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VECTOR AND MULTIVARIABLE CALCULUS
ANIMATIONS
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The figure on the left shows the level
curves for f(x,y) = 64 - x2 - 2y2. The level
curves correspond to c = 0 (red),
c = 8 (green), c
= 16, 24, 32, 40, 48, 56, c = 63 (yellow),
and c = 63.9 (black).
Click on the figure to see an animation. In the animation you will
see a level curve in red changing as you move "up" the surface,
i.e., as z = c increases. You will also see an animated point and
vector in black moving with the animated red level curve. The length
of this black vector gives the relative speed at which the black point on
the animated red level curve is moving along the x-axis. One can
observe that the speed of the black point is increasing as c
increases. One can also observe that the distance between the level
curves is increasing as c increases when the change in c is constant (as
it is from red to orange). This indicates that the surface is
becoming less "steep". In fact at the origin we are
directly under a relative maximum point where the tangent plane would be
horizontal. Click here to see
an animation with c going from 32 to 64 with a step size of 4 and click
here to see an animation with c going from 48 to 64 with a step size
of 2. |
Here
is a graph with animated (moving) points showing the four position functions
demonstrated in class using the TI graphing calculator viewscreen. The
functions are given below with t going from 0 to 1. Each point has a
different color tangent vector (not velocity vector) attached to
it. See if you can match each color to the functions given below. xt1
= -2 + 4t, yt1 = (-2 + 4t)2 xt2
= -2 + 4(sin(pi*t/2)) yt2 = (-2 + 4(sin(pi*t/2)))2 xt3
= -2 + 4(tan(pi*t/2)) yt3 = (-2 + 4(tan(pi*t/2)))2 xt4
= (-2 + 4t)3 yt4 = (-2 + 4t)6
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For
a variation on the animation at the left click here to
see the graphs of r1(t) = < t , t2
> and r2(t) = < sin(pi*t/2) , (sin(pi*t/2))2
> over the t interval [-1,1]. r1(t) will include a
velocity vector in green
and r2(t) will include a velocity vector in red.
Click here to see the same animation over a t
interval of [-2,2]. |
| Here is my own animation
of projectile motion neglecting air resistance with an initial velocity of 100 ft/sec, initial height of 20 ft, and launch
angle of 45o. The figure on the right shows paths with launch angles of 30o
(red),
45o (green),
and 60o, (blue),
each with initial velocity of 100 ft/sec and initial height of zero ft.
Click on the picture to see animated points move along each path
simultaneously. |
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| Click
on each vector valued function below to see its graph and an animation showing
the changing velocity vectors (green)
and acceleration vectors (red).
The numbers inside the brackets indicate the t interval for the graph. r(t)
= < t , t2 >
[-2,2] r(t)
= < t , t3/3
- t > [-2,2] r(t)
= < 3cos(t) , 2sin(t) > [0,2pi] r(t)
= < t3 , 2sin(t) > [-2pi,2pi] r(t)
= < tsin(t) , tcos(t) >
[0,6pi] r(t)
= < 3sin(t) , 3cos(t) , t > [0,2pi]
Click here to see it spin as well. r(t)
= < t2 , et , ln(t + 1) > [0.2]
Click here to see it spin as well. r(t)
= < t2 , 2sin(t) +
cos(t) > [-2pi,2pi] r(t)
= < t3 , 2sin(t) + cos(t)
> [-2pi,2pi] r(t)
= < t3 , 2sin(t) + cos(t)
> [-4pi,4pi] r(t)
= < t , sin(t) , t2 > [0,2pi]
(includes N in magenta) Click
here to see it spin as well. |
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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| Neglecting
air resistance, if we shoot a ball at another ball falling straight down from
above and in front of us, we will hit the falling ball if our aim is such that
we are pointing directly at the falling ball at the moment it is dropped and we
shoot immediately. Click
here or the small icon on the right to see my animation with the shooter 200 feet downrange, the initial
height of the ball 150 feet, and initial velocities (actually
speeds) of 88, 100, 112, 125, 150, 175, and 250 ft/sec. |
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| The picture on the right shows part
of the graph of f(x,y) = 9 - x2 - y2 and the plane
tangent to the surface at (1,1,7). Be able to determine the equation
of this tangent plane. Click on the picture to see an animation. |
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| Here is an animation to introduce the concept
of the Riemann Sum as applied to double integrals. Click on the
picture on the right to see an animation. |
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| Set up an iterated triple integral to model the volume of the region of
space bounded by the graphs of z=x2+2y2, z=2, and z=8. Approximate this
volume. See the graph on the right. Click on the picture to see an
animation. Click here to see an
animation of y2 + z2 = 1/(s*x2+1) with s
varying from 0 to 1. State the name of the surface that results when s=0
and determine (show why) whether or not s=1 yields the graph of a unit sphere
centered at the origin. |
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| The picture on the left
below gives the
circle of curvature at the point (0,0) on the graph of the parabola whose
equation is y = x2. The other two pictures show the
circle of curvature at the point (1,1) on the graph of the same parabola. 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. |
| Find the function of t that would
give the curvature along the sine wave described by the position function r(t)
= <t , sin(t)>. Click here to see an animation over the
interval [0,2pi] and click here to see an animation over the interval [0,4pi].
