<Text-field style="_pstyle258" layout="_pstyle258">Parametric and Polar Equations with a Figure Skater</Text-field>Note: You may notice differences between this Maple worksheet and the equivalent Mathematica notebook. These differences were introduced to preserve the content of these modules and were necessary because of major functional differences between Maple and Mathematica.
<Text-field style="Heading 1" layout="Heading 1">Introduction</Text-field>OBJECTIVE: Represent curves and analyze motion in parametric and polar form.Parametric equations are very powerful, and the purpose of this module is to help you get used to the idea of expressing curves using parametric equations and analyzing motion in the plane using these parametric equations. Polar plots can also be expressed parametrically, and that can be translated easily into Cartesian coordinates.
<Text-field style="_cstyle256" layout="Heading 1"><Font size="18">Technology Guidelines</Font></Text-field>NOTE: If you have just finished a worksheet, restart Maple before executing a new worksheet.
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EXPERIENCING MAJOR PROBLEMS
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<Text-field style="Heading 1" layout="Heading 1">Part I: Parametric Equations of a Curve in Two-Dimentional Space</Text-field>
<Text-field style="Heading 1" layout="Heading 1">Defining the Function</Text-field>First, we define the x and y coordinates parametrically. Suppose that time, t, is the independent variable. Once x and y are defined, we can write the position vector r(t). You omay ignore the error messages that are displayed after these packages are loaded.restart:
with(plots):
with(plottools):
with(linalg):x:=t->cos(t)^3:
y:=t->1-exp(sin(t)):
r:=t-> [x(t),y(t)]:
print(`The position vector is ` ,[x(t),y(t)]);
Now we plot the resulting curve in blue. In what direction are you moving on the curve as t increases?
p1:=plot([x(t),y(t),t=0..2*Pi],labels=["x","y"], color=blue):
display(p1,title=cat(`The position vector is `,convert([x(t),y(t)],string)));
<Text-field style="Heading 1" layout="Heading 1">Taking the Velocity and Acceleration into Account</Text-field>
If you consider the parametric equation as a vector equation for the motion of a particle, the velocity vector is found by differentiating each component of the position vector. Similarly, the acceleration vector is found by differentiating the components of the position vector twice. We do this with Maple and plot the velocity in red and the acceleration in green.p2:=plot([diff(x(t),t),diff(y(t),t),t=0..2*Pi],color=red,labels=["x","y"],title=cat(`The velocity function is `,convert(diff(r(t),t),string))):
p3:=plot([diff(x(t),t$2),diff(y(t),t$2),t=0..2*Pi],color=green,labels=["x","y"],title=cat(`The acceleration function is `,convert(simplify(diff(r(t),t$2),trig),string))):
display(p2);display(p3);Note that the components of the velocity and acceleration functions are more complicated than the components of the postion function.Let's look at the speed function and see what it tells us. The following plot shows the speed in black, the x-coordinate in orange, and the y-coordinate in plum. Contrasting that to your parametric plot, identify the places where the speed function is 0.speed:=t->sqrt(multiply(diff(r(t),t),diff(r(t),t))):
print(`The speed is `,speed(t));
p4:=plot([speed(t),x(t),y(t)],t=0..2*Pi,color=[black,red,plum]):
display(p4);
print(`The speed is in black, the x-coordinate of the path of motion is in orange and the y-coordinate of the path of motion is in plum`);
display(p1);
Identify the places on your path where the speed is 0 and the speed is a maximum.
<Text-field style="Heading 1" layout="Heading 1">Computing the Distance Traveled on a Curved Path</Text-field>Suppose that you are walking along the path given above. The distance traveled can be found by integrating the speed function over a particular interval.display(p1, color=blue);
plot(speed(t),t=0..2*Pi,labels=[t,speed], color=black);
distance:=int(speed(t),t=0..2*Pi):
print(cat(`The distance traveled around the closed path is `,evalf(distance),` units`));
Think of this answer as either the distance around the curve or as the area under the speed function over the interval t from 0 to LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1JI21uR0YkNiRRIjJGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRictSSNtb0dGJDYtUTEmSW52aXNpYmxlVGltZXM7RidGNS8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGPi8lKXN0cmV0Y2h5R0Y+LyUqc3ltbWV0cmljR0Y+LyUobGFyZ2VvcEdGPi8lLm1vdmFibGVsaW1pdHNHRj4vJSdhY2NlbnRHRj4vJSdsc3BhY2VHUSYwLjBlbUYnLyUncnNwYWNlR0ZNLUYsNiVRJSZwaTtGJy8lJ2l0YWxpY0dGPkY1RjVGK0Y1.
