Math B6C – Chapter 15 Quiz – Spring 2007
* Solutions* __________________________________________________________________________________
1. Find the limit as x approaches 0 along the x-axis: ( ) ( )
8 8
8 8,0 0,0
15 4lim ?5 2x
x yx y→
− =+
( ) ( )
8 8 8 8
8 8 8 80 0,0 0,0315 4 15 4 0 15lim lim lim lim
5 2 5 2 0 5x xx
x y x xx y x x→ →→
− − ⋅= =+ + ⋅ 0x→
=
( ) ( )
8 8
8 8,0 0,0
15 4lim 35 2x
x yx y→
− =+
__________________________________________________________________________________
2. Find the limit as y approaches 0 along the y-axis: ( ) ( )
8 8
8 80, 0,0
15 4lim ?5 2y
x yx y→
− =+
( ) ( )
8 8 8 8
8 8 8 80 00, 0,0
15 4 15 0 4 4 4lim lim lim lim5 2 5 0 2 2 2y yy
x y y yx y y y→ →→
− ⋅ − −= = =+ ⋅ + 0y→
−
( ) ( )
8 8
8 80, 0,0
15 4lim 25 2y
x yx y→
− = −+
__________________________________________________________________________________
3. Find the limit: ( ) ( )
8 8
8 8, 0,0
15 4lim ?5 2x y
x yx y→
− =+
( ) ( )
(by problem #18 8
8 8,0 0,0 15 4lim 3
5 2x
x yx y→
− =+
) , but
( ) ( )
( )by problem #28 8
8 80, 0,0 15 4lim 2
5 2y
x yx y→
− = −+
.
Since these two limits are unequal, ( ) ( )
8 8
8, 0,0
15 4lim5 2x y 8
x yx y→
−+
does not exist.
__________________________________________________________________________________ 4. If ( ) ( )( )3, ln cosy yg x y x π= + , then ( ) ?2,2xg =
( ) ( )( )( )( )
( )( )( )
( ) ( )( )
3 23
3 3
2
3
cos 3 cos, ln cos
cos cos3 cos
,cos
x
x
y yy y y
y y
y x y
yx yxg x y x
x xx y
g x yx
πxπ π
π
π π
π
∂ +∂ ∂= + = =∂ +
=+
+
Hence,
( ) ( )( )
2
32 2 23 2 cos 2 3 4 3 4 62,2
2 cos 8 10 5xgππ
⋅⋅ ⋅ ⋅ ⋅= = =+ +
= .
__________________________________________________________________________________
5. If ( ) ( )cos, x yh x y e −= , then ( ) ?1,1xyh =
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( )( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
cos
cos cos
cos cos cos
cos cos
cos sin
sin sin sin
sin
, , , x yxy x
x y x y
x y x y x y
x y x
y y x y x
x y x y x yy x y x
x y x yy y
x y x yy
h x y h x y h x y e
e e
e e e
e e
−
− −
− − −
−
∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠
∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞− − −⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠
⎛ ⎞∂ ∂ x yy∂
− − − −⎜ ⎟∂ ∂ ∂⎝ ⎠⎛ ⎞∂
− − −⎜ ⎟∂⎝ ⎠
= = =
= = −
= − = −
= − ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )cos cos
cos
sin cos,
y
x y x yxy
x y x yy
x y x yh x y e e
−
− −
∂− −
∂
− + −=
Hence, ( ) ( ) ( ) ( ) ( )cos 1 1 cos 1 1 cos0 cos0 1sin 1 1 cos 1 1 0 cos0 0 11,1xyh e e e e e− −− + − ⋅ + + ⋅= = = , so
( )1,1xyh e=
__________________________________________________________________________________ 6. Find the linearization of ( ) 5 2, 5 2f x y x y x y= − + at ( )1, 2 . ( ) ?,L x y =
( ) ( )
( ) [ ]
( ) ( ) [ ]
( ) ( )
5 2 5 2
4 2 5
'( , ) 5 2 5 2
'( , ) 5 5 2 2
' 1, 2 15 6
1 1( , ) 1, 2 ' 1, 2 3 15 6
2 2
( , ) 3 15 1 6 2 3 15 15 6 12
x yf x y f f x y x y x y x yx y
f x y x y x y
f
x xL x y f f
y y
