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EXAMPLES FOR EXAM II Sections 12.4 and 12.5
| If it is not already on your hard drive, you will need to download
the free DPGraph Viewer to view some of
the pictures linked to on this page. |
QuickTime
7 free download. |
Powerpoint Introduction
to Sections 12.4 and 12.5
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Hints for
Section 12.5 Homework Problem 10 in WebAssign
SphereAndParaboloid1
SphereAndParaboloid2
SphereParaboloidAnimation
A sphere of
radius 4 is dropped into the paraboloid given by z = x2 + y2.
(a)
How close will the sphere come to the vertex of the paraboloid?
(b)
What is the radius of the largest sphere that will touch the vertex?
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In
SphereAndParaboloid1 the value for "a"
gives the z-coordinate of the center of the sphere. By activating "a" on
the scrollbar you can "scroll a" to move the sphere to the position where it
will rest against the inside of the paraboloid. The resulting value of "a"
can be used to determine how close the bottom of the sphere comes to the vertex
of the paraboloid. It can also be fun to look at "X slice" or "Y slice" on
the scrollbar. This can show you a picture of a cross section (circle and
parabola). You might want to change the background color to white and the
object color to blue. Here is a Quicktime animation of the
sphere and paraboloid ("A" gives the
z-coordinate of the center of the sphere) and a Quicktime animation of a cross
section, a circle and parabola, ("a"
gives the y-coordinate of the center of the circle) that also includes the
largest circle that could rest at the vertex of the parabola (part b, largest
sphere to rest at the vertex of the paraboloid). Here is the
Maple worksheet used to create the Quicktime
animations.
In
SphereAndParaboloid2 the value for "b"
gives the radius of the sphere. By activating "b" on the scrollbar you can
"scroll b" to change the radius of the sphere. The value for "a" again
gives the z-coordinate of the center of the sphere. By activating "a" on
the scrollbar you can "scroll a" to move the sphere up and down. Thus by
changing "a" and "b" you can attempt to visually estimate the radius of the
largest sphere that will rest inside the paraboloid at its vertex.
In
SphereParaboloidAnimation you can see an
animated sphere of radius 4 dropping into the z = x2 + y2
paraboloid as far as it will go, "bouncing back up" and then dropping back down
again and repeating this over and over. You might want to turn off the
animated colorization by changing the color to one color (blue works well).
For
Extra Credit analytically
prove the results that could be obtained visually in the DPGraphs above.
| Finding a Line Tangent to a Space
Curve
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Click on the picture to see an animation.
Larger Picture |
| Finding v, a, T, N, k
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Click on the picture above to see an
animation. The velocity vector is in green,
acceleration vector in red,
and the principle unit normal vector in purple.
Notice the relationship between a(t) and N(t) as t gets
larger. Can you see analytically why this is happening?
Quicktime
animation of the picture above |
| Finding T, N, k, aN, aT
Note that since we are working in R2
we could have narrowed down the possibilities for N(t) to two based
on N(t) being orthogonal to T(t) and then determined N(t)
if we knew enough about the concavity of r(t) (N is in the
"direction of concavity"). So we could determine N
from T if the nature of the graph at the right was known to us.
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The figure
below shows the velocity vectors (green)
and acceleration vectors (red)
at t = -1, t = 0, and t = 1 along the path described by the vector valued
function r(t) = <t,4-t2>. Click on the picture to
see an animation that will also include the principal unit normal vector N
(magenta).
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| Finding T, N, k, aN, aT
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| Arc Length Example
Find
the length of the curve described by r(t) = < t , t2 , t3
-3t2 + 2t > over the t interval [0,2]. See the figure at the
right (click to spin). In the figure the arc (blue)
is being approximated by four line segments (red).
Maple Worksheet
showing line segment approximations and pictures
Quicktime animation
showing approximations using 4, 8, 12, 16, 20, 24, 28, 32 line segments
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We could approximate the arc length by adding the
lengths of the four approximating red line segments.
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Roller Coaster
Problem The position function r(t)
= < 10sin(2t) , 10cos(2t) , 3t >, t going from 0 to 4pi, describes part of
the motion
of a roller coaster car along a spiral track at an amusement park (the path
back to the bottom of the ride is not being described here). The
mass of the roller coaster car is 400kg, distance is in meters, and time in
seconds. Find the force along N required to keep the roller coaster car on its
path. There
is more than one way to do this so I will find more than is necessary.
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| Ferris Wheel Problem
A circular Ferris wheel has a radius of 20
feet. The center of the Ferris wheel is 26 feet above the
ground. There is one hanging seat hooked up and this seat always
hangs straight down 4 feet from a point on the circumference of the Ferris
wheel. When running the Ferris wheel makes one revolution every 20
seconds and turns counterclockwise. Construct a position function
for the point at the bottom of the hanging seat (the point always 4 feet
directly below a point on the circumference of the Ferris wheel). In
doing this assume the Ferris wheel reaches full speed in less than 1/4
revolution and model your position function such that the point on the
circumference of the Ferris wheel directly above the hanging seat is at
three o'clock at time t = 0. From your position function find a
velocity function for the point on the bottom of the hanging seat.
With the Ferris wheel at full speed, find the magnitude of the velocity of
the point on the bottom of the hanging seat 5/3 seconds after it reaches
its lowest point. What is its lowest point? Click
here to see an animation.
Solution The
path of the point will be a circle with a radius of 20 feet and a center
that is 22 feet above the ground. The lowest point on this circle
will be 2 feet above the ground. Take the ground to be at y = 0 and
introduce an xy-coordinate system such that the center of the circular
path to be described is at (0,22). Thus the equation of the path in
rectangular coordinates is
 Now
we need to find A such that the point completes one revolution in 20
seconds. To do this we need the fundamental period of cos(At) and
sin(At) to be 20. The fundamental period of cos(At) is
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| Circle of Curvature
Find the curvature, radius of curvature, center
of the circle of curvature, and equation of the circle of curvature at the
point (0,0) on the graph of the function given below.
The radius of curvature at (0,0) is 1/2. To
find the center of curvature we must go 1/2 unit from (0,0) in the
direction of N. In this case that would be 1/2 unit straight
up to the point (0,1/2). Using the standard form for the equation of
a circle
we find that the equation of the circle of
curvature is
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For extra
credit you may find the curvature, center of curvature, and equation of the
circle of curvature at the point (1,1) on the graph of the same parabola (see
the figures below). Click here
for an animation of many of the circles of curvature for this parabola and click
here for the same animation but also showing the changing radius of
curvature. Quicktime Version 
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| Circle of Curvature Find the curvature, radius of curvature, center
of the circle of curvature, and equation of the circle of curvature at the
point (1,1/6) on the graph of the function given below.
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Click the Picture for an Animation |
return |
