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UrbanXrisis
Nov8-04, 10:35 PM
v=vi*e^(-ct)
The question asks to differentiate that equation for v(t) and thus show that the acceleration is porportional to the speed at any time.
I have no clue how to differentiate. Could someone start me off or give some clues? I'll find th rest in my text book. Thanks
Justin Lazear
Nov8-04, 10:38 PM
http://mathworld.wolfram.com/Derivative.html
--J
UrbanXrisis
Nov8-04, 10:45 PM
differentiate = take the derivative?
wow...I must not be paying attention in math class
UrbanXrisis
Nov8-04, 10:47 PM
I dont understand... the derivative of e^x = e^x?
then the derivative of v=vi*e^(-ct) is:
a=c*e^(-ct)+e^(-ct) * Vi?
Justin Lazear
Nov8-04, 10:49 PM
More like
a = -c*vi*e^(-ct)
Don't forget the vi.
Now, anything in that expression look familiar? Could you fit v(t) in there somewhere?
--J
UrbanXrisis
Nov8-04, 10:53 PM
Where did the -c come from? Doesn't the derivative of Vi become c?
Justin Lazear
Nov8-04, 10:57 PM
\frac{d}{dt}v(t) = \frac{d}{dt} \left( v_i e^{-ct} \right)
v_i is a constant, so
\frac{d}{dt} v_i e^{-ct} = v_i \frac{d}{dt}e^{-ct}
Make a substitution u = -ct and apply the chain rule
v_i \frac{d}{dt}e^{-ct} = v_i \left( \frac{d}{du} e^u \right) \left( \frac{du}{dt} \right) = v_i e^u \frac{d}{dt} \left( -ct \right) = v_i e^u \cdot (-c) = -c v_i e^{-ct}
--J
UrbanXrisis
Nov8-04, 11:01 PM
thank you, that helps a lot. I was just wondering, my book gives an answer of a=-cv, where does the e go?
Justin Lazear
Nov8-04, 11:02 PM
v = v_i e^{-ct}
Plug v into the book's expression for a and see if it matches up with what you got. Find the e?
--J
UrbanXrisis
Nov8-04, 11:07 PM
v = v_i e^{-ct}
a = -c v_i
v_i e^{-ct}= -c v_i
e^{-ct}= -c
?
Justin Lazear
Nov8-04, 11:09 PM
You're confusing v_i and v(t).
v_i is the initial velocity, i.e. the velocity at time = 0, or v(0).
v is the velocity as a function of time, i.e. v_i e^{-ct}, or v(t).
Remember that v_i is a constant.
--J
UrbanXrisis
Nov8-04, 11:16 PM
okay, I understand, thanks for all your help.
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