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Vector Calculus

  • Thread starter Ted123
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Homework Statement



[PLAIN]http://img576.imageshack.us/img576/1710/vectorp.png [Broken]

Homework Equations





The Attempt at a Solution



I've done part (a) but how do I do (b)?
 
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Answers and Replies

  • #2
Char. Limit
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Well, do you know how to find a tangent vector to the surface G?
 
  • #3
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Well, do you know how to find a tangent vector to the surface G?
No...
 
  • #4
Char. Limit
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This site may help you. Since your equation is defined in two variables u and v, I would try taking the partial derivatives with respect to u and v, and you should get two tangent vectors. Then you should check to see if they are linearly independent.
 
  • #5
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This site may help you. Since your equation is defined in two variables u and v, I would try taking the partial derivatives with respect to u and v, and you should get two tangent vectors. Then you should check to see if they are linearly independent.
Thanks that's really helpful.

Taking partial derivatives wrt [itex]u[/itex] I get:

[itex]\mathbf{r} '(u,v) = sinh(u)cos(v) \mathbf{i} + sinh(u)sin(v) \mathbf{j} + cosh(u) \mathbf{k}[/itex]

and [itex]\| \mathbf{r} '(u,v) \| = \sqrt{cosh(2u)}[/itex]

and so a tangent vector to G is

[itex]\mathbf{T}(u,v) = \frac{sinh(u)cos(v) \mathbf{i} + sinh(u)sin(v) \mathbf{j} + cosh(u) \mathbf{k}}{\sqrt{cosh(2u)}}[/itex]

So at [itex]\mathbf{r}_1[/itex] from part (a) we know [itex]u=sinh^{-1}(1)[/itex] and [itex]v=\frac{\pi}{4}[/itex] .

Hence [itex]\mathbf{T}(sinh^{-1}(1) , \frac{\pi}{4}) = \frac{1}{\sqrt{6} \mathbf{i} + \frac{1}{\sqrt{6} \mathbf{j} + \sqrt{\frac{2}{3}} \mathbf{k}[/itex] is a tangent vector to G at [itex]\mathbf{r}_1[/itex] .

This time, taking partial derivates wrt [itex]v[/itex] ,

[itex]\mathbf{r} '(u,v) = -cosh(u)sin(v) \mathbf{i} + cosh(u)cos(v) \mathbf{j}[/itex]

and [itex]\| \mathbf{r} '(u,v) \| = \sqrt{cosh^2(u)}[/itex]

Evaluating at [itex]\mathbf{r}_1[/itex] I get 0???
 
  • #6
Char. Limit
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That's odd. I'm not sure you're doing it right, then... remember that:

1. [tex]\sqrt{cosh^2(u)} = cosh(u)[/tex]. This will simplify both of your calculations when finding the second unit tangent vector.

2. sin(pi/4) and cos(pi/4) are not zero.
 
  • #7
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That's odd. I'm not sure you're doing it right, then... remember that:

1. [tex]\sqrt{cosh^2(u)} = cosh(u)[/tex]. This will simplify both of your calculations when finding the second unit tangent vector.

2. sin(pi/4) and cos(pi/4) are not zero.
For the first tangent vector:

[itex]\mathbf{T} (sinh^{-1}(1) , \frac{\pi}{4}) = \frac{1}{\sqrt{6} \mathbf{i} + \frac{1}{\sqrt{6} \mathbf{j} + \sqrt{\frac{2}{3}} \mathbf{k}}[/itex] is a tangent vector to G at [itex]\mathbf{r}_1[/itex]

For the 2nd:

[itex]\mathbf{T}(u , v) = \frac{\mathbf{r} '(u,v) = -cosh(u)sin(v) \mathbf{i} + cosh(u)cos(v) \mathbf{j}}{cosh(u)}[/itex]

[itex]\mathbf{T}(sinh^{-1}(1) , \frac{\pi}{4}) = \frac{-1+1}{\sqrt{2}} = 0[/itex]
 
  • #8
Char. Limit
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What happened to the unit vectors i and j?
 
  • #9
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What happened to the unit vectors i and j?
Sorry, ballsed up the latex.

For the first tangent vector:

[itex]\mathbf{T}(u,v) = \frac{sinh(u)cos(v) \mathbf{i} + sinh(u)sin(v) \mathbf{j} + cosh(u) \mathbf{k}}{\sqrt{cosh(2u)}}[/itex]

[itex]\mathbf{T} (sinh^{-1} (1) , \frac{\pi}{4}) = \frac{1}{\sqrt{6}} \mathbf{i} + \frac{1}{\sqrt{6}} \mathbf{j} + \sqrt{\frac{2}{3}} \mathbf{k}}[/itex] is a tangent vector to G at [itex]\mathbf{r}_1[/itex]

For the 2nd:

[itex]\mathbf{T}(u , v) = \frac{\mathbf{r} '(u,v) = -cosh(u)sin(v) \mathbf{i} + cosh(u)cos(v) \mathbf{j}}{cosh(u)}[/itex]

[itex]\mathbf{T}(sinh^{-1}(1) , \frac{\pi}{4}) = -\frac{1}{\sqrt{2}} \mathbf{i} +\frac{1}{\sqrt{2}}\mathbf{j}[/itex]
 
  • #10
Char. Limit
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But that last line is just not true. -i+j does not equal 0.
 
  • #11
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But that last line is just not true. -i+j does not equal 0.
Yeah of course not.

So presumably if I now calculate a normal vector of G at [itex]\mathbf{r}_1[/itex] it will be a multiple of the vector [itex]\nabla g(\mathbf{r}_1) =(2,2-2)[/itex]
 
  • #12
Char. Limit
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Presumably, yes. And the link I provided above also shows you how to find the normal vector.
 
  • #13
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Presumably, yes. And the link I provided above also shows you how to find the normal vector.
Trouble is if I try to find a normal vector from the 2nd tangent vector (by differentiating wrt v),

[itex]\mathbf{T}(u , v) = -sin(v) \mathbf{i} + cos(v) \mathbf{j}[/itex]

Then the resulting normal vector [itex]\frac{\mathbf{T}'(u , v)}{\|\mathbf{T} '(u , v) \|}[/itex] certainly isn't a multiple of [itex](2,2,-2)[/itex] as there's no k component!

However it works for the 1st one as [itex]\frac{\mathbf{T}'(u , v)}{\|\mathbf{T} '(u , v) \|} = \frac{1}{3 \sqrt{3}} \mathbf{i} + \frac{1}{3\sqrt{3}} \mathbf{j} - \frac{1}{3\sqrt{3}} \mathbf{k}[/itex]
 
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