Question

solve dv/dt=32-.8v

Answer 1

\(\large\color{black}{ \displaystyle \frac{dv}{dt} =32-0.8v }\) ?

Answer 2

yup

Answer 3

we gotta seperate

Answer 4

dv/(32-.8v)=dt

Answer 5

then integrate

Answer 6

yes, that is fine.

Answer 7

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.

Answer 8

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.

Answer 9

how do i do it my way

Answer 10

like integrating dv/(32-.8v)

Answer 11

\(\large\color{black}{ \displaystyle \int\frac{1}{32-0.8v}dv =\int dt}\)

Do a u substitution on the left side, and the right side should be obvious obvious at this point.

Answer 12

int dt = t but idk wat int dv/32-.8v is

Answer 13

u usb

Answer 14

oh ok

Answer 15

u=32-.8v?

Answer 16

ok, go ahead and solve the integral on the left side, and tell me what you get:)

Answer 17

should u = 32-.8v

Answer 18

yes, that is the correct u-sub.

Answer 19

ln u

Answer 20

is tat correct

Answer 21

you had du=?

Answer 22

du= v?

Answer 23

no dv

Answer 24

what is the derivative of: 32-0.8v (with respect to v?

Answer 25

-.8

Answer 26

Yes, so you have:
du/dv=-0.8

and then,
du=-0.8 dv ---> du=-(8/10) dv

then,
(-10/8) du=dv

Answer 27

-10/8(lnU)

Answer 28

=t