# Differentiate that equation for v(t)

1. Nov 8, 2004

### 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 text book. Thanks

2. Nov 8, 2004

3. Nov 8, 2004

### UrbanXrisis

differentiate = take the derivative?

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

4. Nov 8, 2004

### UrbanXrisis

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?

Last edited: Nov 8, 2004
5. Nov 8, 2004

### 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

6. Nov 8, 2004

### UrbanXrisis

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

7. Nov 8, 2004

### 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

Last edited: Nov 8, 2004
8. Nov 8, 2004

### UrbanXrisis

thank you, that helps a lot. I was just wondering, my book gives an answer of a=-cv, where does the e go?

9. Nov 8, 2004

### 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

10. Nov 8, 2004

### UrbanXrisis

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

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

11. Nov 8, 2004

### 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

12. Nov 8, 2004

### UrbanXrisis

okay, I understand, thanks for all your help.

Last edited: Nov 8, 2004