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EXAMPLES FOR
SECTION 13.7
Here is an
animation of planes tangent to the
paraboloid whose equation is z = x2 + y2.
More planes
tangent to the paraboloid whose equation is z = x2 + y2 (slower
loading).
DPGraph Version
Quicktime Version
Here are some
of my notes on normal lines and tangent planes.
PowerPoint
Presentation and Examples (#10 and #46 and #43)
Example--Finding The Equation Of A
Tangent Plane
Here is a
link to a
video presentation with audio of me describing how to find a tangent
plane and normal line to the surface shown here.
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 (zoom) 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 13.7
#10
Find a unit
normal vector to the surface at the indicated point.
Let's
look at the tangent plane and normal line at the same point. A
normal to the tangent plane would be < 4 , 3 , 12 >. Thus an
equation for the tangent plane would be
4x +
3y + 12z = d
and if
we substitute in the point (2,-1,2) we find that
d =
4(2) + 3(-1) + 12(2) = 29.
Tangent
Plane: 4x + 3y + 12z = 29
A
direction vector for the normal line would be < 4 , 3 , 12 >.
The
line with this direction vector passing through the point
(2,-1,2)
would be given parametrically by
x = 2
+ 4t
y = -1
+ 3t
z = 2
+ 12t |
Picture
of the Surface and the Unit Normal Vector
Picture of the Surface with the Tangent Plane,
Normal Line, and Unit Normal Vector
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| Section 13.7 #43 (a) Find symmetric equations of the
tangent line to the curve of intersection of the surfaces at the indicated
point, and (b) find the cosine of the angle between the gradient
vectors. State whether or not the surfaces are orthogonal at the
given point.
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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.
Click here
to see a picture that includes the two normal (gradient) vectors that were
found at the left not to be orthogonal. (The angle between them is
about 50.2o.) Click
here to view them from a different angle.
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| Section 13.7 #46
(a) Find symmetric equations of the
tangent line to the curve of intersection of the surfaces at the indicated
point, and (b) find the cosine of the angle between the gradient
vectors. State whether or not the surfaces are orthogonal at the
given point.
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Pictures
of the Surfaces, Curve of Intersection, Tangent Line at (1,2,5), and Unit
Normal Vectors at (1,2,5)
Larger Pictures with Audio
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| Section 13.7 #50 
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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. |
| Finding Tangent Planes
Find the equation of the plane tangent to the
given surface at the point (1,0,1) and find the points where the tangent
plane is horizontal.
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DPGraphpicture
including the tangent planes (tilt the graph to see them
clearly) DPGraphpicture with blue planes
Click on the
Maple picture to see a different view. The blue
traces intersect at (0,0,0) and the red
traces intersect at (1,1,-1).
Maple Picture of the tangent plane at
(1,0,1) along with the normal line at the same point.
Maple Picture, different view.
This Quicktime movie shows many views
of the surface.
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