Solve String Theory Problem: dX/dx=∂X/∂x?

[ehrenfest]

Homework Statement

http://books.google.com/books?id=Xm...5Dw&sig=6cUrZKqmPMoe0QBRTSYNnipNRw4#PPA114,M1

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}?

Homework Equations

The Attempt at a Solution

 

[Jimmy Snyder]

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}?

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.

 

[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?

 

[Jimmy Snyder]

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?

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

 

[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?

 

[Jimmy Snyder]

The first equality in equation 3 is the chain rule for that, isn't it?

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