L x y x y x y
⎡ ⎤∂ ∂⎡ ⎤= = − + − +⎢ ⎥⎣ ⎦ ∂ ∂⎣ ⎦
⎡ ⎤= − +⎣ ⎦
=
−⎛ ⎞⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + − = +⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎝ ⎠
= + − + − = + − + −
( , ) 15 6 24L x y x y= + − A graph of ( ) 5 2, 5 2f x y x y x y= − + along with its linearization at ( )1, 2 , : ( , ) 15 6 24L x y x y= + −
__________________________________________________________________________________
7. Find the linearization of ( ) 3 2 4, , sinh( )f x y z x z y xy= − at ( )1,0,2 . ( ) ?, ,L x y z =
( ) ( ) ( )
( ) ( ) ( )( ) ( ) ( )
( ) [ ]
3 2 4 3 2 4 3 2 4
2 2 4 3 4 3
2 2 5 3 4 3
'( , , )
sinh( ) sinh( ) sinh( )
3 cosh 4 sinh cosh 2
'( , , ) 3 cosh 4 sinh cosh 2
' 1,0,2 12 0 4
( , , ) 1
x y zf x y z f f f
x z y xy x z y xy x z y xyx y z
x z y xy y y xy y xy x x z
f x y z x z y xy y xy xy xy x z
f
L x y z f
⎡ ⎤= ⎣ ⎦⎡ ⎤∂ ∂ ∂
= − − −⎢ ⎥∂ ∂ ∂⎣ ⎦⎡ ⎤= − ⋅ + ⋅⎣ ⎦⎡ ⎤= − +⎣ ⎦
=
= ( ) ( ) [ ]
( ) ( )
1 1,0, 2 ' 1,0, 2 0 4 12 0 4
2 2
( , , ) 4 12 1 0 4 2
x xf y y
z z
L x y z x z
⎛ ⎞ −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ − = +⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎝ ⎠
= + − + + −
( , , ) 12 4 16L x y z x z= + −
The level surface
3 2 4 sinh( ) 4x z y xy− = :
__________________________________________________________________________________
8. Find the gradient of ( ) ( )3 3 2 2cos, x y x yg x y = at the point ( . )2,1 Rounded off to the nearest thousandth, what is the x-component of this gradient?
( )3 3 2 2cosz x y x y=
( ) ( ) ( ) ( )( ) ( )( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )( )
3 3 2 2 3 3 2 2
3 3 2 2 3 3 2 2 3 3 2 2 3 3 2 2
2 3 2 2 3 3 2 2 2 3 2 2 2 3 3 2 2 2
2 3 2 2 4 5
, , , , cos , cos
cos cos , cos cos
3 cos sin 2 , 3 cos sin 2
3 cos 2 si
x yg x y g x y g x y x y x y x y x yx y
x y x y x y x y x y x y x y x yx x y y
x y x y x y x y xy x y x y x y x y x y
x y x y x y
∂ ∂∇ = =
∂ ∂
⎛ ⎞∂ ∂ ∂ ∂⎛ ⎞= + +⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠
= − ⋅ − ⋅
= − ( ) ( ) ( )( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )( )
2 2 3 2 2 2 5 4 2 2
2 3 2 2 4 5 2 2 3 2 2 2 5 4 2 2
n , 3 cos 2 sin
2,1 3 2 1 cos 2 1 2 2 1 sin 2 1 , 3 2 1 cos 2 1 2 2 1 sin 2 1
12,1 12cos 4 32sin 4 , 24cos 4 64sin 4 12cos 4 32sin 4
2
x y x y x y x y x y
g
g
−
∇ = ⋅ ⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅
⎡ ⎤∇ = − − = − ⎢ ⎥
⎣ ⎦
The x-component of this gradient is thus ( ) ( )12cos 4 32sin 4 16.373956399− ≅ , about 16.374 . __________________________________________________________________________________
9. Find the gradient of ( ) 4 4 72, , y zf x y z x −= at the point ( )1,1,1 . Rounded off to the nearest thousandth, the z-component of this gradient is _____ .
( ) ( ) ( ) ( )
( )
( )
4 4 7 4 4 7 4 4 7
3 4 4 3 6
, , 2 , 2 , 2
, , 4 , 4 , 14
1,1,1 4, 4, 14
f x y z y z y z y zx y z
f x y z x y x y z
f
x x x∂ ∂ ∂∇ = − − −
∂ ∂ ∂
∇ = −
∇ = −
The z-component of this gradient is thus 14− .
