- #1
fizzacist
- 2
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While taking an AP physics practice exam, I encountered a difference in the way I solve a differential equation and the way the exam's rubric solves it.
The equation is as follows:
[itex]\frac{dv}{dt}[/itex] = [itex]\frac{F-KV}{m}[/itex]
My solution:
[itex]\int[/itex][itex]\frac{dv}{F-KV}[/itex] = [itex]\int[/itex] [itex]\frac{dt}{m}[/itex]
u = F-KV
[itex]\frac{du}{-K}[/itex] = dv
[itex]\frac{-1}{K}[/itex] [itex]\int[/itex][itex]\frac{1}{u}[/itex]du = [itex]\int[/itex][itex]\frac{dt}{m}[/itex]
Integrate that to find
ln|F-KV|+C = -K[itex]\frac{t}{m}[/itex]
But before I go any further, the 1993 Exam's Rubric shows that by integrating [itex]\int[/itex][itex]\frac{dv}{F-KV}[/itex] should yield ln|F-KV|-lnC
To me, this makes no sense. The constant of integration should be ln|u| + C, not ln|u|-lnC
Here's what I'm talking about:
http://imgur.com/c83p1
I've also attached the '93's rubric to this post. The problem I'm referring to is problem #2.
Can any of the math/physics gurus out there help me out? :P
Thanks
The equation is as follows:
[itex]\frac{dv}{dt}[/itex] = [itex]\frac{F-KV}{m}[/itex]
My solution:
[itex]\int[/itex][itex]\frac{dv}{F-KV}[/itex] = [itex]\int[/itex] [itex]\frac{dt}{m}[/itex]
u = F-KV
[itex]\frac{du}{-K}[/itex] = dv
[itex]\frac{-1}{K}[/itex] [itex]\int[/itex][itex]\frac{1}{u}[/itex]du = [itex]\int[/itex][itex]\frac{dt}{m}[/itex]
Integrate that to find
ln|F-KV|+C = -K[itex]\frac{t}{m}[/itex]
But before I go any further, the 1993 Exam's Rubric shows that by integrating [itex]\int[/itex][itex]\frac{dv}{F-KV}[/itex] should yield ln|F-KV|-lnC
To me, this makes no sense. The constant of integration should be ln|u| + C, not ln|u|-lnC
Here's what I'm talking about:
http://imgur.com/c83p1
I've also attached the '93's rubric to this post. The problem I'm referring to is problem #2.
Can any of the math/physics gurus out there help me out? :P
Thanks