Understanding Derivative of Position Function: Is Velocity Wrong?

[mopit_011]

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!

 

[Dragon27]

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.

 

[robphy]

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.)