<Text-field style="Heading 1" layout="Heading 1">You Try It: Part I</Text-field>
<Text-field style="Heading 1" layout="Heading 1">Defining a Function </Text-field>
Select your own functions for x(t) and for y(t) in the cell below, and then execute the command below. You need not select a closed path, and you may wish to change the bounds for the parameter to something other than [t = 0..10]. x:=t->sin(t/3.2):
y:=t->exp(-t)*t^2:
r:=t->[x(t),y(t)]:
[x(t),y(t)];
plotf:=plot([x(t),y(t),t=0..10],color=blue,labels=["x","y"]):
print(plotf);
<Text-field style="Heading 1" layout="Heading 1">Computing the Velocity and Acceleration Vectors and Analyzing the Speed</Text-field>print(cat(`The position function is `,r(t)));
print(cat(`The velocity function is `,collect(diff(r(t),t),exp(-t))));
print(cat(`The acceleration function is `,collect(diff(r(t),t$2),exp(-t))));
print(cat(`The speed is `,simplify(speed(t),exp)));
plot([speed(t),x(t),y(t)],t=0..10,color=[black,orange,plum],labels=[t,function]);
print(`The speed is in black; the x-coordinate and the y-coordinate of the path of motion are in orange and violet respectively.`);
print(plotf);
Identify the places on your path where the speed is 0 and the speed is a maximum.
<Text-field style="Heading 1" layout="Heading 1">Find the Distance Along the Curved Path</Text-field>
Adjust the values of the parameter in the integration if you did so earlier.print(plotf);
plot(speed(t), t=0..10, labels=["t", "speed"], color=black);
distance:=2.46223:
print(cat(`The distance traveled along the path is `,distance,` units`));As before, think of this answer as either the distance around the curve or as the area under the speed function.
<Text-field style="Heading 1" layout="Heading 1">Part II: A Figure Skater Tracing a Polar Plot</Text-field>
<Text-field style="Heading 1" layout="Heading 1">Four-Petal Pattern</Text-field>
Think of a figure skater who is tracing out a four-petal flower on the ice. The first set of commands gives the parametric equations of the figure skater in terms of a path that would be traced in the x-y plane. It is easiest to start with the equations in polar form.unassign(r,theta,t,x,y);
r:=t -> 16*sin(t)^2;
theta:= t -> t/2;p1:=polarplot(r(2*theta),theta=0..2*Pi,color=blue):
display(p1,labels=["East/West","North/South"]);parx:=t-> r(t) * cos(theta(t)):
pary:=t-> r(t) * sin(theta(t)):
position:= [parx(t),pary(t)]:
velocity:= diff(position,t):
speed:=t->simplify(sqrt(multiply(velocity,velocity))):
with(plottools):
arr1:=arrow([parx(1), pary(1)], [parx(1.1), pary(1.1)], .3, .6, 1, color=black):
arr2:=arrow([parx(2.5), pary(2.5)], [parx(2.6), pary(2.6)], .3, .6, 1, color=black):
arr3:=arrow([parx(4), pary(4)], [parx(4.1), pary(4.1)], .3, .6, 1, color=black):
arr4:=arrow([parx(5.5), pary(5.5)], [parx(5.6), pary(5.6)], .3, .6, 1, color=black):
arr5:=arrow([parx(7), pary(7)], [parx(7.1), pary(7.1)], .3, .6, 1, color=black):
arr6:=arrow([parx(8.5), pary(8.5)], [parx(8.6), pary(8.6)], .3, .6, 1, color=black):
arr7:=arrow([parx(10), pary(10)], [parx(10.1), pary(10.1)], .3, .6, 1, color=black):
arr8:=arrow([parx(11.5), pary(11.5)], [parx(11.6), pary(11.6)], .3, .6, 1, color=black):
print(cat(`Position vector = `,position));
print(cat(`speed =`, speed(t)));
print(display({p1,arr1, arr2, arr3, arr4, arr5, arr6, arr7, arr8}));
Can you tell in what direction the skater is moving right at the origin?
