dI/dt ? math error?

[J4e8a16n]

Hi,

I checked an integration 'error?'  in my book . See the red  !  .
Unfortunately the book is in french.  Anyhow . . .

[c4757p]

Yeah, should be -1/3 ln |5-I|. Somebody missed a factor of three...

In my experience you don't usually make it far into a textbook with a bunch of math before you find an arithmetic error. Maybe my school just picks shitty books, though...

[IanB]

Hmm? Let's work it backwards:

f(I) = -1/3 ln(15 - 3I)

Let u = 15 - 3I, then df/dI = (df/du)(du/dI), where

f(u) = -1/3 ln u

So:

df/du = (-1/3)(1/u)

du/dI = -3

df/dI = (-1/3)(1/u)(-3) = 1/u = 1 / (15 - 3I)

Which is what the book has.

So what did I get wrong?

[c4757p]

Actually, you're right, thank you!

The catch here is that we came to different answers through different methods, but both are correct because of the constant that is being ignored.

ln(5 - i) = ln(15 - 3i) - ln(3)

And ln(3) is of course a constant.

(I used a lowercase i for legibility)

Kind of ashamed I missed that, I used to grade math papers...

[J4e8a16n]

Why not that way?

 integration sign ((1/ (15-3i) ) dI) =

 ln|(1/ (15-3i) | =
-ln|(15-3i) | =
-|15-3i|  = e^(t-C)


and be stuck with no 1/3  :-/

[IanB]

Why not that way?

 integration sign ((1/ (15-3i) ) dI) =

ln|(1/ (15-3i) | =
-ln|(15-3i) | =
-|15-3i|  = e^(t-C)


and be stuck with no 1/3  :-/

Because your working is wrong at the line indicated in red. The antiderivative of 1/x is ln x, and you are missing a factor of (-1/3):

http://www.wolframalpha.com/input/?i=integrate+1%2F%2815-3i%29+di

[J4e8a16n]

substitute u = 15-3 i
du/di = 0-3
and  du = -3  di:
du/-3  =  di
1/-3 *du  =  di

ok

-1/3 log(15+3i)) constant  = t- C
-1/3 log(15+3i))  = t- (C +constant)
log(15+3i))  =-3( t- C)

...

Why Wolfgang say?
Substitute back for u = 15-3 i: = -1/3 log(-3 (-5+i))+constant
Which is equivalent for restricted i values to:
Answer: | 
 |  = -1/3 log(5-i)+constant

[IanB]

substitute u = 15-3 i
du/di = 0-3
and  du = -3  di:
du/-3  =  di
1/-3 *du  =  di

ok

-1/3 log(15+3i)) constant  = t- C
-1/3 log(15+3i))  = t- (C +constant)
log(15+3i))  =-3( t- C)

Be careful with your signs. In several places there you have changed a "-" into a "+". If you do that in a real problem you will lose marks in a test, or build something that doesn't work the way you expect...

[KJDS]

For a long time, almost every paper and book I read had errors in the equations, probably so that those who don't understand the maths can't make use of them.