|
EXAMPLES FOR EXAM II
| 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. |
| Limit and Continuity Example
The graph of the position function is pictured at
the right. |
Click on
the picture above to see an animation. |
See the picture at the right. |
 |
See the picture at the right. |
 |
See the picture at the right. |
 |
Note: The answer above is a
VECTOR. Even though this limit exists, we can observe
that r(t) is not defined at t = 1 because both x(t) and z(t) are
not defined at t = 1. Thus r(t) is not continuous at t =
1. For r(t) to be continuous at t = 1 all three of its
component functions, x(t), y(t), and z(t) must be continuous at t
=1. Recall that the definition of continuity for a function of one
variable, say f(x), requires that for f to be continuous at x = a it must
be true that
 .
This means that the limit must exist at a, the
function must be defined at a, and the two values must be equal.
This boils down to saying that you must be able to compute the limit by
direct substitution. |
| Example--the Definite Integral of
a Vector Valued Function
I'll say that again. The answer is a
VECTOR. |
| Finding a Line Tangent to a Space
Curve
|
Click on the picture to see an animation.
Larger Picture |
Projectile Motion Example
A projectile is launched from ground level at an angle of 45o
with the horizontal and with an initial velocity of 64 feet per second. A
television camera is located in the plane of the path of the projectile 50 feet
behind the launch site.
Parametric equations for the path of the projectile in
terms of the parameter t representing time are
Some of the basic things we can easily compute are the
maximum height attained by the projectile and the range of the projectile.
The graph in red below represents the path of the
projectile and the blue point moving along the graph in red represents
the projectile. The length of the vertical blue line segment at x = -50
represents the measure of angle a (the angle the camera makes with the
horizontal) in degrees
animation of the
projectile motion and changing angle a
The angle a that the camera makes with the
horizontal is given by
Below is a graph of the measure of angle a in
degrees as a function of time (t).
Notice that a is not a maximum at the same time
that y is a maximum.
| Animation For Section 11.3, Example
6
A baseball is hit 3 feet above the ground at 100
feet per second and at an angle of 45o with respect to the
ground. Find the maximum height reached by the baseball. Will
it clear a 10 foot high fence located 300 feet from home plate? The solution can be found in your textbook.
Click on the picture at the right to see the animation. Quicktime
Animation Quicktime
Animation Extended
|
 |
| Section 11.3 #34 (6e 27) (similar)
The quarterback of a football team releases a
pass at the height of 7 feet above the playing field, and the football is
caught by a receiver 30 feet
directly downfield at a height of 4 feet. The pass is released at an
angle of 35o with the horizontal.
Click on the picture to see an animation.
In my animation the receiver is 20 feet
directly downfield when the
quarterback releases the football.
Quicktime
version
(a) Find the speed of the football when it
is released.
(b) Find the maximum height of the
football.
(c) Find the time the receiver has to reach
the proper position after the quarterback releases the football.
The receiver has approximately 1.2249 seconds to
reach the proper position after the quarterback releases the football.
Highlights of the Solution to Section 11.3 #34
(6e 27)
(The difference is that the 30 feet
is 30 yards.)
Animation
for the original problem
Quicktime Version
|
|
| Section 11.3 #36
(6e 30)
|
| The Path of the Bomb with Air
Resistance
The picture below shows the path the bomb would
take if we do not neglect air resistance but rather take air resistance as
proportional to velocity (of the bomb), i.e., r = kv. The time
interval is the length of time it takes the bomb to hit the ground
neglecting air resistance (approximately 43.3 seconds). The successive values for r in the graph,
from left to right, are: 0.1, 0.08, 0.06, 0.04, 0.02, 0.01, 0. Click
on the picture below to see a similar picture that also includes the paths
corresponding to r = 0.4 and r =0.2. The bomb falls very slowly if r
= 0.4. |
|
| The graph
below is a graph of the endpoints of paths such as those shown above as r
goes from 0.000001 to 3. |
 |
Lawn Sprinkler
Here is an example of the lawn sprinkler problem found
in the exercises (#56) for Section 3.1 (It is a Calculus III problem.). In
the example here the speed of the water
is 16 ft/sec so the distance the water travels horizontally is given by
and the path the water takes through the air is given
by
Can you see why (neglecting air resistance)? Answer
Click here to see
an animation for this problem and click here
for an animation with scales. Would this sort of lawn sprinkler water the
lawn uniformly? Answer For more
information on the "calculus of lawn sprinklers" see the article
"Design of an Oscillating Sprinkler" by Bart Braden in Mathematics
Magazine. You can view the article at matharticles.com.
| EC
The picture at the right shows the paths of a
projectile launched from sea level with an initial speed of 48
ft/sec. The projectile has been launched in the direction of a gully
whose flat bottom is 192 feet lower than the spot from which the
projectile was launched. The blue
path corresponds to a launch angle of 60 degrees. The red
path corresponds to a launch angle of 45 degrees. The green
path corresponds to the launch angle that maximizes the horizontal
distance traveled by the projectile. The black
path corresponds to a launch angle of 0 degrees. The extra credit is to
approximate the radian measure of the launch angle for the green path (6
significant digits), give the approximate degree measure (4 significant
digits), and approximate the horizontal distance traveled (6 significant
digits) for the green path. Air resistance is being ignored.
Click on the picture to see an animation showing the path of the
projectile as the launch angle varies from zero to almost ninety
degrees. Quicktime version of
the animation |
Maple
Version of the Animation Quicktime
Maple Version
Maple
Worksheet for the Maple Animation
|
| Finding v, a, T, N, k
|
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.
|
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).
.
|
| Finding T, N, k, aN, aT
|
 |
| 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).
|
We could approximate the arc length by adding the
lengths of the four approximating red line segments.
|
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.
 |
| 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
 |
| 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
|
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.
|
| 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.
 |
 |
return |
