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Examples For Exam III
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the pictures linked to on this page. |
QuickTime
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Domain,
Range Examples
| Section 12.3 #36 (similar to 6e 32)
DPGraph
Picture of the surface and the plane. |
 Graph of the curve of intersection in the xz-coordinate
plane |
DPGraph
Picture of the surface and the plane. |
 Graph of the curve of intersection in the yz-coordinate
plane |
This DPGraph
Picture shows the surface intersecting the graph of the plane y = (4/3)x.
What might this lead to?
Total
Differential Example
Implicit
Differentiation Example DPGraph
of the Surface
| Section 12.5 # 42 (6e 36) Implicit
Differentiation Example
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| Chain Rule Application
Here is a DPGraph
Picture of the expanding and contracting frustum.
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| Directional Derivative Example
If your answer to the question posed at the end
of the Section 12.3 #36 example related to the directional derivative, or if you have
not yet heard about directional derivatives but your answer somehow
related to a derivative as a rate of change relative to change along the
line y = (4/3)x or change relative to change in the direction of a vector,
say <3,4>, then good show.
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| Section 12.6 #32 (6e 26)
Maximizing the Directional Derivative
DPGraph
Picture of the Surface DPGraph
Picture with the Point
DPGraph
Picture of the surface, the point, and a plane containing the point
and a line in the xy-coordinate plane with direction vector u where
the x- and y-components of u equal the x- and y-components of the
direction vector for the directional derivative. See the analysis
below for more details.
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Heat
Seeking Particle Examples
T(x,y) = 100 - x2 - 2y2
Starting Point (2,4)
Let the path be described by r(t) = <
x(t) , y(t) >
Then r'(t) = < dx/dt , dy/dt >
gives the direction of motion.
Since this is a heat seeking particle it must move at
all times in the direction of maximum increase in temperature. This must
be the direction of the gradient of the temperature function T(x,y).
grad T(x,y) = < -2x , -4y >
In general then it must be the case that grad T =
k r'(t) where k is some constant. Let us assume a parameterization
of the path such that k = 1 so that grad T = r'(t). Note
that we are not claiming that r'(t) gives the velocity of the heat
seeking particle, only that it gives the direction.
From grad T = r'(t) we equate vector
components to get dx/dt = -2x and dy/dt = -4y. Let 0 be the value of t at
the starting point so that x(0) = 2 and y(0) = 4. Thus we have two
differential equations to solve.
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dx/dt =
-2x x(0) = 2
(1/x)dx =
-2dt
ln | x |
= -2t + c
| x | = e-2t
+ c = e-2t ec = c1e-2t
x = c2e-2t
x = 2e-2t
since x(0) = 2 yields c2 = 2
In a
similar fashion we would find that from
dy/dt =
-4y and y(0) = 4 we get
y = 4e-4t
Thus the
path is given by r(t) = < 2e-2t , 4e-4t
>.
Since y =
(2e-2t)2 we see that the path is also given by y = x2.
The
picture on the right shows the path. Animation |
 |
A similar problem involves finding the path followed by
a heat seeking object in space where the temperature distribution is given by
T(x,y,z) = 100 - 3x - y - z2 and the starting point is (2,2,5).
Let the path be described by r(t) = <
x(t) , y(t) , z(t) >
Then r'(t) = < dx/dt , dy/dt , dz/dt >
gives the direction of motion.
Since this is a heat seeking particle it must move at
all times in the direction of maximum increase in temperature. This must
be the direction of the gradient of the temperature function T(x,y,z).
grad T(x,y,z) = < -3 , -1 , -2z >
In general then it must be the case that grad T =
k r'(t) where k is some constant. Let us assume a parameterization
of the path such that k = 1 so that grad T = r'(t). Note
that we are not claiming that r'(t) gives the velocity of the heat
seeking particle, only that it gives the direction.
From grad T = r'(t) we equate vector
components to get dx/dt = -3, dy/dt = -1, and dz/dt = -2z.
Let 0 be the value of t at
the starting point so that x(0) = 2, y(0) = 2, and z(0) = 5.
We have
three
differential equations to solve and their solutions will yield x = -3t + 2, y =
-t + 2, and z = 5e-2t.
Thus r(t) = < -3t + 2 , -t + 2 , 5e-2t
>
Click here to see a
picture of a portion of the path.
| Example--Finding The Equation Of A
Tangent Plane
The picture on the right shows part
of the graph of
z = f(x,y) = 9 - x2 - y2
and the plane
tangent to the surface at (1,1,7). Click on the picture to see an animation.
Write the equation of the surface as
F(x,y,z)
= x2 + y2 + z - 9 = 0
grad F
= < 2x , 2y , 1 >
grad F(1,1,7)
= n = < 2, 2 , 1 > so the equation of the tangent plane will
be
2x + 2y + z = d and substituting in (1,1,7) yields d =
11.