The graph depicts the gradient field,
( ) 3 4 4 3 6, , 4 , 4 , 14f x y z x y x y z∇ = −
which is always perpendicular to f ’s level surfaces, one of which is also shown:
( ) 4 4 72 1, , y zf x y z x − = −= __________________________________________________________________________________
10. Find , for the f featured in problem #9, with (ˆ 1,1,1uD f )1
1ˆ 214
3u
⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
.
( ) ( )
( )
ˆ
4 1 2 11 14 2 2 2 214 1414 3 7 3
ˆ1,1,1 1,1,1
2 12 2 30 30 142 2 2 4 21 14
1 1414 14 147 3 4
uD f f u⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
= ∇
− −= = + − = =−
i ii
i
( )ˆ15 141,1,1
7uD f −=
__________________________________________________________________________________
11. In which direction is the temperature function ( ) 2arctan, , xzy
T x y z ⎛⎜⎝ ⎠
= ⎞⎟
)
increasing most rapidly, from
the point ( ? Find a vector in this desired direction, and with a magnitude equal to the rate of 1, 2, 4 increase in this direction. The gradient points in the direction in which the scalar temperature field T is increasing most rapidly, and also has the desired magnitude. So we’ll calculate this gradient, and evaluate it at the point .
T∇( )1, 2, 4
( ) ( ) ( ) ( )
( ) ( ) ( )
( )( )
( )( )
( )( )
2 2 2
22 2
2 2 22 2 2
2 3 2
2 2 4 2 2 4 2 2 4
2 2 4 2 3 2
, , , , , , , , , ,
arctan , arctan , arctan
, ,1 1 1
2, ,1 1 1
1 2 1, ,1
x y zT x y z T x y z T x y z T x y z
xzy xzy xzyx y z
xzyxzy xzyyx zxzy xzy xzy
zy xzy xyx z y x z y x z y
z xz xx z y y y y y
− − −
−− −
− − −
− − −
− − −
−
∇ =
∂ ∂ ∂=
∂ ∂ ∂
∂∂ ∂∂∂ ∂=
+ + +
−=
+ + +
−= =
+
( )
( )
2 2 2 2
2
4 2 2
2
4 2 2 2 2
2, ,
2, , , ,
2 2 1 4 1 11, 2, 4 4, , 1 4, 4, 1 4, 4, 12 1 4 2 2 1 4 8
xzz xx z y y
y xzT x y z z xy x z y
T
−
−+
−∇ =
+
− ⋅ ⋅∇ = = − = −
+ ⋅ + ⋅
( ) 1 1 11,2,4 , ,2 2 8
T∇ = −
The gradient field,
( )2
4 2 2
2, , , ,y xzT x y z z xy x z y
−∇ =
+
__________________________________________________________________________________
12. How fast is the temperature function ( ) 2arctan, , xzy
T x y z ⎛⎜⎝ ⎠
= ⎞⎟ increasing at the point ,
in its direction of maximum increase? Round off your answer to the nearest hundred-thousandth.
( )1, 2, 4
The gradient points in the direction in which the temperature T is increasing most rapidly, and its magnitude is the rate of increase in this direction. We calculated this gradient in problem #11, so now we need but find this vector’s length:
T∇
( )2 2 21 1 1 1 1 1 1 1 1 331,2,4 , ,
2 2 8 2 2 8 4 4 64 8T ⎛ ⎞ ⎛ ⎞ ⎛ ⎞∇ = − = + − + = + + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
( )1,2,4 0.71807T∇ ≅
NOTE: If distance is measured in centimeters and temperature measured in degrees Celsius, then the quantity we found has units of degrees Celsius per centimeter.
