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11.1 You will need to be able to evaluate a vector valued function, find its domain, compute the limit of a vector valued function (p793: 57-69-74), and identify any discontinuities (p793: 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. Can you identify the graphs of the vector valued functions above? The spiral on the left is in 2-D and the spiral on the right is 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. Somebody online found the butterfly. The vector valued function graphed below on the left is
Try this ExploreScience activity in which you can adjust the frequency, amplitude, and phase shift of the x- and y- inputs to an oscilloscope to create various paths including a Lissajous figure 8. Here is a Lissajous applet whose default settings yield a figure 8.
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].
Check out this projectile motion / center of mass applet where the object in motion is dumbbell shaped (a massless rod with a mass at each end) and twirling around its center of mass.
Powerpoint Introduction to Sections 11.4 and 11.5 Maple Worksheet to accompany powerpoint presentation 11.4 Be able to find a set of parametric equations for the line tangent to a space curve at a given point (p817: 5-10). 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 (p818: 25-30, 37-40).
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, p825, and be able to give the equation of the circle of curvature as well). Find the curvature of a space curve (p829: 32-36). Find the force of friction necessary to keep a car on a given path (see example 8, p827). 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.
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
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