2x + 2y + z = 11
A line
normal to the surface at (1,1,7) would be given by
x = 1 +
2t, y = 1 + 2t, z = 7 + t.
DPGraphPicture
Blow-up of DPGraphPicture
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Click here
to zoom in on the point of tangency.
Maple
picture of the surface, plane, and normal line
Maple
picture, different view
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| Section 12.7 #41 (6e 35)
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Click on the picture above to see an animation.
DPGraph
Picture of the surfaces.
DPGraph
picture of the surfaces similar to the picture above.
Click on the picture above to see an
animation. The picture shows the curve of intersection of the two
surfaces in red and the tangent line at the point (3,3,4) in blue.
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| Section 12.7 #48 (6e 42 similar)
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Click on the picture above to see an
animation. The surface is in red, the tangent plane in blue, and the
xy-coordinate plane in green.
DPGraph
Picture 1
DPGraph
Picture 2
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| Example--Finding Where the Tangent
Plane Is Horizontal
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DPGraph
Picture of the surface and the tangent plane. |
EXAMPLE--Tangent
Planes and Max-Min
Find the points on the graph of the function below
where the tangent plane is horizontal.
f(x,y) = x3 - 3xy + y2
DPGraphPicture (z-axis
scale: 1 unit = 10) DPGraphPicture
with the 2 horizontal tangent planes
Since the surface can be represented by F(x,y,z) =
f(x,y) - z = x3 - 3xy + y2 - z = 0 and grad F will
give a vector normal to the surface we need to find points where grad F =
< 0 , 0 , c > where c is any nonzero constant.
grad F = < fx , fy , -1
> so we are looking for points on the surface where fx = fy
= 0.
fx =
3x2 - 3y = 0
fy =
-3x + 2y = 0
The solutions
to this system are (0,0), and (3/2,9/4).
f(0,0) =
0 and f(3/2,9/4) = -27/16 so the points on the surface where the
tangent plane is horizontal are (0,0,0) and (3/2,9/4,-27/16). The picture
on the left below shows the graph of the surface around the point (0,0,0) and
the picture on the right shows the graph around the point (3/2,9/4,-27/16).
Click on each graph to see an animation that will include the tangent plane at
the critical point on that graph.
 
To determine the nature of these
points where the tangent plane is horizontal we can apply the Second Partials
Test.
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d = fxxfyy
- (fxy)2 fxx =
6x, fyy = 2, fxy
= -3
d(x,y) = 12x - 9
d(0,0) = -9 so (0,0,0) is a
saddle point.
d(3/2,9/4) = 9 and fxx
and fyy are positive at (3/2,9/4) so (3/2,9/4,-27/16) is a
relative minimum point.
On the right is a graph of the
surface over the x-interval [-1,3] and y-interval [-1,3]. Click on
the picture at the right to see an animation. It would be difficult
to visually determine the nature of the critical points from the graph on
the right but less difficult from the graphs above. The analytical
approach tells us for sure. It is also pretty obvious from this DPGraph
Picture with the two critical points (z-axis scale: 1 unit = 1)
and this DPGraph Picture of
the surface with the two horizontal tangent planes. Use the cursor
keys to look at the pictures from different views. |
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Now let's consider the problem of
finding the absolute minimum and absolute maximum value of the function over the
square region of the xy-coordinate plane described by the x-interval [-1,3] and
y-interval [-1,3]. See the picture on the right above or this DPGraph
Picture (z-axis
scale: 1 unit = 10). The two
critical points found above yield -27/16 as a possible minimum and 0 as a
possible maximum. We must now evaluate the function on the boundary of the
square region being investigated.
Check of the boundary x = -1 with y
in [-1,3]
f(-1,y) = g(y) = -1 + 3y + y2
g'(y) = 3 + 2y = 0 if y =
-3/2 which is not in [-1,3]
g(-1) = -3
g(3) = 17
Check of the boundary x = 3 with y in
[-1,3]
f(3,y) = h(y) = 27 - 9y + y2
h'(y) = -9 + 2y = 0 if y =
9/2 which is not in [-1,3]
h(-1) = 37
h(3) = 9
Check of the boundary y = -1 with x
in [-1,3]
f(x,-1) = j(x) = x3 + 3x +
1
j'(x) = 3x2 + 3 cannot
equal 0
j(-1) and j(3) would put us at
corners of the square already evaluated above.
Check of the boundary y = 3 with x in
[-1,3]
f(x,3) = k(x) = x3 - 9x +
9
k'(x) = 3x2 - 9 = 0 if x =
31/2 and -31/2 -31/2
is not in [-1,3]
k(31/2) = 9 - 6(31/2)
which is greater than -3
k(-1) and k(3) would put us at
corners of the square already evaluated above.
