CALCULUS III EXAM II NOTES

CALCULUS III EXAM II NOTES AND LINKS

Examples For Sections 12.1 and 12.2 Examples For Section 12.3 Examples For 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 free download.

Seventh Edition Notes and Links

Allegany College of Maryland has a Calculus III course online. Unit 2 is on vector valued functions. Click on Unit 2 at the top of the page when you get there.

Here are some terrific class notes for Calculus III from Paul Dawkins at Lamar University.

Online Calculus III from the Springfield Technical Community College distance learning project is another place you might find help.

Here is a 3-D parametric grapher found by Denis Untilov.

Traces Animes A really nice set of 2D and 3D WIMS graphing tools by XIAO Gang which will produce pictures and graphs that can then be saved and printed or used on a web site.

Hotmath You can look at solutions to problems in exercise sets from a wide variety of mathematics textbooks including Calculus by Larson, Hostetler, Edwards, sixth , seventh, and eighth editions. They have chapters 1 - 12 available. Only a few solutions are still free (solutions to problems 15, 25, 35 in each section are free).

Examples

Lake Tahoe CC Examples

DPGraph Pictures: Two Spheres, Two Spheres Plus, and Two Balls

Here is a Maple Worksheet with differentiation and integration examples.

Here is a Maple Worksheet illustrating graphing space curves.

Here is a Maple Worksheet on L'Hopital's Rule and limits

12.1 You will need to be able to evaluate a vector valued function, find its domain, compute the limit of a vector valued function (pp837-9: 1-12, 69-74), and identify any discontinuities (p839: 75-80). The following introduction to Parametric Vector Equations might be helpful. Here is a Parametric Plotter (parametric grapher that draws tangent vectors at the same time) for curves in the plane. Harvey Mudd College has a tutorial on parametric equations. Here is one more parametric grapher.

Fast Large Animation

Slow Large Animation QT

Can you identify the graphs of the vector valued functions above? The spiral in the center is in 2-D and the spirals on the left and right are the same and in 3-D. Click on each graph to see an animation.

Below on the left is a simple variation of the butterfly curve developed by Temple H. Fay. Click on the picture to see a picture with an animated point tracing around the curve. Quicktime Version Click here for three animated point butterflies with a black background. Below on the right is a butterfly constructed from four variations of the Fay butterfly curve. The vector valued function graphed below on the left is

Here is a bit on the French physicist Jules Antoine Lissajous and why ABC chose a Lissajous figure for its logo. This Lissajous applet (click Applet Menu upper left then Oscillations>Lissajous Figures) has default settings that yield a figure 8. Here is my own Quicktime animation of a Lissajous figure 8 path described by the position function r(t) = < sin(2t) , cos(t) > with the velocity vector given in green, acceleration vector given in blue, and the principal unit normal vector given in red. Here is another of my Lissajous figure demonstrations including arrows and the position function.

12.2 You will need to be able to differentiate a vector valued function and compute indefinite and definite integrals (pp846-7: 11-26, 49-62) of a vector valued function. You will need to be able to determine open intervals where a vector valued function is smooth (see the definition at the top of page 842 in your text and Example 3 on page 842). At the right is a picture of the epicycloid described in example 3 on page 842. Click on the picture to animate the point.

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. Quicktime version

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

For a variation on the animation above 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].

12.3 Here is a nice little applet relating to the instantaneous velocity of a particle moving around a circle. You will need to be able to find the velocity and position (vector valued) functions given the acceleration function, initial position, and initial velocity (p854: 19-22). Be able to apply theorem 12.3 (Position Function for a Projectile) to solve application problems like pp854-5: 25-38. Here is a projectile motion applet. Here is my own animation with an initial velocity of 100 ft/sec, initial height of 20 ft, and launch angle of 45^o. The figure on the right shows paths with launch angles of 30^o (red), 45^o (green), and 60^o, (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.
Click here to see a path animated with its position function vectors, velocity function vectors, and acceleration function vectors. Here is a second, similar demonstration. Here is a third such demonstration involving vectors in space. This fourth demonstrates a spiral in space. Here is one demonstrating a Lissajous figure and a 3-D demo combining Lissajous and spiral.

Check out this projectile motion / center of mass applet (click Applet Menu upper left then Dynamics>Center of Mass) where the object in motion is dumbbell shaped (a massless rod with a mass at each end) and twirling around its center of mass.

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 45^o 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 30^o to 45^o. Notice that without air resistance the range of the projectile continues to increase all the way up to a launch angle of 45^o but with air resistance the range of the projectile reaches a maximum before the launch angle reaches 45^o 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 45^owith 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 30^o to 45^o. You will see the range of the path shown in red creep past that in blue before the launch angle value reaches 45^o.

