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

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