Vector Calculus

[Ted123]

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)?

 

[Char. Limit]

Well, do you know how to find a tangent vector to the surface G?

 

[Ted123]

Well, do you know how to find a tangent vector to the surface G?

No...

 

[Char. Limit]

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.

 

[Ted123]

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 $u$ I get:

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

and $\| \mathbf{r} '(u,v) \| = \sqrt{cosh(2u)}$

and so a tangent vector to G is

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

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

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

This time, taking partial derivates wrt $v$ ,

$\mathbf{r} '(u,v) = -cosh(u)sin(v) \mathbf{i} + cosh(u)cos(v) \mathbf{j}$

and $\| \mathbf{r} '(u,v) \| = \sqrt{cosh^2(u)}$

Evaluating at $\mathbf{r}_1$ I get 0???

 

[Char. Limit]

That's odd. I'm not sure you're doing it right, then... remember that:

1. $$\sqrt{cosh^2(u)} = cosh(u)$$. 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.

 

[Ted123]

That's odd. I'm not sure you're doing it right, then... remember that:

1. $$\sqrt{cosh^2(u)} = cosh(u)$$. 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:

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

For the 2nd:

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

$\mathbf{T}(sinh^{-1}(1) , \frac{\pi}{4}) = \frac{-1+1}{\sqrt{2}} = 0$

 

[Char. Limit]

What happened to the unit vectors i and j?

 

[Ted123]

What happened to the unit vectors i and j?

Sorry, ballsed up the latex.

For the first tangent vector:

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

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

For the 2nd:

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

$\mathbf{T}(sinh^{-1}(1) , \frac{\pi}{4}) = -\frac{1}{\sqrt{2}} \mathbf{i} +\frac{1}{\sqrt{2}}\mathbf{j}$

 

[Char. Limit]

But that last line is just not true. -i+j does not equal 0.

 

[Ted123]

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 $\mathbf{r}_1$ it will be a multiple of the vector $\nabla g(\mathbf{r}_1) =(2,2-2)$

 

[Char. Limit]

Presumably, yes. And the link I provided above also shows you how to find the normal vector.

 

[Ted123]

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),

$\mathbf{T}(u , v) = -sin(v) \mathbf{i} + cos(v) \mathbf{j}$

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

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