A few problems on parametric equations and multivariables

[Daniel_Feldman]

Help is, as always, much appreciated.

1. I figured this one out.

2. Solved. Thanks.

3. r(t)=<3cos(t)+sin(t)+4t,2cos(t)-sin(t)-6t,2sin(t)-5t>, t in [pi,4pi]

I need to identify the type of curve that this is, and to find all of its fundamental geometric elements. I believe that it is a helix, which means that I need to find the radius and axis, but I'm not sure how to go about doing that.

4. Solved. Thanks.

 

[soroban]

Re: A few problems on parametric equations and multivariable

Hello, Daniel!

I think I've got #2 . . .

2. If the curve has the property that the position vector \(\displaystyle r(t)\)
is always perpendicular to the tangent vector \(\displaystyle r'(t)\),
show that the curve lies on a sphere with center at the origin.

Let the position vector be: \(\displaystyle \:r(t) \;=\;\langle\,x(t),\:y(t),\:z(t)\,\rangle\)

The tangent vector is: \(\displaystyle \:r'(t)\;=\;\left\langle\,\frac{dx}{dt},\:\frac{dy}{dt},\:\frac{dz}{dt}\,\right\rangle\)


Since \(\displaystyle r(t)\,\perp\,r'(t)\), we have: \(\displaystyle \:r(t)\,\cdot\,r'(t)\:=\:0\)

. . That is: \(\displaystyle \,x\left(\frac{dx}{dt}\right)\,+\,y\left(\frac{dy}{dt}\right)\,+\,z\left(\frac{dz}{dt}\right)\:=\:0\)


Multiply by \(\displaystyle dt:\;\;x\,dx\,+\,y\,dy\,+\,z\,dz\;=\;0\,dt\)

Integrate: \(\displaystyle \:\int x\,dx \:+\:\int y\,dy \:=\:\int dz \;=\;\int 0\,dt\)

. . and we have: \(\displaystyle \:\frac{1}{2}x^2\,+\,\frac{1}{2}y^2\,+\,\frac{1}{2}z^2\;=\;c\)

Multiply by \(\displaystyle 2:\;\:x^2\,+\,y^2\,+\,z^2\;=\;C\)

. . This is a sphere with its center at the origin.

 

[pka]

There is really a neat way to do #2.
Recall \(\displaystyle \frac{{d\left( {\left\| {R(t)} \right\|} \right)}}{{dt}} = \frac{{R(t) \cdot R'(t)}}{{\left\| {R(t)} \right\|}}.\)
Define the distance from the origin to a point \(\displaystyle R(t)\) on the curve as \(\displaystyle D(t)={\left\| {R(t)} \right\|}.\)
But that gives us \(\displaystyle D'(t) = \frac{{R(t) \cdot R'(t)}}{{\left\| {R(t)} \right\|}} = 0.\)
Thus because the distance from the origin to points on \(\displaystyle R(t)\) is constant then the curve is on the surface of a sphere with center at the origin.\

 

[Daniel_Feldman]

Ok, thanks. It actually makes sense, both algebraically and theoretically.

 

[soroban]

Re: A few problems on parametric equations and multivariable

Hello again, Daniel!

Given two curves: \(\displaystyle \:\begin{array}{cc}16x^2\,+\,9y^2\,+\.z^2\:=\:84 \\ 16x^2\,+\,9y^2\,-\,8z\:=\:0\end{array}\)

Find parametric equations for the surfaces,
then a parametrization for the curve which is the intersection of the two surfaces.


So far I have this:

First surface:
This is an ellipsoid, so the parametric equations should be in spherical coordinates.

Since the surface can be rewritten as: \(\displaystyle \L\:\frac{x^2}{\frac{1}{16}}\,+\,\frac{y^2}{\frac{1}{9}}\,+\,z^2\:=\:84\)

so the parametric equations are: \(\displaystyle \:\begin{array}{ccc}x\:=\:&\frac{1}{4}\cos\theta\sin\phi \\
y\:=\:&\frac{1}{3}\sin\theta\sin\phi \\
z\:=\:&\cos\phi\end{array}\;\;\) . . . You didn't include the "84"

The first surface is: \(\displaystyle \:16x^2\,+\,9y^2\,+\,x^2\:=\:84\)

Divide by 84: \(\displaystyle \L\:\frac{x^2}{\frac{21}{4}} \,+\,\frac{y^2}{\frac{84}{9}}\,+\,\frac{z^2}{84}\;=\;1\)

Then:\(\displaystyle \:\begin{Bmatrix}x\:=\:&\frac{21}{2}\cos\theta\sin\phi \\ y\:=\:&\frac{2\sqrt{21}}{3}\sin\theta\sin\phi \\ z\:=\:&2\sqrt{21}\cos\phi \end{Bmatrix}\)



We have: \(\displaystyle \:\begin{array}{cc}16x^2\,+\,9y^2\,+\.z^2&\:=\:84 \\ 16x^2\,+\,9y^2\,-\,8z&\:=\;0\end{array}\)

Subtract: \(\displaystyle \:z^2\,+8z\:=\:84\;\;\Rightarrow\;\;z^2\,+\,8z\,-\,84\:=\:0\)

. . which factors: \(\displaystyle \z\,-\,6)(z\,+\,14)\:=\:0\)

. . and has roots: \(\displaystyle \:z\:=\;6,\,-14\)


If \(\displaystyle z\,=\,6\), we have: \(\displaystyle \:16x^2\,+\,9y^2\,+\,36\:=\:84\;\;\Rightarrow\;\;16x^2\,+\,9y^2\:=\:48\)

Divide by 48: \(\displaystyle \L\:\frac{x^2}{3}\,+\,\frac{y^2}{\frac{48}{9}}\:=\:1\)
. . We have an ellipse: \(\displaystyle \:\begin{Bmatrix}x\:=\:&\sqrt{3}\cos\theta \\ y\:=\:&\frac{4\sqrt{3}}{3}\sin\theta \\ z\:=\:&6\end{Bmatrix}\)


If \(\displaystyle z\,=\,-14\), we have: \(\displaystyle \:16x^2\,+\,9y^2\,+\,196\:=\:84\;\;\Rightarrow\;\;16x^2\,+\,9y^2\:=\:-112\)
. . . . . not a "real" ellipse.

 

[Daniel_Feldman]

Thanks soroban. So that is the curve of intersection for the two surfaces. So was my parametrization for the second surface correct?