = 2y Solve 2 z 7 = 2z 2 2 2 x + y + z 1 = 0
Solve the first three equations for . Observe that x = 0 , y = 0 , and z = 0 cannot solve the first three equations, so we may assume that x, y, and z are nonzero.
( Eq.1)
=
( Eq.2)
=
( Eq.3)
=
(
)
2 x3
2x 2 y5
(
(
)
2y
2 z7 2z
)
=
x3 x
=
y5 y
=
z7 z
Equate our expressions for .
=
x3 y5 z 7 = = x y z
We may then solve for y and z in terms of x. Remember that we may assume that x, y, and z are nonzero. x3 y5 = y x y 5 x 3 x y = x y x y
x3 z7 = z x x 3 z 7 x z =xz x z
y x3 = x y5
z x3 = x z7
xy 3y = xy 5x
xz 3z = xz 7x 3z = 7x 7 z= x 3
(
) (
(
)
3y = 5x 5 y= x 3
) (
)
5 7 Substitute y = x and z = x into (Eq.C). 3 3 5 7 y = x and z = x 3 3
x 2 + y 2 + z 2 1 = 0, 2
2
5 7 x + x + x 1 = 0 3 3 2
25 2 49 2 x + x 1= 0 9 9 9 2 25 2 49 2 x + x + x =1 9 9 9 83 2 x =1 9 9 x2 = 83
x2 +
x=± x=± x=± Take x =
9 83 3 83 3 83 83
3 83 . 83
( We restrict our attention to Octant I.)
5 7 Find our candidate point. Remember: y = x and z = x . 3 3 x=
3 83 83
y=
5 3 83 5 83 = 83 3 83
and
z=
7 3 83 7 83 . = 83 3 83
3 83 5 83 7 83 Candidate point: , , . 83 83 83
Our candidate is in the domain of f , which is R 3 . If we had not restricted our attention to Octant I, we would have obtained two 3 83 5 83 7 83 3 83 5 83 7 83 candidate points: , , , , and . 83 83 83 83 83 83 We know that our candidates for local extrema are absolute extrema, because there are only two, the graph of the constraint is a closed surface (in this case, a sphere), and f is continuous on it. We may identify which is the absolute maximum and which is the absolute minimum by evaluating and comparing values of f at these candidate points. It turns out that: 3 83 5 83 7 83 f , , = 84 2 83 65.8 , while 83 83 83 3 83 5 83 7 83 f , , = 84 + 2 83 102.2 . 83 83 83 3 83 5 83 7 83 Therefore, , , is the absolute minimum of f on the given 83 83 83 unit sphere. In other words, it is the point on the unit sphere that is closest to the 3 83 5 83 7 83 point 3, 5, 7 ; the point , , is the point on the sphere that has 83 83 83 the absolute minimum squared distance (and therefore the absolute minimum distance) from the point 3, 5, 7 .
(
)
(
Answer:
)
3 83 5 83 7 83 , , 83 83 83
(
)
Note: Observe that the origin, this point on the sphere, and the point 3, 5, 7 all lie on a straight line. Our visual intuition leads us to believe that we have, in fact, found the correct point.
