Vector Functions

Example 13.1.14

r[t_]={Exp[-t],2Cos[3t],2Sin[3t]}
v[t_]=r'[t]
a[t_]=v'[t]
v[0]
a[0]
speed=(Sqrt[v[0].v[0]])

Example 13.1.24

r[t_]={Sec[t]Tan[t],Tan[t],2Sin[t]Cos[t]}
Integrate[r[t],{t,0,Pi/3}]

Example 13.1.38 (alternate solution with visualization)

Solution

r[t_]={2,2,1}+Cos[t]*{(1/(Sqrt[2])),-(1/(Sqrt[2])),0}+Sin[t]*{(1/(Sqrt[3])),(1/(Sqrt[3])),(1/(Sqrt[3]))}

Show that for the vector between any two points that r(t) passes through, the normal to that vector is <1,1,-2>, which means the space curve is confined to a plane.

General vector between any two points:

r[t1]-r[t2]

The normal must satisfy:

{n1,n2,n3}.(r[t1]-r[t2])==0

Pick values of n1 and n2, and solve for n3:

Solve[%,n3]/.n1->1/.n2->1

Simplify[%]

Now, the equation of the plane the space curve moves in is found from (notice I am using r(t1) to get any point in the plane:

{1,1,-2}.({x,y,z}-r[t1])==0

Simplify[%]

So the particle moves in the plane x+y-2z=2.

To show it moves in a circle centered at (2,2,1), we just need to show the distance from any point along r(t) to (2,2,1) is constant (this will actually show it moves on a sphere).

(Sqrt[(((r[t][[1]]-2))^2)+(((r[t][[2]]-2))^2)+(((r[t][[3]]-1))^2)])

Simplify[%]

So, it moves on a sphere centered at (2,2,1), but we already showed it moves on a plane--so it moves on the intersection between a sphere and a plane, which is a circle in this case.

Visualization

r[t_]={2,2,1}+Cos[t]*{(1/(Sqrt[2])),-(1/(Sqrt[2])),0}+Sin[t]*{(1/(Sqrt[3])),(1/(Sqrt[3])),(1/(Sqrt[3]))}
curve=ParametricPlot3D[r[t],{t,0,2Pi},PlotStyle->{Red,Thick}]

plane=Plot3D[((x+y-2)/2),{x,0,5},{y,0,5}]

sphere=Show[Graphics3D[{Opacity[0.5],Sphere[{2,2,1},1]}]]

Show[curve,plane,sphere,Graphics3D[{Green,Sphere[{2,2,1},0.05]}],PlotRange->{{0,3},{0,3},{0,2}}]

Example 13.1.43

Visualizations

r[t_]={0,3Cos[t],2Sin[t]}
curve=ParametricPlot3D[r[t],{t,0,2Pi},PlotStyle->{Red,Thick}]

absv[t_]=(Sqrt[r'[t].r'[t]])
curve=Plot[absv[t],{t,0,2Pi},PlotStyle->{Red,Thick}]

absa[t_]=(Sqrt[r''[t].r''[t]])
curve=Plot[absa[t],{t,0,2Pi},PlotStyle->{Red,Thick}]

Example 13.1.54c

r[t_]={x[t],y[t],z[t]}
c={c1,c2,c3}

Integrate[c.r[t],{t,a,b}]
c.Integrate[r[t],{t,a,b}]

r[t_]={x[t],y[t],z[t]}
c={c1,c2,c3}

Integrate[Cross[c,r[t]],{t,a,b}]
Cross[c,Integrate[r[t],{t,a,b}]]

Example 13.1.61

Part a

r[t_]={Log[t^2+2],ArcTan[3t],(Sqrt[(t^2)+1])}
curve=ParametricPlot3D[r[t],{t,-3,5},PlotStyle->{Red,Thick},AxesLabel->{"x","y","z"}]

Part b

v[t_]=r'[t]

Partc

t0=3
v[t0]

Let's use the vector form of a line for the tangent line:

rtangent[t_]=r[t0]+s*v[t0]
tangent=ParametricPlot3D[rtangent[t],{s,-5,5},PlotStyle->{Blue,Thick},AxesLabel->{"x","y","z"}]

point=Graphics3D[{Green,Sphere[r[t0],0.1]}];
Show[curve,tangent,point]

Animation (not asked for, but easy to get)

To get an animation, all we have to do is make some things a function of the poitn (Subscript[t, 0]) that we are getting the tangent line at. Here's one way to do it:

Clear[t0,curve,point,tangent]
r[t_]:={Log[t^2+2],ArcTan[3t],(Sqrt[(t^2)+1])};
v[t_]:=r'[t];

rtangent[t_,t0_]:=r[t0]+s*v[t0];

curve=ParametricPlot3D[r[t],{t,-3,5},PlotStyle->{Red,Thick},AxesLabel->{"x","y","z"}];
point[t0_]:=Graphics3D[{Green,Sphere[r[t0],0.1]}];
tangent[t0_]:=ParametricPlot3D[rtangent[t,t0],{s,-5,5},PlotStyle->{Blue,Thick},AxesLabel->{"x","y","z"}];

Manipulate[Show[curve,tangent[t0],point[t0]],{t0,-3,5}]