Question

Use seperation of variables to find an expression for V in terms of t where t is measured in seconds for the equation dv/dt=-2v-32

Answer 1

So when u bring that -2v to the other side it doesnt go under the dv?

Answer 2

I misread the equation at first, I believe this is the right one. \[\frac{dv}{dt}=(-2v)(-32) \]
\[\frac{1}{-2v}\frac{dv}{dt}=-32 \rightarrow -\frac{1}{2v}dv=-32dt\] And to answer your question, yes, it does go under. :)

Answer 3

ok when i was doing it i had dt-32 i didnt put those together theres my mistake

Answer 4

You didn't have parenthesis so I was not sure if it was -2v * -32 or if it was -2v-32, they would be rather different.

Answer 5

it was the -2v-32 they were subtracted

Answer 6

This way you would end up with \[\frac{1}{64v}dv = dt\]

Answer 7

Then my answer is incorrect.

Answer 8

\[\frac{1}{-2v-32}dv=dt\]

Answer 9

ok then so far i have brought the dt to the other side and divided dv by -2v so my equation looks like dv/-2v=dt-32

Answer 10

Except that if you bring the -2v to the other side you would add it, so you would have \[2v+\frac{dv}{dt}=-32\]

Answer 11

ok then so when i get that i have to bring the dt to the other side and so now i have 2v+dv=dt-32. how do i clean that up to where it is a v= equation

Answer 12

You cannot divide out your dt from a single term because you have an addition, you would end up with \[\frac{2v}{dt}+dv=-32dt\]

Answer 13

well i startede out with a dv/dt equation so i brought the dt to the other side unless im making a mistake

Answer 14

You are on the right track, by multiplying dt to the other side you end up with \[dv=(-2v-32)dt\]

Answer 15

ok so then i bring the -2v to the other side right? because i need to get the equation for velocity

Answer 16

The only thing you can do now is to distribute your dt, or, move the (-2v-32). \[\frac{1}{-2v-32}dv=dt\]

Answer 17

ok thank you