Terminal Velocity: Solving a Puzzling Problem

[Cyrus]

looking back at a problem that puzzled the heck out of me.


integrate this: dv/(v-vt) = -k/m (dt).

The book says you get: ln ( (vt-v)/Vt ) = -k/m t

the -k/mt part is fine, because you integrate from 0 to t. The ln part is tricky though. When I integrate, I get ln(v-vt), vt is just a constant. If you do the derivative of my anwser or the books they are the same. How did the book arrive at this anwser and not mine.

 

[cepheid]

Can you rewrite the problem more carefully? I don't understand how vt can be a constant if it has 'v' in it, or why a lowercase v suddenly becomes a capital V somewhere, or why v-vt suddenly becomes vt-v. Thanks.

 

[Tide]

Just evaluate the definite integral to get the desired result:

$$\int_{0}^{v} \frac {dv'}{v'-v_t} = -\int_{0}^{v} \frac {dv'}{v_t-v'} = \ln{(v_t-v)} - \ln{v_t} = \ln \frac {v_t-v}{v_t}$$

 

[Cyrus]

Just evaluate the definite integral to get the desired result:

$$\int_{0}^{v} \frac {dv'}{v'-v_t} = -\int_{0}^{v} \frac {dv'}{v_t-v'} = \ln{(v_t-v)} - \ln{v_t} = \ln \frac {v_t-v}{v_t}$$

but why even to bothing factoring out a minus sign tide? It just complicateds the integral. Its much simpler to just integrate and get ln(v-vt), no?

when you do the limits you get, ln(v-vt)-ln(v). so that's equal to ln ((v-vt)/v).


Oh my gosh. I completely poo-pooed that integral. I worked it out my way and got the same anwser. Boy do I feel like an idiot. it should work out to be,
ln(v-vt) from 0 to v, which is ln(v-vt)-ln(-vt) = to ln(1-v/vt). Thats just Embarrassing.

 

[Tide]

Cyrus,

You factor out the -1 because (a) v < vt and (b) you avoid the nastiness of dealing with logarithms of negative numbers.