Expert: Paul Klarreich - 4/29/2008

Question
A particle that moves along a straight line has velocity
v(t)=t^(2)*e^(-3t)
meters per second after t seconds. How many meters will it travel during the first t seconds?

I did the integration by parts two times. I then got a huge thing MINUS when t=0. My work looks correct.  I don't know why it's wrong according to webwork. I just want to compare my work.  I hope you can help!!  

Answer
Questioner:   Rahma
Category:  Calculus
Private:  No
 
Subject:  integration by parts
Question:  A particle that moves along a straight line has velocity
v(t)=t^(2)*e^(-3t)
meters per second after t seconds. How many meters will it travel during the first t seconds?

I did the integration by parts two times. I then got a huge thing MINUS when t=0. My work looks correct.  I don't know why it's wrong according to webwork. I just want to compare my work.  I hope you can help!!
............................................
Hi, Rahma,

Your velocity is:

v = t^2 exp(-3t)

Doing IBP:  u = t^2, du = 2t dt
dv = exp(-3t),   v = -1/3 exp(-3t)

uv - INT(v du)

-t^2/3 exp(-3t) - INT(-2t/3 exp(-3t) dt)

-t^2/3 exp(-3t) + 2/3 INT(t exp(-3t) dt)

Do it again (as you noted)

u = t, du = dt
dv = exp(-3t),   v = -1/3 exp(-3t)

-t^2/3 exp(-3t) + 2/3[-1/3 t exp(-3t) - INT(-1/3 exp(-3t) dt)]

-t^2/3 exp(-3t) + 2/3[-1/3 t exp(-3t) + 1/3 INT(exp(-3t) dt)]

-t^2/3 exp(-3t) + 2/3[-1/3 t exp(-3t) - 1/9 exp(-3t)]

-t^2/3 exp(-3t) - 2/9 t exp(-3t) - 2/27 exp(-3t)

OK, now, do that from 0 to t.

-------------------- at t = t -----------------  --- t=0 ---
-t^2/3 exp(-3t) - 2/9 t exp(-3t) - 2/27 exp(-3t) - [-2/27]

[at x = 0, the first two terms vanish, and the last exp() is 1.]

-t^2/3 exp(-3t) - 2/9 t exp(-3t) - 2/27 exp(-3t) + 2/27

How does that look?

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If you blew the sign of the last term (as you indicated), you could have discovered your error as follows:

x(t) = -t^2/3 exp(-3t) - 2/9 t exp(-3t) - 2/27 exp(-3t) - 2/27
-------------------wrong sign here--------------------->|

Now find x(0):

x(0) = -0^2/3 exp(-3(0)) - 2/9 (0) exp(0) - 2/27 exp(0) - 2/27

x(0) =  - 2/27 - 2/27 = - 4/27

But x(0) should be zero.  [In t = 0 to t = 0, how could the particle move anywhere?]

In many situations, it is good to have a way to find your own errors.  There are few who are perfect. (besides me, of course.)