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.

OBJECTIVE: Learn to use Maple to perform calculations to analyze motion in the equations that are given parameterically. Moving in three dimensions is something with which we are familiar, but visualizing equations of motion and computations involved in analyzing that motion can be cumbersome. This module gives a broad overview on the analysis of motion in the equations that are given parametrically.

NOTE: If you have just finished a worksheet, restart Maple before executing a new worksheet. TO OPEN SECTIONS, Click on the Arrow sign at the left hand side of the screen or select Expand All Sections from the View drop down menu. TO STOP AN EXECUTION Click on STOP button from the toolbar. ORDER OF EXECUTION Execute commands in the order given. Do not skip any Maple Input lines within a given worksheet Alternatively, you can execute the entire worksheet by selecting the Execute Worksheet command from the Edit drop down menu. SAVING WORKSHEETS. You can save anytime to any directory you choose, and it is wise to save often. EXPERIENCING MAJOR PROBLEMS Save if appropriate, and then shut down Maple and start it up again.

First, we load the plots and plottools packages and the define x, t, y, and z coordinates for motion parametrically. restart: with(plots): with(plottools): x:=t-> cos(t):x(t); y:=t-> sin(t):y(t); z:=t-> 4 - t^2/25; Next, we plot the resulting curve in blue. Can you tell in which direction you are moving on the curve as t increases? Once the graph is plotted, click and drag the plot to view it from different angles. You can do this with any Maple 3D plot. p:=spacecurve([x(t),y(t),z(t),t=0..10],axes=boxed,labels=["x","y","z"],orientation=[-15,65],color=blue): display(p); If you consider the parametric equation as a vector equation for the motion of a particle, the derivative of that vector is the velocity vector that we form by differentiating each component. The commands below plot both the first and second derivative of the vector position function. p1:=spacecurve([diff(x(t),t),diff(y(t),t),diff(z(t),t),t=0..10],axes=boxed,labels=["x","y","z"],orientation=[-15,65],color=COLOR(RGB,1,0,0)): p2:=spacecurve([diff(x(t),t$2),diff(y(t),t$2),diff(z(t),t$2),t=0..10],axes=boxed,labels=["x","y","z"],orientation=[-15,65],color=COLOR(RGB,0,1,1)): print(cat(`The velocity vector is `,[diff(x(t),t),diff(y(t),t),diff(z(t),t)])); display(p1); print(cat(`The acceleration vector is `,[diff(x(t),t$2),diff(y(t),t$2),diff(z(t),t$2)])); display(p2); display(p,p1,p2);

Define your own x, y, and z coordinates for motion by changing the entries for x, y, z. You may or may not want to change the time interval (timeinterval) over which you plot your functions. x:=t-> 10* exp(cos(t)): y:=t-> 10* sin(t^2/15): z:=t-> t^2/2: timeinterval:= 5: p:=spacecurve([x(t),y(t),z(t)],t=0..10,axes=boxed,color=COLOR(RGB,0,0,1),orientation=[-50,80], labels=["x","y","z"]):display(p); Now check out the velocity and acceleration p1:=spacecurve([diff(x(t),t),diff(y(t),t),diff(z(t),t),t=0..10],axes=boxed,labels=["x","y","z"],orientation=[-15,65],color=COLOR(RGB,1,0,0)): p2:=spacecurve([diff(x(t),t$2),diff(y(t),t$2),diff(z(t),t$2),t=0..10],axes=boxed,labels=["x","y","z"],orientation=[-15,65],color=COLOR(RGB,0,1,1)): print(cat(`The velocity vector is `,[diff(x(t),t),diff(y(t),t),diff(z(t),t)])); display(p1); print(cat(`The acceleration vector is `,[diff(x(t),t$2),diff(y(t),t$2),diff(z(t),t$2)])); display(p2); display(p,p1,p2); The plot of all three together may or may not be instructive. Compare your results to those of your classmates.