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 , e^t , ln(t + 1) >    [0.2]     Click here to see it spin as well.
r(t) = < t² , 2sin(t) + cos(t) >     [-2pi,2pi] QT
r(t) = < t³ , 2sin(t) + cos(t) >     [-2pi,2pi] QT
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. QT QTspin
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).
.
The links below animate the velocity vector (green), acceleration vector (blue), and principal unit normal vector (red), along the sine wave given by
r(t) = < t , sin(t) >.
sinewave    sinewaveZoom
Position/Velocity Animation QT Version

Powerpoint Introduction to Sections 12.4 and 12.5

12.4 Be able to find a set of parametric equations for the line tangent to a space curve at a given point (p863: 11-16). Be able to find T, N, and the components of acceleration in the direction of T and N for a given vector valued position function (p864: 35-44, 53-56).

12.5 Given an object's mass and position function, be able to find the force along N necessary to keep the object on its path (for example, the roller coaster problem I will do in class--see below). Here is a terrific interactive visual relating to the roller coaster problem. It's fun to play although it does not do the math for you. Find the length of a given curve over a parameter interval (p875: 9-14). An example of this would be to 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).

Find the curvature, radius of curvature, and equation of the circle of curvature of the graph of a given function at an indicated point (see example 6, p872, and be able to give the equation of the circle of curvature as well). Find the curvature of a space curve (p876: 31-40). Find the force of friction necessary to keep a car on a given path (see example 8, p874). In Class I will find the curvature, center of curvature, and equation of the circle of curvature at the point (0,0) on the graph of the parabola whose equation is y = x² (see the figure on the left below). Solution 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 other two 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.

For yet another couple of bonus points, 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. Quicktime animation

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. Solution

Here is a fun little dot moving along a looping roller coaster and here is a dot moving along another roller coaster with loops. Below is the equation for the second looping roller coaster. Here is the second roller coaster spinning.

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

BONUS PROBLEMS

A child standing 20 feet from the base of a silo attempts to throw a ball into an opening 40 feet above the level of the point of release. This refers to (A) and (B) below.

(A) Find the minimum initial speed and the corresponding angle at which the ball must be thrown to go into the opening.

(B) Find the initial speed and corresponding angle at which the ball must be thrown to go into the opening at the instant when it has reached its maximum height.

(C) A 1000 pound object is moving along a parabolic path at 30 mph. The path is modeled in a Cartesian coordinate system and goes through the points (-150,-30), (0,0), and (150,-30) with distance measured in feet. Find the force necessary to keep the object on the described path at (0,0). Use g=32 in converting to mass and use feet and seconds.

SUPER EC Derive the position function for a projectile if, rather than neglecting air resistance, we represent air resistance (R) as proportional to velocity (v). That is take R = -cv, some constant times velocity. What this leads to, rather then starting out with r"(t) = <0,-g> where g is a positive constant (related to gravity) and y-positive is up, is a somewhat more complicated force equation. For a freely falling body (or more safely perhaps a parachutist), if m stands for mass, a for acceleration, W for weight, then since net force F = ma equating forces yields ma = -W + R, taking the weight force to be in the negative direction and the air resistance force to be in the positive direction. This could be written as ma = -mg - cv. If k = c/m we get a = -g - kv. For a position function r(t) = <x(t),y(t)> this translates into the following:

r"(t) = <-kx'(t),-g-ky'(t)>

Starting with r"(t) above and using the same initial conditions that I used in class in deriving the position function shown in Theorem 11.3, derive the position function, r(t).

A slightly more accurate model for projectile motion (particularly when the drag is not too large, for example free fall compared to using a parachute or a ball traveling through the air compared to traveling through shampoo) is found when we take air resistance (or in general the drag factor) as proportional to the square of velocity. In this case the resulting differential equations cannot be solved analytically but accurate approximate solutions can be found. See flight of a baseball with R = kv^2. A wonderful animation illustrating the path of a batted ball for different speeds of the pitch, bat speeds, and launch angles can be found at Science of Baseball. There is lots of fun here for a baseball fan. Click on Scientific Slugger when you get to the site to see the batted ball demonstration. You can set the pitch speed, bat speed, and launch angle yourself and compare results from various settings. The animation requires Shockwave which can be downloaded for free from this site. All of these wonderful site finds were provided by Terri Schein. Another nice animation involving the path of a golf ball with air resistance taken as proportional to velocity can be found at Golf Range. You might also enjoy Freefall Lab-Terminal Velocity.

ONE MORE EC PROBLEM

Prove that (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. See Shoot the Monkey for an illustration of this. You may also 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 (or monkey) 150 feet, and initial velocities (actually speeds) of 88, 100, 112, 125, 150, 175, and 250 ft/sec. Quicktime animation

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Lane Vosbury, Mathematics, Seminole State College email: vosburyl@seminolestate.edu