Expert: Paul Klarreich - 1/18/2008

Question
QUESTION: if I have a function s(t) =t^3-3t^2+t+3 that describes distance of a particle from the origin on the time interval 0<=t<=3
How do I find the total distance traveled by the particle on the time interval
thanks!

ANSWER: Questioner:   vince
Category:  Calculus
Private:  No
 
Subject:  calculus
Question:  if I have a function s(t) = t^3 - 3t^2 + t + 3 that describes distance of a particle from the origin on the time interval 0<=t<=3
How do I find the total distance traveled by the particle on the time interval
thanks!
..........................
Hi, Vince,

You can't just do  s(3) - s(0) because the particle may have reversed direction. (As you probably figured.)  A quick-and-dirty graph shows a couple of turning points, one near t = 1/3 and one near t = 2. So we should find them

s'(t) = 3t^2 - 6t + 1

Set that = 0 and solve:

3t^2 - 6t + 1 = 0
   6 +- sqrt(24)
t = ---------------
        6

   6 +- 2 sqrt(6)
t = ---------------
        6

   3 +- sqrt(6)
t = ---------------
        3

Here's the drill, which I'll leave to you.

1. Compute those (Use Excel?) and call them t1, t2.

2. Substitute to find:

s(0), s(t1), s(t2), s(3)

3a. Find abs(s(t1) - s(0))
3b. Find abs(s(t2) - s(t1))
3c. Find abs(s(3) - s(t2))

4. Add those three numbers.


---------- FOLLOW-UP ----------

QUESTION: I have a question on l'hospitals rule

how can I use it to find lim (1/x -cotx) as x->0

Also lim x/(ln(1/x))as x-> 0

Answer
Questioner:   vince
Category:  Calculus
Private:  No
 
Subject:  calculus

QUESTION: I have a question on l'hospitals rule

how can I use it to find lim (1/x -cotx) as x->0

Also lim x/(ln(1/x))as x-> 0
.........................................
HI, Vince,
Thank you for the kind comments on the last one.  
This is not a followup -- it is a completely new question. (But you can just send it; as long as I am not 'maxed' out you can send new questions any time.)

Using l'Hospital's Rule, you are required to have something of the form:
     f(x)
lim   -----
x->?? g(x)

and you have something like:

lim  (f(x) - g(x))
x->??

So you apply the Clint Eastwood Rule. (Man's gotta do what a man's gotta do.)

You find a way to change it to a single term

lim   (1/x - cot x)
x->0

    1 - x cot x
lim (------------)  << LCD = x
         x

Now that will approach 0/0, because
         x cos x     x
x cot x = ------- = ------ cos x --> 1 * 1 = 1
         sin x   sin x

Use the rule now:

   0 - x(-csc^2(x)) - cot x
lim ------------------------ =
         1

     x/sin^2(x)) - cos x/sin x
lim ------------------------------- =
         1


    x - sin x cos x
lim ------------------ =
        sin^2(x)

--> 0/0 again, so one more go-round with the rule:
   1 - [- sin^2(x) + cos^2(x)]
lim ---------------------------
    2 sin x cos x

   1 - cos 2x
lim -----------
     sin 2x

We're getting there. [said the dentist to the patient, pulling harder on the tooth.]

    2 sin 2x         0
lim ---------- --> ------- = 0
    2 cos 2x         1
..............................................

lim x/(ln(1/x))as x-> 0
      x
lim ---------
    ln(1/x)
         zero
This looks like   ------
         infinity

which is not an indeterminate form.  The limit is zero.