__________________________________________________________________________________ 13. Find the directional derivative of ( ), , yz xzW x y z xe ye= − at the point ( )3,2,0
in the direction of vector 5, 1, 1v = − . We need a unit vector in the direction of v :
( )22 2 3
51
31
5, 1,1 5, 1,1 5, 1,1ˆ27 35 1 1
1ˆ3
vuv
u⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦
− − −= = = =
+ − +
=
Next we need the gradient of W at the point ( )3,2,0 :
( ) ( ) ( ) ( )
( ) ( ) ( )
( ) 0 0 0 0
, , , , , , , , , ,
, ,
, ,
3,2,0 0, 0 , 6 6 1, 1, 0
x y z
yz xz yz xz yz xz
yz xz yz xz yz xz
x y z
x y z
W W x y z W x y z W x y z
xe ye xe ye xe ye
e yze xze e xye yxe
W e e e e
∂ ∂ ∂=
∂ ∂ ∂
−
−
∇ =
− − −
= − −
∇ = − − = −
Now we can compute the desired directional derivative:
( ) ( )
( )
ˆ 3, 2,0 3, 2,0
2 35 1 03
ˆ1 5 1 5
1 11 1 1 13 3 3 30 1 0 1
1 6 6 33 33 3 3
333
uD W W u⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
+ +
=∇
= − − = − −
= = = =⋅
i
i i
The level surface ( ) 1, , yz xzW x y z xe ye == −
containing the point ( ) . 3,2,0
14. Find the equation of the tangent plane to 2 3 42 2x z xyz y− + = at the point ( )1,1,1 . Let ( ) 2 3 4, , 2x y zF x z xyz y= − +
( ), , 2x y z =
, so we seek an equation for a tangent plane to the implicitly defined surface
. Since F F∇ is normal to the level surface, it will be normal to the tangent plane as well.
( )
( )
3 3 2 2, , 4 , 4 , 6
1,1,1 3, 3, 5
F x y z xz yz xz y x z xy
F n
∇ = − − + −
∇ = =
The point of tangency, , is on both the implicit surface(1,1,1) ( ), , 2x y zF = and the tangent plane to the graph of F at that point. So the equation of the tangent plane is given by:
, , 1,1,1
3, 3, 5 , , 3, 3, 5 1,1,1
3 3 5 3 3 5
n x y z n
x y z
x y z
=
=
+ + = + +
i i
i i
3 3 5 11x y z+ + =
The implicitly defined surface,
2 3 42 2x z xyz y− + = ,
and it’s tangent plane at the point ( ) , 1,1,1
3 3 5 11x y z+ + = :
__________________________________________________________________________________ 15. Find the equation of the tangent plane to the implicitly defined surface
( )cos sin cos 0xyz x z− − = at the point ( ),1,0π .
Let ( ) ( ), , cos sin cosx y z x zF xyz= − −
(, so we seek an equation for a tangent plane to the implicitly defined
surface ), , 0x y zF = . Since F∇ is normal to the level surface, it will be normal to the tangent plane as well.
( ) ( ) ( ) ( )
( )
, , sin cos , sin , sin sin
,1,0 0 cos , 0, 0 1, 0, 0
F x y z yz xyz x xz xyz xy xyz z
F nπ π
∇ = − − − − +
∇ = − = =
The point of tangency, , is on both the implicit surface( ,1,0π ) ( ), , 0x y zF = and the tangent plane to the graph of F at that point. So the equation of the tangent plane is given by:
,1,0, ,1, 0, 0 , , 1, 0, 0 ,1,0
n x y z nx y z
π
π
=
=
i ii i
x π=
The implicitly defined surface,
( )cos sin cos 0x zxyz − − = ,
and it’s tangent plane at the point ( ) , ,1,0π
x π=
Be grateful I didn’t ask you to graph it !
__________________________________________________________________________________ 16. Find any local extrema and saddle points of 2 2( , ) 5 2 3 4f x y x xy y x y= − + + − . Extrema and saddle points occur at any points where '( , ) 0f x y = :
'( , ) 10 3 4 4 0 0f x y x y x y⎡ ⎤⎣ ⎦= − + − + − = ⎡ ⎤⎣ ⎦
10 1 31 4 4
xy
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
− −=
−
Which has solution 8 ,39 39
x y= − = 37 . So a possible extreme value or saddle point is 8 37,39 39
⎛ ⎞⎜ ⎟⎝ ⎠− . To see
which it is, we invoke the multivariable second derivative test. We need the second derivative matrix, the Hessian:
10 11 4
xx xy
yx yy
f ff f
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥
⎣ ⎦⎣ ⎦
−=
−
The determinant of this Hessian matrix is 40 – 2 = 38 > 0, and 10 0xxf = > , hence our point 8 37,39 39
⎛ ⎞⎜ ⎟⎝ ⎠− is a
relative minimum.
