Question

I'm thinking you take the derivative, but I'm not getting any of the answers.

http://i.imgur.com/B3R6bhPh.jpg

Answer 1

Yes, derivative.
Exponential gives you the same thing back, but with chain rule.\[\large\rm \left(v_o e^{-\beta t}\right)'\quad=\quad v_o e^{-\beta t}(-\beta t)'\]

Answer 2

So,\[\large\rm \left(v_o e^{-\beta t}\right)'\quad=\quad v_o e^{-\beta t}(-\beta)\]And then just replace the thing you started with, as v,\[\large\rm \left(\color{orangered}{v_o e^{-\beta t}}\right)'\quad=\quad \color{orangered}{v_o e^{-\beta t}}(-\beta)\]

Answer 3

@zepdrix I don't understand the last step.

Answer 4

\[\large\rm \left(\color{orangered}{v_o e^{-\beta t}}\right)'\quad=\quad \color{orangered}{v_o e^{-\beta t}}(-\beta)\]This thing in orange is what you started with,
it's called v,\[\large\rm \left(\color{orangered}{v}\right)'\quad=\quad \color{orangered}{v}(-\beta)\]

Answer 5

\(a = \dot v = - \beta v_o e^{- \beta ~ t}\)

Answer 6

Ah. Of course.

Answer 7

:)