If you are given the velocity, you can differentiate to find the acceleration and integrate to find the displacement. Since we do three integrations, we need three arbitrary constants to get the general solution. unassign('t,a,b,c'); velocity:= t-> [3*sqrt(t+1)/2,exp(-t),1/(t+1)]: acceleration:= t-> diff(velocity(t),t): ([seq(int(velocity(t)[i],t),i=1..nops(velocity(t)))] ): % + [a,b,c]: rgeneral:= %: print(cat(`The velocity vector is defined to be `,velocity(t))); print(cat(`The acceleration vector is `,acceleration(t))); print(cat(`The position vector is `,rgeneral)); s:=eval(rgeneral, t=0); initial:=[1,1,0]: eqns:={seq(s[i]=initial[i],i=1..3)}; assign(solve(eqns,{a,b,c})); r:=rgeneral;

Change the entries in your velocity vector and your initial position (velocity and initial), and re-execute the section. unassign('a,b,c,t'); velocity:=[sqrt(t+1)*3/2,exp(-t),1/(t+1)]: initial:=[1,1,0]: acceleration:=diff(velocity,t): [seq(int(velocity[i],t),i=1..nops(velocity))]: %+[a,b,c]: rgeneral:=%: s:=subs(t=0,rgeneral): eqns:={seq(subs(s[i]=initial[i]),i=1..3)}: assign(solve(eqns,{a,b,c})): r:=rgeneral;

First we define the x, y, and z coordinates for motion parametrically. unassign('r,t,v'); r:=[3*cos(2*t),t^(10/3)/50,t]: v:=diff(r,t): speed:=(sqrt(linalg[dotprod](v,v))): p:=spacecurve(r(t),t=0..10,axes=boxed,labels=["x","y","z"],orientation=[-15,65],color=blue,orientation=[-30,33],scaling=constrained): distance:=evalf(int(speed,t=0..10)): print(`The position given in feet is `,r); display(p); print(`The velocity vector given in feet per minute is `,v); print(`The speed `,ds/dt,` given in feet per minute is `,speed); print(`The distance traveled in feet in 10 minutes is `,distance); In the above example, if you had tried to integrate symbolically, instead of numerically, what would have happened? (The f after eval signifies numerical integration.) Many integrals arising in the computation of arc length are too complicated to evaluate symbolically. int(speed,t=0..10); print(`Speed at 5 minutes is `,evalf(subs(t=5,speed),6),` feet per minute`); plot(speed,t=0..10,labels=[t,"speed"]); If t represents time in minutes, what does this say about the speed of the hiker? Is it possible?

Choose your own position function (r) and the interval over which you wish to integrate. unassign('r,t,v,speed,distance'); r:=[2*t^(.3),cos(t-1),exp(-t*t)]: v:=diff(r,t): speed:=simplify((sqrt(linalg[dotprod](v,v)))): p:=spacecurve(r(t),t=0..10,axes=boxed,labels=["x","y","z"],orientation=[-15,65],color=blue,orientation=[-30,33],scaling=constrained): distance:=int(speed,t=0..10): print(`The position given in feet is `,r); display(p); print(`The velocity vector given in feet per minute is `,v); print(`The speed `,ds/dt,` given in feet per minute is `,speed); print(`The distance traveled in feet in 10 minutes is `,distance); You will notice that Maple was unable to evaluate the integral for the distance function symbolically. In Version 6, it also fails to evaluate the integral numerically. This could be a glitch that will be fixed soon. You can check it out by executing the following command, but you will probably have to hit the STOP button. distance:=evalf(int(speed,t=0..10));