The surface
2 25 2 3z x xy y x= − + + − 4y ,
along with its relative minimum at the point
8 37 86, ,39 39 39
⎛ ⎞⎜ ⎟⎝ ⎠− − :
__________________________________________________________________________________
17. Find the absolute maximum and the absolute minimum of the function 3 2 2( , ) y 3f x y x x xy y−= + − over the unit disk 2 2 1x y ≤+ . First we’ll look for any critical points of f within the disk:
2 2
2 2
23 0
2 3 0x
y
yf x x y
f x xy y
−
−
= + =
= + − =
Adding these equations eliminates that pesky middle term, and allows us to solve this system of simultaneous nonlinear equations. We get:
( )( ) ( )
2 2
2 2
2 22 0
2 0
x yx y
x y x y+ −
0− =
− =
=
This equation has for it’s solutions ALL points on the lines y x= − and y x= , so the “X” in the unit disk on these lines consists entirely of critical points. We have an infinite number of critical points! ☺ First lets look at the values of f on the line y x= :
( )
3 2 2 3
3 2 2 3 3 3 3 3, 0( , ) y
x x xx xf x y x x xy yf x x x x x x x
−
− − +
= + −= + − = − =
So for all the input points on the line y x= , f outputs 0. Next consider the values of f on the line y x= − :
( ) ( ) ( ) ( )
3 2 2 3
2 33 2 3 3 3 3, 4
( , )
x x x x
y
x
f x y x x xy y3f x x x x x x x x− −
−
− +− =−
= + −
= + − = + +
So for the input point on the line ( , )a a− y x= − , f outputs . Thus along the line 34a y x= − , f attains its
minimum at the point 2 22 2
,⎛ ⎞⎜⎝ ⎠− ⎟ on the unit circle, where
324 4
2⎞
= =⎟⎠
2 2 2 22 2 8
,⎛ ⎞ ⎛ ⎛ ⎞
= −⎜ ⎟ ⎜ ⎜ ⎟⎝ ⎠ ⎝ ⎝ ⎠− − − 2f .
And along the line y x= − , f attains its maximum at the point 2 22 2
,⎛ ⎞⎜ ⎟⎝ ⎠
− , where
32 2 2 2 24 4
2 2 2 8,f
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠2= .
We still must check for possible extrema on the circular boundary of our disk. This boundary is parametrized by:
cos( )
sint x
r tt y
⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
Evaluating f on this circle gives us a real valued function of t , whose extrema we seek.
Letting and cosc = t sins t= :
( ) ( )
( ) ( ) ( )( ) ( )
3 2 2 3
3 2 2 3
2 2 2 2
( , )( ) ( ) ,
1
( )
3 7'( ) 0 sin cos4 4
f x y x x y xy yg t f r t f c s
c c s cs s
c c s s c s c s c s c s
g t c s
g t s c t t t π π
= − + −
= =
= − + −
= − + − = + − = ⋅ −
= −
= − − = ⇒ = − ⇒ = or
These values of t yield the points 2 2,2 2
⎛ ⎞−⎜ ⎟⎝ ⎠
and 2 2,2 2
⎛ ⎞−⎜
⎝ ⎠⎟ , which we already found earlier.
To summarize, the possible absolute maximum (and minimum) must be one of the following:
( ) 0,
2 2,2 2
2 2
2
2,2 2
f a a
f
f
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
=
− =
←=
−
−
← absolute minimum
absolute maximum
The absolute minimum of f over the disk is 2− ,
and the absolute maximum is 2 .
3 2 2( , ) y 3f x y x x xy y−= + − __________________________________________________________________________________
Problems #18 - 21 refer to the level curve of a function, ( , ) 3f x y = , depicted in the following graph (along with some other level curves of this function:
18. At what point is it true that ( , )f x y∇ has a negative x-component and a small negative y-component? A 19. At what point is it true that ( , )f x y∇ has a negative x-component and a positive (though small) y-component? D 20. At what point is it true that ( , )f x y∇ has zero x-component and a negative y-component? B 21. At what point is it true that ( , )f x y∇ has zero x-component and a positive y-component? E
__________________________________________________________________________________
22. Suppose that and that ( , , )
( , , )( , , )
u x y zg x y z
v x y z⎡
= ⎢⎣ ⎦
⎤⎥
( , )( , )
( , )m u v
h u vn u v⎡ ⎤
= ⎢ ⎥⎣ ⎦
. Let f h g= .
