Differentiate that equation for v(t)

[UrbanXrisis]

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 textbook. Thanks

 

[Justin Lazear]

http://mathworld.wolfram.com/Derivative.html

--J

 

[UrbanXrisis]

differentiate = take the derivative?

wow...I must not be paying attention in math class

 

[UrbanXrisis]

I don't 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]

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]

Where did the -c come from? Doesn't the derivative of Vi become c?

 

[Justin Lazear]

$$\frac{d}{dt}v(t) = \frac{d}{dt} \left( v_i e^{-ct} \right)$$

$v_i$ 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]

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]

$$v = v_i e^{-ct}$$

Plug v into the book's expression for a and see if it matches up with what you got. Find the e?

--J

 

[UrbanXrisis]

$$v = v_i e^{-ct}$$
$$a = -c v_i$$

$$v_i e^{-ct}= -c v_i$$
$$e^{-ct}= -c$$
?

 

[Justin Lazear]

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]

okay, I understand, thanks for all your help.