Maple simplifies the process of finding curvature and torsion. The computations below use formulas directly from your text, and we graphically explore the interpretations of the curvature and torsion functions. with(linalg): unassign('r,v,t,speed'); mag:=proc(v) RETURN(sqrt(linalg[dotprod](v,v))): end : r:=[10*sin(t),10*exp(-t),t^(10/3)/100]: v:=diff(r,t): a:=diff(v,t): speed:=mag(v): utan:=[v[1]/speed,v[2]/speed,v[3]/speed]: curvature:=simplify(mag(linalg[crossprod](v,a))/speed^3): torsion:=simplify(det(matrix(3,3,[v,a,diff(a,t)]))/simplify(multiply(crossprod(v,a),crossprod(v,a)))): print(`The position vector is `, r): print(`The velocity vector is `, v): print(`The acceleration vector is `, a): print(`The speed is `, speed): print(`The unit tangent vector is `, utan): print(`The curvature is `, curvature): print(`The torsion is `, torsion): Let's visualize some of these quantities. We will begin by drawing the path of motion: p:=spacecurve(r,t=0..10,axes=boxed,orientation=[-70,80],labels=[x,y,z], scaling=constrained): display(p); print(`The speed is `,speed); ps:=plot(speed,t=0..10,labels=["t","speed"]): display(ps); print(`The curvature is `,curvature); pc:=plot(curvature,t=0..10,color=green, labels=["t","curvature"]): display(pc); print(`The torsion is `,torsion); pt:=plot(torsion,t=0..10,color=blue, labels=["t","torsion"]): display(pt); display(ps,pc,pt); print(`Speed in Red, curvature in Green, torsion in Blue`); Look at your space curve and see if you can identify the points on your space curve at which the curvature or torsion spike. What has caused these events? Are they related at all to the speed? Recall from the definition of curvature that relates it to the cross product of the velocity and acceleration that the curvature is smallest when the velocity and acceleration are in the same direction. The curvature is largest when the velocity and acceleration are perpendicular to one another. Looking at the torsion as the dot product of the derivative of the unit binormal and the normal vectors. Torsion will peak when those two vectors are in the same direction and be close to 0 when those two vectors are perpendicular. display(p); You can see the TNB frame for a curve in action in Part VI of this module.

Redefine your position function (r) and compute. It should be noted that the computations here are not trivial, so more complicated functions could lead to long waits. unassign('r,v,t,speed'); r:=[2*t^.3,cos(t-1),exp(-t*t)]: mag:=proc(v) RETURN(sqrt(dotprod(v,v))): end : v:=diff(r,t): a:=diff(v,t): speed:=mag(v): utan:=[v[1]/speed,v[2]/speed,v[3]/speed]: curvature:=simplify(mag(linalg[crossprod](v,a))/speed^3):torsion:=simplify(det(matrix(3,3,[v,a,diff(a,t)]))/simplify(multiply(crossprod(v,a),crossprod(v,a)))): print(`The position vector is `, r): print(`The velocity vector is `, v): print(`The acceleration vector is `, a): print(`The speed is `, speed): print(`The unit tangent vector is `, utan): print(`The curvature is `, curvature): print(`The torsion is `, torsion): p:=spacecurve(r,t=0..10,axes=boxed,orientation=[-70,80],labels=[x,y,z], scaling=constrained): display(p); print(`The speed is `,speed); ps:=plot(speed,t=0..10,labels=["t","speed"]): display(ps); print(`The curvature is `,curvature); pc:=plot(curvature,t=0..10,color=green, labels=["t","curvature"]): display(pc); print(`The torsion is `,torsion); pt:=plot(torsion,t=0..10,color=blue,labels=["t","torsion"]): display(pt); display(ps,pc,pt,view=[0..10,-3..3]); print(`Speed in Red, curvature in Green, torsion in Blue`);

