# Error in Young/Freedman 13th? (Integral)

1. Aug 10, 2012

### SiennaTheGr8

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 $v_{y}$ 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.

Last edited: Aug 10, 2012
2. Aug 10, 2012

### DonAntonio

Don't apologize for the lack of Latex. Better learn how to use it in this forum and thus make yourself a good

service as the way it is your message has less chance to be read and addressed by other people than it'd have

if you'd write it with Latex.

DonAntonio

3. Aug 10, 2012

### SiennaTheGr8

Thanks -- added some basic Latex.

4. Aug 10, 2012

### DonAntonio

If the expression $\,v_t\,$ is really a constant wrt $\,t\,$ (and this is odd for me because it

has a subindex depending on t), then you're right and the book's wrong.

Nevertheless I'd try to be dead sure about that $\,v_t\,$ since it is not usual so huge mistakes in known text books.

DonAntonio

5. Aug 10, 2012

### voko

Differentiating 5.12, we get 5.10. No problem there.

What you miss is that any indefinite integral is defined up to an additive constant. Which is what is different between your solution and the book's. Why the constant in the book has that particular form, should be evident from other conditions on the function sought.

6. Aug 10, 2012

### SiennaTheGr8

Yes, the additive constant crossed my mind, but I think in this case it should be zero, since the initial position seems to be the origin on the given y-t graph that we are told represents Eq. 5.12. I very well might be missing something, but I read nothing in the text that would indicate that the term $-\frac{m}{k}v_t$ should be the initial condition.

(Eq. 5.12 is only given in passing to show mathematically what is happening in that particular figure. The equation is never brought up again, as far as I can tell, so I figured it's the kind of error that might easily go undetected, if it's an error at all.)

7. Aug 10, 2012

### SiennaTheGr8

I'd also add that the terminal speed $v_t$ of a low-mass, slow-moving object falling through a fluid is defined in the text as $v_t = \frac{mg}{k}$, and so if this mysterious $-\frac{m}{k}v_t$ term is really the additive constant, then wouldn't that imply that the initial y-position is $-\frac{m^2}{k^2}g$? I just don't understand how a term like that could have anything to do with the initial position of a small object slowly falling through a fluid.

(In the given frame of reference, by the way, the negative y-direction is up, and the positive y-direction is down).

Sorry if I'm just being dense.

8. Aug 10, 2012

### SiennaTheGr8

Still, the mystery term does yield the correct unit (meters), since the constant of proportionality $k$ has units kg/s. I still have no idea where the term comes from, and its existence seems to contradict the given y-t graph, which clearly shows the origin as the initial condition.

9. Aug 10, 2012

### SiennaTheGr8

Ah, figured it out.

Classic rookie mistake: it was a definite integral from time 0 to t, not an indefinite integral. Substituting 0 for t in my indefinite integral equation indeed yields the mystery term.

Thanks for the help!