Click here to see a
slower animation (more circles) over [0,2pi] that also includes the
changing radius of curvature. This animation may take a while to load. |
Roller Coaster
Problem The position function r(t)
= < 10sin(2t) , 10cos(2t) , 3t >, t going from 0 to 4pi, describes the motion
of a roller coaster car along a spiral track at an amusement park. 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. |
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In the problem pictured below, determine the values for x and y
that will minimize total construction costs. The idea is to lay pipe from
point P to Point Q. It costs 3 million dollars per mile to lay the pipe
through the blue area, 2 million dollars per mile to lay the pipe through the
green area, and 1 million dollars per mile to lay the pipe along the boundary
between the green area and the brown area. Consider the colored regions to
be rectangles, x to represent the horizontal distance for the pipe in the blue
region and y to be the horizontal distance for the pipe in the green
region. The blue region and the green region are each 1 mile wide and the
horizontal distance from P to Q is 5 miles. You must thoroughly
investigate the costs on the boundary of the region over which you would be
applying the cost function. Click on the picture at the right to see an
animation of some of the possible paths.
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| Click
here to go to two multivariable calculus optimization examples with
graphs. |
| At time t = 0 one plane is directly
above a second plane at an altitude of 1 mile. The second plane is
at an altitude of 1/2 mile. The first plane is flying along a
somewhat elliptical path described by r(t) = < 4 - 4cos(t) ,
3sin(t) , 1 + t/(2pi) > and the second plane is flying along a
hyperbolic path described by r(t) = < (8/31/2)tan(t/3)
, 0 , -1/2 + sec(t/3) >. These do represent the actual position
functions for the airplanes with distance in miles and time in
minutes. In how many minutes after t = 0 will the planes
collide? Approximate the distance between the planes and the rate of
change of the distance between the planes 30 seconds before they
collide. Approximate the rate of change of the distance between the
planes at the instant just before they collide.
Animation
with scales
Animation
without scales
Animation
without scales and with rotation |
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Lawn Sprinkler
Here is an example of the lawn sprinkler problem
found in the exercises for Section 3.1. In this example 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
Click here
to see an animation for this problem and click
here for an animation with scales. Can you see the answer to the
questions posed in the text and can you support your answer analytically?
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.
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| Here are three little animations
depicting projectile motion with air resistance taken as proportional to
velocity, i.e., R = kv. In each animation the initial height is 0
and the initial velocity is 100 ft/sec. In the first
animation the launch angle is 45o and you will see the path
of the projectile in blue with air resistance taken to be 0 and the
changing path of the projectile in red as air resistance varies in value
from 0.02 to 0.27. In the second
animation you will see the changing path of the projectile in blue
with air resistance taken to be 0 and the changing path of the projectile
in red as air resistance is taken to be 0.16. The path changes will
be due to the launch angle varying from 30o to 45o.
Notice that without air resistance the range of the projectile continues
to increase all the way up to a launch angle of 45o but with
air resistance the range of the projectile reaches a maximum before the
launch angle reaches 45o and then decreases. To see this
more clearly look at the third animation
in which the fixed path in blue corresponds to a launch angle of 45o with
air resistance set at 0.16 and initial velocity 100 ft/sec. and the
variable path in red corresponds to the same initial velocity and air
resistance but with the launch angle varying from 30o to 45o.
You will see the range of the path shown in red creep past that in blue
before the launch angle value reaches 45o. |
| 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. |
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A projectile is launched 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
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.
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| Section 11.3 #36
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