The steps here resemble the ones above, except that we choose a simpler function and extend the formulas to the unit normal and binormal vectors. restart: with(linalg): with(plots): mag:=proc(v) RETURN(sqrt(multiply(v,v))): end : r:=[sin(2*t),cos(2*t),t^2/10]: v:=diff(r,t): a:=diff(v,t): speed:=simplify(mag(v)): utan:=simplify(expand(v/speed)): curvature:=simplify(mag(linalg[crossprod](v,a))/speed^3): torsion:=simplify(det(matrix(3,3,[v,a,diff(a,t)]))/simplify(multiply(crossprod(v,a),crossprod(v,a)))): top:=diff(utan,t): bottom:=simplify((mag(top))): un:=simplify(expand(simplify(top/bottom))): ubn:=convert(simplify(crossprod(utan,un)),list): print(`The position vector is `, r): print(`The velocity vector is `, v): print(`The acceleration vector is `, a): print(`The speed is `, speed): print(`The unit tangent vector is `, utan): print(`The curvature is `, curvature): print(`The torsion is `, torsion): print(`The unit normal is `,un): print(`The unit binormal is `,ubn): Let's find the tangential and normal components of the accelerations. at:=combine(dotprod(a,utan)): an:=radsimp(simplify(dotprod(a,un))): print(`The tangential component of acceleration is `,at); print(`The normal component of acceleration is `,an); You can get a visual perspective of the tangential and normal components of the acceleration in two dimensions by exploring the Java applet, "Tangent and Normal Vectors." If you compare these components found by the dot products to the formulas given in your book, you will see that they agree, once the absolute value sign is removed from the curvature formula. print(`tangential component of acceleration is`,diff(speed,t)); print(`normal component of acceleration is`,curvature*speed^2); The equality of the tangential components can be verified easily if you notice that the first two terms of the result found previously by dotting the acceleration vector into the unit tangent vector subtract out with one another. The equality of the normal components is not as obvious. You can see the advantage in using the formulas used in the latter part, due to the complexity of the unit normal vector. To see the results in more simplified form, check out the Mathematica version on your CD. The following set of commands will plot the unit tangent, unit normal, and unit binormal vectors as you move along the curve. After the plots are generated, click anywhere on the graph then right-click to bring up the toolbar. From the toolbar, select Animate and click on Play to view the animation. Remember, you can click and drag the plot to view it from different angles. with(plottools):temp:=1: p1:=spacecurve(r,t=0..2*Pi,view=[-1.5..1.5,-1.5..1.5,0..4],axes=boxed): for i from 0 to evalf(2*Pi) by evalf(Pi/4) do p1:=spacecurve(r,t=0..2*Pi,view=[-1.5..1.5,-1.5..1.5,0..4],axes=boxed): p2:=plottools[line](evalf(subs(t=i,r)),evalf(subs(t=i,r+utan)),color=red,thickness=3): p3:=plottools[line](evalf(subs(t=i,r)),evalf(subs(t=i,r+un)),color=green,thickness=3): p4:=plottools[line](evalf(subs(t=i,r)),subs(t=i,r+ubn),color=blue,thickness=3): p5:=display(line(evalf(subs(t=i,r+0.2*utan)),evalf(subs(t=i,r+0.2*utan+0.2*ubn))), line(evalf(subs(t=i,r+0.2*ubn)),evalf(subs(t=i,r+0.2*utan+0.2*ubn))), line(evalf(subs(t=i,r+0.2*utan)),evalf(subs(t=i,r+0.2*utan+0.2*un))), line(evalf(subs(t=i,r+0.2*un)),evalf(subs(t=i,r+0.2*utan+0.2*un))), line(evalf(subs(t=i,r+0.2*un)),evalf(subs(t=i,r+0.2*un+0.2*ubn))), line(evalf(subs(t=i,r+0.2*ubn)),evalf(subs(t=i,r+0.2*un+0.2*ubn))), line(evalf(subs(t=i,r+0.2*utan+0.2*ubn)),evalf(subs(t=i,r+0.2*utan+0.2*ubn+0.2*un))), line(evalf(subs(t=i,r+0.2*utan+0.2*un)),evalf(subs(t=i,r+0.2*utan+0.2*ubn+0.2*un))), line(evalf(subs(t=i,r+0.2*ubn+0.2*un)),evalf(subs(t=i,r+0.2*utan+0.2*ubn+0.2*un))),color=black): P[temp]:=display(p1,p2,p3,p4,p5): temp:=temp+1: od: P:=convert(P,list): display(P,insequence=true); print(`Tangent in RED, unit normal in GREEN, unit binormal in BLUE.`); After the plots are generated, click anywhere on the graph then right-click to bring up the toolbar. From the toolbar, select Animate and click on Play to view the animation.

