Classical Understanding Derivative of Position Function: Is Velocity Wrong?

AI Thread Summary
The discussion revolves around a misunderstanding of the derivative of the position function in Kleppner and Kolenkow's textbook. A participant questions the accuracy of the velocity expression derived from the position function, suggesting it should be A(alpha squared)(e^2x). Responses clarify that the correct approach involves deriving the x-component of velocity from the full position expression. Key points include the importance of checking units for consistency, noting that A must have units of position, while e^(alpha t) is dimensionless. Additionally, alpha is identified as having units of inverse-time, emphasizing that the proposed expression for velocity contains dimensional inconsistencies. The conversation highlights the need for careful interpretation of mathematical notation, particularly in physics contexts.
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Hello! So, I was beginning to skim Kleppner and Kolenkow for an upcoming course I’m taking over the summer. I saw this on pg. 17 and was wondering if I’m making a silly mistake in understanding what the book is saying. When they take the derivative of the position function, isn’t the velocity wrong? For vx, shouldn’t the function be A(alpha squared)(e^2x)? Thank you!
 

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Not sure how you came to that conclusion. Using the full expression for position write out the expression for the x-component of the position and derive from it the expression for the x-component of velocity.
 
It is useful to check units.
##A## must have units of position ##e^{\alpha t}## is dimensionless.
##\alpha## has units of inverse-time since the argument of the exponential function ##{\alpha t}## is dimensionless.

Thus, in your proposed expression,
##A\alpha^2## has units of position per time-squared (an acceleration)
and ##2x## (with units of position) can't be the necessarily-dimensionless argument of the exponential function.
As @Dragon27 says, it's unclear how you arrived at your expression.

(Note: ##v_x## is the ##x##-component of a vector ##\vec v##.
Once, when I taught a math-methods class for physics students,
one of the students who was a mathematics major interpreted
"##v_x##" as the partial-derivative of a function ##v##...
since some math books use that notation.)
 
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