# Homework Help: String theory question

1. Sep 19, 2007

### ehrenfest

1. The problem statement, all variables and given/known data

In the solution to problem 6.3 shown in the attachment, can someone explain to me why $$d\vec{X}/dx$$ was implicitly set equal to $$\partial \vec{X}/\partial{x}$$?

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### 63solution.jpg
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2. Sep 20, 2007

### Jimmy Snyder

Actually, the trouble begins before that. The unnumbered equation is wrong. They have
$$d\vec{X} = (dx, y' dx) = (1, y') dx$$
but in the statement of the problem on page 114, Zwiebach has $y' = \partial y / \partial x$. So the unnumbered equation should read:
$$d\vec{X} = (dx, y' dx + \dot{y} dt)$$
From this, added to the fact that as you said, he needs $\partial \vec{X}/\partial{x}$ not $d\vec{X}/dx$, I think you can see how to finish up.

Last edited: Sep 20, 2007
3. Sep 20, 2007

### ehrenfest

You are right. Then how do you get dx/ds when the expression in $$d\vec{X}$$ now has a time differential in it so you cannot use equation 2?

4. Sep 21, 2007

### Jimmy Snyder

Don't use equation 2. Don't use equation 3 either. Use the chain rule to find
$$\frac{\partial\vec{X}}{\partial s}$$

5. Sep 21, 2007

### ehrenfest

The first equality in equation 3 is the chain rule for that, isn't it? So, I still need dx/ds, don't I?

6. Sep 21, 2007

### Jimmy Snyder

It isn't. It doesn't into account the fact that
$$\vec{X}$$
also depends on time.

Last edited: Sep 21, 2007