anonymous
  • anonymous
solve dv/dt=32-.8v
Mathematics
schrodinger
  • schrodinger
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SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \frac{dv}{dt} =32-0.8v }\) ?
anonymous
  • anonymous
yup
anonymous
  • anonymous
we gotta seperate

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anonymous
  • anonymous
dv/(32-.8v)=dt
anonymous
  • anonymous
then integrate
SolomonZelman
  • SolomonZelman
yes, that is fine.
SolomonZelman
  • SolomonZelman
What I would tend to do is: \(\large\color{black}{ \displaystyle \frac{dv}{dt} =32-0.8v }\) \(\large\color{black}{ \displaystyle \frac{1}{32-0.8v}\frac{dv}{dt} =1 }\) \(\large\color{black}{ \displaystyle \color{red}{\int}\frac{1}{32-0.8v}\frac{dv}{dt}\color{red}{dt} =\color{red}{\int}1\color{red}{dt} }\) and then cancel the differentials on the left side.
SolomonZelman
  • SolomonZelman
So that you aren't just adding an integral sign (which is technically meaningless), but integrating with respect to a variable. But, what you did is fine too.
anonymous
  • anonymous
how do i do it my way
anonymous
  • anonymous
like integrating dv/(32-.8v)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \int\frac{1}{32-0.8v}dv =\int dt}\) Do a u substitution on the left side, and the right side should be obvious obvious at this point.
anonymous
  • anonymous
int dt = t but idk wat int dv/32-.8v is
anonymous
  • anonymous
u usb
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
u=32-.8v?
SolomonZelman
  • SolomonZelman
ok, go ahead and solve the integral on the left side, and tell me what you get:)
anonymous
  • anonymous
should u = 32-.8v
SolomonZelman
  • SolomonZelman
yes, that is the correct u-sub.
anonymous
  • anonymous
ln u
anonymous
  • anonymous
is tat correct
SolomonZelman
  • SolomonZelman
you had du=?
anonymous
  • anonymous
du= v?
anonymous
  • anonymous
no dv
SolomonZelman
  • SolomonZelman
what is the derivative of: 32-0.8v (with respect to v?
anonymous
  • anonymous
-.8
SolomonZelman
  • SolomonZelman
Yes, so you have: du/dv=-0.8 and then, du=-0.8 dv ---> du=-(8/10) dv then, (-10/8) du=dv
anonymous
  • anonymous
-10/8(lnU)
anonymous
  • anonymous
=t
SolomonZelman
  • SolomonZelman
|dw:1443636158125:dw|
SolomonZelman
  • SolomonZelman
yes, *CORRECT* !
SolomonZelman
  • SolomonZelman
-10/8 ln(u) = t Now, solve for u.
anonymous
  • anonymous
so -10/8 ln (32-.8v)=t
SolomonZelman
  • SolomonZelman
yes, right, now solve for v
SolomonZelman
  • SolomonZelman
(you can solve for u and then subsitute the v back, or as you want to do - to sub back the v first and then to solve for v)
anonymous
  • anonymous
v= e^(8t/-10)-32 all over -.8
anonymous
  • anonymous
? right
SolomonZelman
  • SolomonZelman
Yes
SolomonZelman
  • SolomonZelman
hold on...
SolomonZelman
  • SolomonZelman
-10/8 ln (32-.8v)=t ln (32-.8v)=(-8/10)t e^ln (32-.8v)=e^(-8/10)t 32-.8v=e^(-8/10)t -0.8v=e^(-8/10)t-32 0.8v=32-e^(-8/10)t v=[ 32-e^(-8/10)t ] 0.8
SolomonZelman
  • SolomonZelman
yes, yes, you are right...
anonymous
  • anonymous
thank u
SolomonZelman
  • SolomonZelman
the result is simplifiable, you can make it into: v = 40 - (5/4)e\(\large ^{-8t/10}\)
SolomonZelman
  • SolomonZelman
Good luck

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