Suppose further that and that 7
(0, 1,2)1
g ⎡ ⎤− = ⎢ ⎥
⎣ ⎦
( )( )
( )( )
7,1 7,1 3 17,1 7,1 1 2
u v
u v
m mn n
−⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥−⎣ ⎦⎣ ⎦
and that
(0, 1,2)(0, 1,2) (0, 1,2) 1 2 3(0, 1,2)(0, 1,2) (0, 1,2) 0 3 0
yx z
yx z
uu uvv v
−− − −⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥−− − −⎣ ⎦⎣ ⎦
. Then '(0, 1,2) ?f − =
Computing the given derivative matrices, we get:
( , , )( , , ) ( , , )'( , , )
( , , )( , , ) ( , , )yx z
yx z
u x y zu x y z u x y zg x y z
v x y zv x y z v x y z⎡ ⎤
= ⎢ ⎥⎣ ⎦
and ( , ) ( , )
'( , )( , ) ( , )
u v
u v
m u v m u vh u v
n u v n u v⎡ ⎤
= ⎢ ⎥⎣ ⎦
Applying the chain rule, we obtain:
( ) ( ) ( )
( ) ( )( )
( )( )
' , , ' ( , , ) ' , ,
( , , )( , , )( , , ) ( , , ) ( , , )' , ,
( , , )( , , )( , , ) ( , , ) ( , , )yxu v z
yxu v z
f x y z h g x y z g x y z
u x y zu x y zm g x y z m g x y z u x y zf x y z
v x y zv x y zn g x y z n g x y z v x y z
=
⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥
⎣ ⎦⎣ ⎦
Thus, ( )( )
( )( )
(0, 1,2)(0, 1,2)7,1 7,1 (0, 1,2)'(0, 1,2)
(0, 1,2)(0, 1,2)7,1 7,1 (0, 1,2)
3 1 1 2 31 2 0 3 0
3 9 9'(0, 1,2)
1 8 3
yxu v z
yxu v z
uum m uf
vvn n v
f
−− −⎡ ⎤ ⎡ ⎤− = ⎢ ⎥ ⎢ ⎥−− −⎣ ⎦⎣ ⎦
− −⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦
−⎡ ⎤− = ⎢ ⎥−⎣ ⎦
__________________________________________________________________________________
23. Same situation described in problem #22. If ( , , )
( , , )( , , )x y z
f x y zx y z
ϕψ⎡ ⎤
= ⎢ ⎥⎣ ⎦
, then at the point (0, 1,2)− ,
?x y z zϕ ψ ϕ ψ∂ ∂ ∂ ∂
− =∂ ∂ ∂ ∂
In problem #22 we found that 3 9 9
'(0, 1,2)1 8 3
f−⎡ ⎤
− = ⎢ ⎥−⎣ ⎦. But this matrix is none other than the matrix:
3 9 91 8 3
yx z
x y z
δϕδϕ δϕδδ δ
δψ δψ δψδ δ δ
⎡ ⎤⎢ ⎥ −⎡ ⎤⎢ ⎥ = ⎢ ⎥−⎢ ⎥ ⎣ ⎦⎢ ⎥⎣ ⎦
Hence, ( ) ( )3 8 9 3 24 27 3x y z zϕ ψ ϕ ψ∂ ∂ ∂ ∂
− = ⋅ − − − = − = −∂ ∂ ∂ ∂
.
__________________________________________________________________________________
24. Find the maximum and minimum values of the function ( , , )f x y z yz xz xy= − + on the unit sphere . 2 2 2 1y zx + + = We’ll use Lagrange multipliers to maximize f subject to the constraint equation . 2 2 2 1( , , ) y zG x y z x= + + =
Then f Gλ∇ = ∇ at the extreme values of f on a level curve of . Calculating the gradients, and setting up the Lagrange multiplier equations, we get:
G
( ) ( ), , 2 , 2 , 2, , & , ,y z z x y x x y zf x y z G x y z− + −∇ = ∇ =
222
y z xz x yy x z
λλλ
− =+ =− =
Adding the first two equations, we get:
( ) ( ) ( )2 2 2 0 2 1x y x y x y x y xλ λ λ λ+ = + + = = −⇒ y+ +⇒
So either or y = −x 12
λ = . We’ll investigate the case where 12
λ ≠ below.
Adding the second two equations, we get:
( ) ( ) ( )2 2 2 0 2 1y z y z y z y z yλ λ λ λ+ = + + = = −⇒ z+ +⇒
So either or y = −z 12
λ = . We’ll investigate the case where 12
λ ≠ below.
