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.