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Show that the vector value function is on the surface of a sphere

  • Thread starter tsamocki
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  • #1
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Homework Statement



r(t) = [cos(t^3)sin(t), sin(t)sin(t^3), cos(t)], where t is an element of (0,2pi).

Show that the graph of this function is on the surface of a sphere. Then find it's radius.

Homework Equations



T(t) = r'(t)/norm[r'(t)]

equation for a sphere in 3-space: r^2=x^2+y^2+z^2

The Attempt at a Solution



My thought is, a tangent vector must lie on the surface of this function (which is a sphere), and therefore the tangent vector at some point (t) would be sufficient proof. But that seems too easy.

Secondly, to find the radius of a sphere, set up the parametric equations in the form of x= cos(t^3)sin(t), y = sin(t^3)sin(t), z = cos(t), in the form of the equation for a sphere:

(cos(t^3)sin(t))^2 + (sin(t^3)sin(t))^2 + (cos(t))^2 = r^2

then substitute one of the elements to find the radius squared?
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement



r(t) = [cos(t^3)sin(t), sin(t)sin(t^3), cos(t)], where t is an element of (0,2pi).

Show that the graph of this function is on the surface of a sphere. Then find it's radius.

Homework Equations



T(t) = r'(t)/norm[r'(t)]

equation for a sphere in 3-space: r^2=x^2+y^2+z^2

The Attempt at a Solution



My thought is, a tangent vector must lie on the surface of this function (which is a sphere), and therefore the tangent vector at some point (t) would be sufficient proof. But that seems too easy.
Too easy and wrong too. The tangent vector is always tangent to any surface.
Secondly, to find the radius of a sphere, set up the parametric equations in the form of x= cos(t^3)sin(t), y = sin(t^3)sin(t), z = cos(t), in the form of the equation for a sphere:

(cos(t^3)sin(t))^2 + (sin(t^3)sin(t))^2 + (cos(t))^2 = r^2

then substitute one of the elements to find the radius squared?
You need to show that is an identity. Just simplify it.
 
  • #3
Dick
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Homework Statement



r(t) = [cos(t^3)sin(t), sin(t)sin(t^3), cos(t)], where t is an element of (0,2pi).

Show that the graph of this function is on the surface of a sphere. Then find it's radius.

Homework Equations



T(t) = r'(t)/norm[r'(t)]

equation for a sphere in 3-space: r^2=x^2+y^2+z^2

The Attempt at a Solution



My thought is, a tangent vector must lie on the surface of this function (which is a sphere), and therefore the tangent vector at some point (t) would be sufficient proof. But that seems too easy.

Secondly, to find the radius of a sphere, set up the parametric equations in the form of x= cos(t^3)sin(t), y = sin(t^3)sin(t), z = cos(t), in the form of the equation for a sphere:

(cos(t^3)sin(t))^2 + (sin(t^3)sin(t))^2 + (cos(t))^2 = r^2

then substitute one of the elements to find the radius squared?
Just expand your expression a little bit and use trig. Like sin(x)^2+cos(x)^2=1.
 
  • #4
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Too easy and wrong too. The tangent vector is always tangent to any surface.
r^2 = x^2 + y^2 + z^2

r^2 = (cos(t^3)sin(t))^2 + (sin(t^3)sin(t))^2 + (cos(t))^2

r^2 = [cos^2(t^3) sin^2(t)] + [sin^2(t^3)sin^2(t)] + [cos^2(t)]

since r^2 = 1, and the other side simplifies to 1, the graph of the function lies on the surface of a sphere?

You need to show that is an identity. Just simplify it.
Since it simplifies to 1, and the root of 1 is one, the radius is 1.

Am i on the right track?
 
  • #5
LCKurtz
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r^2 = x^2 + y^2 + z^2

r^2 = (cos(t^3)sin(t))^2 + (sin(t^3)sin(t))^2 + (cos(t))^2

r^2 = [cos^2(t^3) sin^2(t)] + [sin^2(t^3)sin^2(t)] + [cos^2(t)]

since r^2 = 1, and the other side simplifies to 1, the graph of the function lies on the surface of a sphere?



Since it simplifies to 1, and the root of 1 is one, the radius is 1.

Am i on the right track?
Yes. If you show |r(t)| = 1 for all t that does it.
 

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