In the case where 12
λ = , the original equations become:
y z xz x yy x z
− =+ =− =
Now simply notice that any point with coordinates such that x=y and z=0 (i.e., any point of the form ( ), ,0a a ), satisfies all three equations. Or if you must, you can always find a parametrization of the line of intersection of these two planes. Oops, I mean three planes, but the first two equations are equations of the same plane, so yea, I really did mean two planes. Take the cross product of the normal vectors to these distinct planes to get a direction vector for the line, which must be through the origin since the planes all pass through that common point. But after all that work, you’ll be convinced that x=y and z=0. Substituting these values into our constraint equation, we get:
2 2 2 2 2 2 11 0 1 2
y z xx x x+ + = +⎯ + =⎯→ ⇒ y= ± =
So possible extrema for f on the unit sphere are the two points
1 1, ,02 2
⎛ ⎞⎜ ⎟⎝ ⎠
, where 1 1, ,022 2
f ⎛ ⎞ 1=⎜ ⎟
⎝ ⎠,
and 1 1, ,2 2
⎛ ⎞− −⎜ ⎟⎝ ⎠
0 , where 1 1, ,022 2
f ⎛ ⎞ 1− − =⎜ ⎟⎝ ⎠
.
Now suppose 12
λ ≠ . Then we know from above that y x= − and that y z= − . But that immediately implies
that x z= . So we can express all the variables in terms of z: , , x z y z z z= = − = . Substituting these values into our constraint equation, we get:
( )22 2 2 11 3 1 3
z zz z⎯⎯→ ⇒+ − + = = z x= ± =
And since , these variables have opposite signs. We’ve found two points, y = −z 1 1 1, ,3 3 3± ±⎛ ⎞
⎜ ⎟⎝ ⎠
∓ .
So possible extrema for f on the unit sphere are the two points
1 1 1, ,3 3 3
⎛ ⎞−⎜ ⎟⎝ ⎠
,
where
1 1 1 1 1 1 1 1 1 1 1, ,3 3 33 3 3 3 3 3 3 3 3
1 1 1, , 13 3 3
f
f
⎛ ⎞ ⎛ ⎞− = − − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞− = −⎜ ⎟⎝ ⎠
1− + − −
and 1 1 1, ,3 3 3
⎛ ⎞− −⎜ ⎟⎝ ⎠
where
1 1 1 1 1 1 1 1 1 1 1 1, ,3 3 33 3 3 3 3 3 3 3 3
1 1 1, , 13 3 3
f
f
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞− − = − − − − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠
⎛ ⎞− − = −⎜ ⎟⎝ ⎠
− + − −
We’ve found two absolute maximums for f on the unit sphere, and two absolute minimums:
absolute maximum
minimum
1 1 1 1 1 1 1, ,0 0 0 22 2 2 2 2 2
1 1 1 1 1 1 1 1 1, absolute
, 1 3 3 3 3 3 3 3
3
3
( , , )
f
f x y z y z x z x y
f
⋅ ⋅ ⋅
⎛ ⎞ ⋅ − ⋅ + ⋅ =⎜ ⎟⎝ ⎠± ± ± ± ± ±⎛ ⎞ ⋅ − ⋅
←
+ ⋅ = ←−⎜ ⎟⎝ ⎠
= − +± ± ± ± ± ±=
=∓ ∓ ∓
__________________________________________________________________________________ 25. Given that a tetrahedron has volume = 1, what is the minimum possible value that A h+ can have, where A is the area of the tetrahedron and is its height?
h
(Hints: The area of a tetrahedron is the one third the area if its base times its height. Also, ignore units for this problem.)
We’ll use Lagrange multipliers to minimize the quantity volume A h+ subject to the constraint equation
113
Ah= . We can obtain a nicer constraint equation by multiplying both sides by 3, obtaining 3Ah = . Let
( , )f A h A h= + be the “objective function” we wish to minimize, and let be our constraint equation. Then
( , ) 3G A h Ah= =f Gλ∇ = ∇ at the extreme values of f on a level curve of . Calculating the gradients, and
setting up the Lagrange multiplier equations, we get: G
( ) ( )1,1 ,, & , h Af A h G A h∇ = ∇ =
11
11
A A
h h
λλ
λλ
⇒
=
=
⇒=
=
Which implies that . Substituting A h= A h= into the constraint equation, we get:
2
33
3
3
A hh hh
h
⋅ =⋅ =
=
= ±
The negative answer makes no sense in this context, so the only extreme point is 3A h= = . For this point:
( )( , ) 3, 3 3 3 2 3f A h f= = + =
This is the only extreme value for f. The function f has arbitrarily large values, for as , , in order for the volume to remain constant, equal to 1. Thus this extreme value must be the minimum we seek.