Let's look at the speed function and see what it tells us. The following plot shows the speed in black, the x-coordinate in orange, and the y-coordinate in violet. Contrasting that to your parametric plot, identify the places where the speed function is 0.
pp4:= plot([parx(t),pary(t),t=0..4*Pi]):tickst:=[seq(n,n=0..15)]:
ticksf:=[seq(5*n,n=-2..3)]:
plot([speed(t),parx(t),pary(t)],t=0..4*Pi,color=[COLOR(RGB,0,0,0),COLOR(RGB,1,.5,0),COLOR(RGB,1,0,1)],xtickmarks=tickst,ytickmarks=ticksf);
print(`The speed is in black; the x-coordinate and the y-coordinate of the path of motion are in orange and violet respectively.`);
print(display({p1,arr1, arr2, arr3, arr4, arr5, arr6, arr7, arr8}));
Where are the places on your four-petal plot that the speed is 0?
<Text-field style="Heading 1" layout="Heading 1">Velocity and Acceleration: When are They Perpendicular? </Text-field>
Suppose we wish to determine for which values of t certain vectors describing the equations of motion are orthogonal. To do this, we will use the dot product, since perpendicular vectors yield a dot product of 0. Here we will examine when the velocity and acceleration vectors are perpendicular to one another. We begin by computing the dot product of the two vectors and we plot the resulting function of t to get an idea of when the dot product might be 0.unassign('acceleration', 'vdota');
acceleration := simplify(diff(velocity,t)):
vdota := simplify(multiply(velocity,acceleration)):
print(cat(`velocity = `,velocity));
print(cat(`acceleration = `,acceleration));
print(cat(`velocity dotted into acceleration = `,vdota));
plot(vdota,t=0..4*Pi,labels=[t, "vdota"], color=black, tickmarks=[8,6]);As you can see from the graph, there are many times when the velocity and acceleration vectors are perpendicular to each other.
The following commands find some of the places where the velocity and acceleration vectors are perpendicular. Notice that we use seed values in solve that seem close to some of the places where our function crosses the horizontal axis.
sol1:=[seq(fsolve(vdota=0,t,(i-1)..(i)),i=1..12)];
Now we evaluate x and y at the values of t we have found.
sol2:=seq([parx(sol1[i]),pary(sol1[i])], i=1..nops(sol1));
matrix([[x,y],sol2]);
We can now see where those points are relative to our petals.
psol:=pointplot([sol2], color=red):
display({psol, p1});
Can you see some points that we missed?
The dots show where the velocity and acceleration vectors are perpendicular to each other. Can you describe what is happening to the figure skater at those points?
<Text-field style="Heading 1" layout="Heading 1">You Try It: Part II</Text-field>
Try your own functions for r(t) and LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEoJnRoZXRhO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGMg==(t). Remember to solve for t as a function of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEoJnRoZXRhO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRicvRjNRJ2l0YWxpY0Yn before you attempt to do a polar plot in the form of r as a function of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEoJnRoZXRhO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGMg==. Replace the terms in r(t) and LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEoJnRoZXRhO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGMg==(t). unassign('t','x','y');
r:=t->(cos(t))^3:
theta(t):=t^2/8:
pp1:=plot(cos(sqrt(8*theta))^3, theta =0..3, color=COLOR(RGB, 0, 0, 1), labels=[`East/West`, `North/South`], coords=polar, view=[-1.5..1.5, -1.5..1.5]):
print(pp1);parx:=t->r(t)*cos(theta(t)):
pary:=t->r(t)*sin(theta(t)):
position:=[parx(t), pary(t)]:
velocity:=diff(position, t):
speed:=sqrt(multiply(velocity, velocity)):
print(cat(`Position Vector`, position));
print(cat(`Speed`, speed));
plot({speed, parx(t), pary(t)}, t=0..5, color=[plum, orange, black], labels=["t" ,"function"]);
print(`The speed is in black, the x-coordinate of the path of motion is in orange and the y-coordinate of the path of motion is in violet.`);print(pp1);What is happening to your speed as t increases?