The computational code is written so that all you have to do is change the initial vector ( r ) function and then re-execute the commands. We suggest that you choose some of the homework problems from your text for the TNB frame, since the computation for the unit normal and unit binormal can get you very bogged down, even in Maple. You may also wish to change the domain over which you extend your function in the second section. If you check out the following example, expect to wait a few minutes for the computations. restart: with(plots): with(linalg): mag:=proc(v) RETURN(sqrt(multiply(v,v))): end : r:=[2*t,cos(t-1),exp(-t)]: v:=diff(r,t): a:=diff(v,t): speed:=simplify(mag(v)): utan:=simplify(v/speed): curvature:=simplify(mag(linalg[crossprod](v,a))/speed^3): torsion:=simplify(det(matrix(3,3,[v,a,diff(a,t)]))/simplify(multiply(crossprod(v,a),crossprod(v,a)))): top:=diff(utan,t): bottom:=simplify((mag(top))): un:=simplify(expand(simplify(top/bottom))): ubn:=convert(simplify(crossprod(utan,un)),list): print(`The position vector is `, r): print(`The velocity vector is `, v): print(`The acceleration vector is `, a): print(`The speed is `, speed): print(`The unit tangent vector is `, utan): print(`The curvature is `, curvature): print(`The torsion is `, torsion): print(`The unit normal is `,un): print(`The unit binormal is `,ubn): The following set of commands will plot the unit tangent, unit normal, and unit binormal vectors as you move along the curve. After the plots are generated, click anywhere on the graph then right-click to bring up the toolbar. From the toolbar, select Animate and click on Play to view the animation Remember, you can click and drag the plot to view it from different angles. with(plottools):temp:=1: for i from 0 to 10 by 1 do p1:=spacecurve(r,t=0..10,view=[-1..21,-2..2,-1..2],axes=boxed,scaling=constrained): p2:=plottools[line](evalf(subs(t=i,r)),evalf(subs(t=i,r+utan)),color=red,thickness=3): p3:=plottools[line](evalf(subs(t=i,r)),evalf(subs(t=i,r+un)),color=green,thickness=3): p4:=plottools[line](evalf(subs(t=i,r)),subs(t=i,r+ubn),color=blue,thickness=3): p5:=display(line(evalf(subs(t=i,r+0.2*utan)),evalf(subs(t=i,r+0.2*utan+0.2*ubn))), line(evalf(subs(t=i,r+0.2*ubn)),evalf(subs(t=i,r+0.2*utan+0.2*ubn))), line(evalf(subs(t=i,r+0.2*utan)),evalf(subs(t=i,r+0.2*utan+0.2*un))), line(evalf(subs(t=i,r+0.2*un)),evalf(subs(t=i,r+0.2*utan+0.2*un))), line(evalf(subs(t=i,r+0.2*un)),evalf(subs(t=i,r+0.2*un+0.2*ubn))), line(evalf(subs(t=i,r+0.2*ubn)),evalf(subs(t=i,r+0.2*un+0.2*ubn))), line(evalf(subs(t=i,r+0.2*utan+0.2*ubn)),evalf(subs(t=i,r+0.2*utan+0.2*ubn+0.2*un))), line(evalf(subs(t=i,r+0.2*utan+0.2*un)),evalf(subs(t=i,r+0.2*utan+0.2*ubn+0.2*un))), line(evalf(subs(t=i,r+0.2*ubn+0.2*un)),evalf(subs(t=i,r+0.2*utan+0.2*ubn+0.2*un))),color=black): P[temp]:=display(p1,p2,p3,p4,p5): temp:=temp+1: od: P:=convert(P,list): display(P,insequence=true); print(`Tangent in RED, unit normal in GREEN, unit binormal in BLUE.`); After the plots are generated, click anywhere on the graph then right-click to bring up the toolbar. From the toolbar, select Animate and click on Play to view the animation