# Vector Calculus

## 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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Char. Limit
Gold Member
Well, do you know how to find a tangent vector to the surface G?

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

No...

Char. Limit
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.

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???

Char. Limit
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.

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
Gold Member
What happened to the unit vectors i and j?

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

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

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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