0h → A→∞
__________________________________________________________________________________
26. You need to construct a tank consisting of a cylinder with two hemispherical ends. Suppose that the material for the cylindrical portion of the tank costs $30 per square meter while the hemispherical ends cost $40 per square meter. Find the length h of the cylindrical part of the tank, and its radius , in order to enclose 10,000 m3 at minimal cost. What is the minimum cost. r Note: None of the given answers were correct for this one, so the correct answer was “n”.
The volume of this tank is 4 34103
V r 2r hπ π= = + . This is our constraint equation, and we let:
( ) 3 24, 13
G r h r r h 40π π= + =
The cost of this tank is ( ) (240 4 30 2C rπ= + )rhπ . This is our objective function, and we let:
( ) 2, 160 60f r h r rhπ π= +
Calculating the gradients, and setting up the Lagrange multiplier equations, we get:
( )
( ) 2 2
320 60 , 60 =20 16 3 , 3
4 2 , 4 2 ,
,&
,
r h r r h
r rh r r r h r
f r h
G r h
π π π π
π π π π
+ +
+ +
∇ =
∇ = =
r
( ) ( ) ( )( )
( )( )
22
20 16 3 10 16 320 16 3 2 2
2 2 260 6060
r h r hr h r r h
r r h r r hrr r
r r
ππ λπ λ
πππ λπ λ
π
+ ++ = + = =
+ +
= =⇒=
⇒
Setting the λ ’s equal, we get:
( )( ) ( )10 16 3 60 16 3 6 2 16 3 12 62
4 4 3 3
r hr h r h r h r
r r h r
r h h r
⇒+
= + = + + =+
= =
⇒
⇒ ⇒
h+
This must yield the minimum we seek, since we could find arbitrarily expensive tanks given the problem constraints. Substituting this value for into our constraint equation, we obtain (in meters): h
3 2 4 3
3
4 4 375010 10.607844179473 3
&
4 4 3750 14.1437922392943 3
r r r r
h r
π ππ
π
⇒⎛ ⎞+ = =⎜ ⎟⎝ ⎠
= =
A diagram of the tank with (close to) the optimal ratio of and h : r
The minimum cost of the tank is thus:
( )2
3 3 3
3 3 3 3 3
2
3
160 60 20 8 3
3750 3750 4 375020 8 33
3750 3750 3750 3750 375020 8 4 20 12
3750240 $84,842.88
Cost r rh r r h
Cost
π π π
ππ π π
π ππ π π π π
ππ
= + = +
⎛ ⎞= + ⋅⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛
= + =⎜ ⎟ ⎜⎝ ⎠ ⎝
⎛ ⎞= ⇒ ≅⎜ ⎟
⎝ ⎠
⎞⎟⎠
__________________________________________________________________________________
27. Find the point in Octant I and on the surface 2 1xy z− = that’s closest to the origin. Let our objective function f be the function that outputs the square of the distance from an arbitrary input point ( , , )x y z to the origin:
2 2( , , ) 2f x y z x y z= + + For our constraint equation, we have 2( , , ) 1G x y z xy z= − = . Calculating the gradients, and setting up the Lagrange multiplier equations, we get:
( , , ) 2 , 2 , 2 =2 , ,&
( , , ) , , 2
f x y z x y z x y z
G x y z y x z
∇ =
∇ = −
22
22
2 2 1 or 0
xx yyyy xx
z z z
λ λ
λ λ
λ λ
= =
= =
= −⇒= −
⇒
=
⇒
Note that if x=0 then also y=0, which can’t be a point on the surface defined by our constraint equation. So we can safely assume that neither x nor y are 0. If 1λ = − , we have a contradiction, since this value substituted
into the previous two equations yields simultaneously 2y x= − and 12
y x= − , which can’t both be true
(unless x =0 which we’ve already eliminated as an option). The only remaining possibility is that z=0. Setting the first two equations for λ equal, we get:
2 22 2 x y x y x y xy x y x= = =⇒ ⇒ = ±⇒ y
Of the points ( , ,0)x x and ( , ,0)x x− , only the first is in Octant I, so we substitute this one into our constraint equation:
2 2( , , ) 1 ( , ,0) 0 1 1G x y z xy z G x x x x= ⇒ ⇒− = = − = = ±
Evaluating our objective function on the point (1 , we get ,1,0) (1,1,0) 2f = . The point we seek is (1,1,0) . __________________________________________________________________________________
Top Related