# Vector Calculus

Ted123

## Homework Statement

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

## The Attempt at a Solution

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

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Gold Member
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...

Gold Member
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.

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???

Gold Member
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:

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

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

Gold Member
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:

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

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

Gold Member
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 $\nabla g(\mathbf{r}_1) =(2,2-2)$

Gold Member
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 $\frac{\mathbf{T}'(u , v)}{\|\mathbf{T} '(u , v) \|} = \frac{1}{3 \sqrt{3}} \mathbf{i} + \frac{1}{3\sqrt{3}} \mathbf{j} - \frac{1}{3\sqrt{3}} \mathbf{k}$

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