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elica85 Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{0}^{t} \frac{ Vm }{ Mmt } dt\]
 one year ago

elica85 Group TitleBest ResponseYou've already chosen the best response.0
take V,m, and M as constants..
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
It might be a little easier to read if pull the constants (in the numerator) outside of the integral, and then apply a `U substitution`.\[\large \int\limits\limits_{0}^{t} \frac{Vm}{Mmt} dt \qquad \rightarrow \qquad Vm\int\limits\limits_{0}^{t} \frac{1}{Mmt}dt\] Are you familiar with U substitution? :)
 one year ago

elica85 Group TitleBest ResponseYou've already chosen the best response.0
thx, yes, what i get is V{ln[(Mmt)/M]}...
 one year ago

elica85 Group TitleBest ResponseYou've already chosen the best response.0
now if M1=Mmt, M=Mb+Mf...and you substitue and simplify, i'm suppose to get Vln[1+(Mf/Mb)]
 one year ago

elica85 Group TitleBest ResponseYou've already chosen the best response.0
but i don't
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
I'm so confused, there are too many M's floating around :( lol
 one year ago

elica85 Group TitleBest ResponseYou've already chosen the best response.0
sorry i will redo with different letters...
 one year ago

elica85 Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{0}^{t} \frac{ ab }{ cbt } dt\] all constants except t
 one year ago

elica85 Group TitleBest ResponseYou've already chosen the best response.0
d=cbt c=e+f
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
So integrating, before doing any fancy simplification should give you,\[\large ab\left(\frac{1}{b}\right)\lncbt_0^t\] Moar letters? :U oh boy lol
 one year ago

elica85 Group TitleBest ResponseYou've already chosen the best response.0
ok so integrate over limits and it should be a[ln(cbt)ln(c)] correct so far?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Yah looks good so far c:
 one year ago

elica85 Group TitleBest ResponseYou've already chosen the best response.0
which is a{ln[(cbt)/c]}
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Yes, you can bring the negative into the log as an exponent. Which will give us this,\[\large a \ln\left[\left(\frac{cbt}{c}\right)^{1}\right] \qquad = \qquad a \ln\left[\left(\frac{c}{cbt}\right)\right]\] Whichever way makes sense to you though :3
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Now what's going on with this d and e and f nonsense? XD lol
 one year ago

elica85 Group TitleBest ResponseYou've already chosen the best response.0
ok, so f has to be in there cuz that's what i'm trying to solve. so plugging in the other letters and simplifying, according to the solution, it's suppose to be a[ln(1+(f/e))] but i can't get it. however, i haven't tried with your second expression up there so i will try it now
 one year ago

elica85 Group TitleBest ResponseYou've already chosen the best response.0
ok, i think one of the variables is messed up on the solution. i got it, thx for your help
 one year ago
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