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SiennaTheGr8
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In chapter 5 of Young/Freedman's University Physics, 13th ed., Eq. 5.10 is given as:
##v_{y} = v_{t}(1 - e^{-\frac{k}{m}t})##,
where [itex]v_{y}[/itex] is the time-dependent speed in the vertical direction of a low-mass, slow-moving object falling through a fluid, and ##v_{t}##, ##k##, and ##m## are all constants (terminal speed, proportionality constant, and mass, respectively).
Eq. 5.12 is supposed to be the time integral of Eq. 5.10, and is given as:
##y = v_{t}\Big(t - \frac{m}{k}(1 - e^{-\frac{k}{m}t})\Big)##,
where ##y## is the time-dependent position of the object in the vertical direction.
The authors do not provide the derivation. When trying to integrate Eq. 5.10 myself, however, I came up with:
##y = v_{t}\Big(t + \frac{m}{k}(e^{-\frac{k}{m}t})\Big)##.Here are my steps:
Distribute ##v_{t}## on the right side of the equation:
##v_{y} = v_{t} - v_{t}e^{-\frac{k}{m}t}##
Integrate term-by-term, treating ##v_{t}## as a multiplicative constant:
##y = v_{t}∫dt - v_{t}∫e^{-\frac{k}{m}t}dt##
##y = v_{t}t - v_{t}∫e^{-\frac{k}{m}t}dt##
Using u-substitution for the last term, treating ##\frac{k}{m}## as a constant ##c##:
##∫e^{-ct}dt##,
where
##u = -ct##
##\frac{du}{dt} = -c##
##dt = \frac{du}{-c}##,
we have:
##\frac{1}{-c}∫e^{u}du##
##\frac{1}{-c}e^{u}##
##\frac{1}{-c}e^{-ct}##.
Subbing back in for ##\frac{k}{m}##:
##-\frac{m}{k}e^{-\frac{k}{m}t}##.Now the right-hand term of the equation becomes:
##(-v_{t})(-\frac{m}{k})e^{-\frac{k}{m}t}##,
and the whole equation:
##y = v_{t}t + v_{t}(\frac{m}{k})e^{-\frac{k}{m}t}##
##y = v_{t}\Big(t + \frac{m}{k}(e^{-\frac{k}{m}t})\Big)##.But again, the book gives
##y = v_{t}\Big(t - \frac{m}{k}(1 - e^{-\frac{k}{m}t})\Big)##
as the answer. Distributing the ##-\frac{m}{k}## in their answer gives:
##y = v_{t}\Big(t - \frac{m}{k} + \frac{m}{k}(e^{-\frac{k}{m}t})\Big)##,
which is the same as my answer, except with an additional term of ##-\frac{m}{k}## in the big parentheses.
I'm not a very competent mathematician, and I've probably made an error somewhere, but I can't spot it, and I don't see where their extra term of ##-\frac{m}{k}## comes from. Am I integrating improperly? Or is this actually an error in the text?
Thanks, and sorry for the lack of LaTeX.
[Edited: basic LaTeX added]
##v_{y} = v_{t}(1 - e^{-\frac{k}{m}t})##,
where [itex]v_{y}[/itex] is the time-dependent speed in the vertical direction of a low-mass, slow-moving object falling through a fluid, and ##v_{t}##, ##k##, and ##m## are all constants (terminal speed, proportionality constant, and mass, respectively).
Eq. 5.12 is supposed to be the time integral of Eq. 5.10, and is given as:
##y = v_{t}\Big(t - \frac{m}{k}(1 - e^{-\frac{k}{m}t})\Big)##,
where ##y## is the time-dependent position of the object in the vertical direction.
The authors do not provide the derivation. When trying to integrate Eq. 5.10 myself, however, I came up with:
##y = v_{t}\Big(t + \frac{m}{k}(e^{-\frac{k}{m}t})\Big)##.Here are my steps:
Distribute ##v_{t}## on the right side of the equation:
##v_{y} = v_{t} - v_{t}e^{-\frac{k}{m}t}##
Integrate term-by-term, treating ##v_{t}## as a multiplicative constant:
##y = v_{t}∫dt - v_{t}∫e^{-\frac{k}{m}t}dt##
##y = v_{t}t - v_{t}∫e^{-\frac{k}{m}t}dt##
Using u-substitution for the last term, treating ##\frac{k}{m}## as a constant ##c##:
##∫e^{-ct}dt##,
where
##u = -ct##
##\frac{du}{dt} = -c##
##dt = \frac{du}{-c}##,
we have:
##\frac{1}{-c}∫e^{u}du##
##\frac{1}{-c}e^{u}##
##\frac{1}{-c}e^{-ct}##.
Subbing back in for ##\frac{k}{m}##:
##-\frac{m}{k}e^{-\frac{k}{m}t}##.Now the right-hand term of the equation becomes:
##(-v_{t})(-\frac{m}{k})e^{-\frac{k}{m}t}##,
and the whole equation:
##y = v_{t}t + v_{t}(\frac{m}{k})e^{-\frac{k}{m}t}##
##y = v_{t}\Big(t + \frac{m}{k}(e^{-\frac{k}{m}t})\Big)##.But again, the book gives
##y = v_{t}\Big(t - \frac{m}{k}(1 - e^{-\frac{k}{m}t})\Big)##
as the answer. Distributing the ##-\frac{m}{k}## in their answer gives:
##y = v_{t}\Big(t - \frac{m}{k} + \frac{m}{k}(e^{-\frac{k}{m}t})\Big)##,
which is the same as my answer, except with an additional term of ##-\frac{m}{k}## in the big parentheses.
I'm not a very competent mathematician, and I've probably made an error somewhere, but I can't spot it, and I don't see where their extra term of ##-\frac{m}{k}## comes from. Am I integrating improperly? Or is this actually an error in the text?
Thanks, and sorry for the lack of LaTeX.
[Edited: basic LaTeX added]
Last edited: