Is There a Value of 'a' That Will Make This Vector Function Form a Circle?

[jp6373]

Homework Statement

2.2) Prove or disprove that there is a value of a for which the image of the following vector function is a circle:
r(t) = < 2cos t + a sin t, a cos t + sin t, 3 cos t - a sin t >.

any help will be appreciated

 

[Mark44]

Is that the exact wording of the problem? I.e., it says circle, and not sphere? My work shows that r(t) generates a sphere for certain values of a.

 

[jp6373]

yes sir, that is the exact wording. copy and pasted

 

[Mark44]

I'm stumped.

I was going to suggest that on a circle given parametrically, the tangent vector is always changing direction as t changes, but the magnitude of the tangent vector is constant. Look at |r'(t)| and see of there is a value of a so that |r'(t)| is a constant.

The problem is, this is also true for a sphere.

 

[jp6373]

ok, tell me if this makes sense:
x= 2 cost + a Sin t
y= a Cos t + Sin t
z = 3 Cos t - a Sin t

then do we have 2 vectors? cost < 2,a,3 > and Sin t < a, 1,-a>...not sure if this can be done. I've never used Cos and Sin as scalar...

 

[Mark44]

That's an interesting idea. Certainly you can break up r(t) into two vectors, and sin t and cos t are scalars for any particular values of t. So you have r(t) = u(t) + v(t) = cos t<2, a, 3> + sin t<a, 1, -a>.

 

[jp6373]

lol, now what,we learned about the derivitive and then the dot product...i don't see how the der. will work here at all, any ideas on the dot product?

 

[jp6373]

if i dot the vectors, what will i get? anthing of value?

 

[Mark44]

If you dot u(t) and v(t) you get zero, which says that they are orthogonal (perpendicular) for any value of a.
u(t) . v(t) = (sin t)(cos t)(2a + a - 3a) = 0.

That means that |r(t)|² = |u(t)|² + |v(t)|², by the Pythagorean Theorem.

I would like to give you more direction on this, but I don't see anything obvious to do.

 

[jp6373]

i think i have got it. when we dot the 2 vctors, they give a Zero answer. this means that they are perpendicular to each other. now, they must be the radius of the circle. so the length of the 2 vectors are equal.
so we end up with a= +_ root 12.

 

[Mark44]

i think i have got it. when we dot the 2 vctors, they give a Zero answer. this means that they are perpendicular to each other.

Right.

now, they must be the radius of the circle. so the length of the 2 vectors are equal.

I'm not sold on this. r(t) is a vector from the origin to a point on whatever figure this is, and r(t) can be decomposed into two perpendicular vectors. Why must these vectors be radii of a circle? Furthermore, the vectors have different magnitudes.
|u(t)| = sqrt(cos^2(t)(13 + a^2))
|v(t)| = sqrt(sin^2(t)(1 + 2a^2))

Even if a = +/- sqrt(12), the magnitudes are different except where cos(t) = +/- sin(t).

so we end up with a= +_ root 12.

If the problem had asked you to prove or disprove that the image of this function was a sphere, not a circle, that would be easy. Is this a problem in a textbook or one that your instructor gave you? The reason I ask is that problems in books are usually checked more carefully than ones that some math profs come up with.

 

[slider142]

Is that the exact wording of the problem? I.e., it says circle, and not sphere? My work shows that r(t) generates a sphere for certain values of a.

I'm not sure how you get a sphere. If the function is continuous with continuous inverse, the image of an interval in R cannot be 2-dimensional (unless it is a pathological space-filling curve). Since this map is smooth, the image is a 1-dimensional smooth curve for each value of a.

 

[Mark44]

I'm not sure how you get a sphere. If the function is continuous with continuous inverse, the image of an interval in R cannot be 2-dimensional (unless it is a pathological space-filling curve). Since this map is smooth, the image is a 1-dimensional smooth curve for each value of a.

What you say makes sense for the individual functions x(t), y(t), and z(t). Each one of them has an image that is a nice smooth curve. They're not invertible, though, unless we restrict the domain to [0, pi/2].

Here's my thinking:

We have r(t) = <x(t), y(t), z(t)>, with x, y, and z given in the OP.

x² + y² + z² = (13 + a²)cos²t + (1 + 2a²)sin²t
If a = +/-sqrt(12), the expression on the right above simplifies to 25.

For these values of a, a point (x(t), y(t), z(t)) satisfies the equation x² + y² + z² = 25, which is a sphere of radius 5, centered at the origin.

 

[slider142]

What you say makes sense for the individual functions x(t), y(t), and z(t). Each one of them has an image that is a nice smooth curve. They're not invertible, though, unless we restrict the domain to [0, pi/2].

This is incorrect. x(t) is not a curve, as it is a function from R -> R. The graph of x(t), which is a subset of RxR, is a curve. r: I -> R³ is a 1-dimensional curve as it continuously maps an interval into R³. See curve. It is not possible for r, a function of one variable, to map an interval to cover a sphere, irrespective of invertibility.

Here's my thinking:

We have r(t) = <x(t), y(t), z(t)>, with x, y, and z given in the OP.

x² + y² + z² = (13 + a²)cos²t + (1 + 2a²)sin²t
If a = +/-sqrt(12), the expression on the right above simplifies to 25.

For these values of a, a point (x(t), y(t), z(t)) satisfies the equation x² + y² + z² = 25, which is a sphere of radius 5, centered at the origin.

This only shows that the curve is a subset of a sphere which is a great start for showing it may be a circle. Your mapping is not invertible though, so it loses other information about the original curve. For example, the point (5, 0, 0) is not the image of any point t, although it is part of the sphere of radius 5.

 

[slider142]

Homework Statement

2.2) Prove or disprove that there is a value of a for which the image of the following vector function is a circle:
r(t) = < 2cos t + a sin t, a cos t + sin t, 3 cos t - a sin t >.

any help will be appreciated

Note first that, as a function of a single variable, r is a curve in R³. The second thing to note is that each component of r is periodic with period 2Pi, so r is periodic of period 2Pi. The periodicity of r implies that r is a closed curve. Thus, we need only consider r on an interval of 2Pi, ie., [-Pi, Pi] or [0, 2Pi], and we get the entire image of the curve.
We need only show now that r is a circle, and not some other crazy closed curve. Mark44 has already shown that r is a subset of a sphere. However, there remains one thing that is always true about circles and is not true for any other curve on the sphere. The second derivative should be everywhere perpendicular to the first derivative and all the points should lie in the same plane. The first criterion shows that there are only two possible values for a that give this, and the second criterion is easy to test for those two values.

 

[firearrow]

You could solve for constant curvature and zero torsion

 

[jp6373]

The question was given by my professor. i am unsure where he got it from. I have no idea what constant curvature and zero torsion is.

 

[slider142]

The question was given by my professor. i am unsure where he got it from. I have no idea what constant curvature and zero torsion is.

In that case, all you need is the fact that the points lie in both a plane and a sphere, or that the second derivative is normal to the first and the points lie in a plane. Curvature and torsion are properties of curves derived from their Frenet-Serret frame, which you will learn about later.