Thus the minimum value of the
function over the region is -3 and occurs at (-1,-1,-3)
and the maximum value of
the function over the region is 37 and occurs at (3,-1,37).
Here is a DPGraph
Picture of the surface
f(x,y) = (x3 - 3xy + y2)/10
and the planes z = -0.3 and z = 3.7
graphed over
the square region of the xy-coordinate plane described by the x-interval [-1,3] and
y-interval [-1,3].
Section 12.8 #25 (6e 17) DPGraphpicture
DPGraphpicture
including the horizontal tangent planes (tilt the graph to see them both
clearly)
| Section 12.8 #28 (6e 20) DPGraphpicture
Here is another DPGraph Picture
(finer mesh). Try looking at z-slices on the scrollbar to roughly
approximate the relative maximum and minimum values to support the answers given
above.
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Absolute Max-Min Example
Similar
to Section 12.8 #59 (6e 41)
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The region R is graphed above in blue
and part of the line y = -x is graphed in red.
DPGraph
Picture of the Surface
DPGraph
Picture of the surface surrounded by the cylinder x2
+ y2 = 8 DPGraph
Picture of the surface and the region using cylindrical coordinates 
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EXAMPLES
WHERE THE SECOND PARTIALS TEST FAILS
| Section 12.8 #45
DPGraph
Picture of the surface Click on the pictures at the
right.
These traces are pictured at the right. |
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| Section 12.8 #49
DPGraph
Picture of the surface
One
should note that in this example the Second Partials Test could not be
applied in any case
since the
partial derivatives never equal zero and do not meet the continuity
requirements.
We see
below what would happen if we did try to
apply it anyway.
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| Optimization Example--Linear
Regression
Find the equation of the linear function (y = ax
+ b) that best fits the data points (1,2), (2,3), (4,6) based on a least
squares regression analysis criteria.
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TWO
MORE OPTIMIZATION EXAMPLES
| Section 12.9 #3
Find the point on the graph of z = x2
+ y2 closest to the point (5,5,0).
In this DPGraph
Picture of z = x2
+ y2 and the point (5,5,0) each unit on the z-axis corresponds
to 10.
d =
[(x-5)2 + (y-5)2 + (z - 0)2]1/2
d(x,y) =
[(x-5)2 + (y-5)2 + (x2+y2)2]1/2
d2
= f(x,y) = (x-5)2 + (y-5)2 + (x2+y2)2
Solve
the system below.
fx
= 2(x-5) + 2(x2+y2)(2x) = 0
fy
= 2(y-5) + 2(x2+y2)(2y) = 0
y(x-5) + y(x2+y2)(2x) = 0
x(y-5) + x(x2+y2)(2y) = 0
Subtraction
produces -5y + 5x = 0 which yields
x = y
Back
substitution produces
x - 5 + 2x(2x2) = 0
4x3 + x - 5 = 0
The
solutions are
x = 1, x
= -1/2 - i and x = -1/2 + 1
For x=1,
y=1, and z=2.
The point
on the graph of z = x2 + y2 closest to the point
(5,5,0) is (1,1,2).
d(1,1) = 6
The graph
at the right is the graph of d(x,y). Click on the graph to see it
rotate. DPGraphPicture of
d(x,y)
Here is a
DPGraph Picture of z = x2
+ y2 and the points (5,5,0) and (1,1,2).
|
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| Similar to Section 12.9 #10
In constructing an open topped box it
costs $3 per square foot for the base and $2 per square foot for the
sides. Find the dimensions of the box of maximum volume that can be
constructed for $100.
Let x =
width of base, y = length of base, z = height
3xy + 4xz
+ 4yz = 100
z = (100
- 3xy)/(4x + 4y) (1)
V = xyz
V(x,y) =
xy(100 - 3xy)/(4x + 4y)
Vx
= (400y2 - 12x2y2 - 24xy3)/(4x
+ 4y)2 = 0
Vy
= (400x2 - 12x2y2 - 24yx3)/(4x
+ 4y)2 = 0
This
leads to the system
100 - 3x2
- 6xy = 0
100 - 3y2
- 6xy = 0
and
subtraction leads to x = y.
Back
substitution produces
100 - 9x2
= 0 so x = 10/3 and y = 10/3
Substituting
into (1) above produces z = 5/2.
Thus the
volume is a maximum if the dimensions are
10/3 ft
by 10/3 ft by 5/2 ft
and the
maximum volume is 250/9 cubic ft.
The graph
at the right is the graph of V(x,y). Click on the graph to see it
rotate. DPgraphPicture of
V(x,y)
You might
enjoy staring at the pictures below of the box of maximum volume for 15
minutes or so if you have no life.
DPGraphPicture
of the box of maximum volume
DPGraphPicture
of the box with shading
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