Expert: Paul Klarreich - 3/7/2008

Question
QUESTION: A particle starts at the point (5,0) at t=0 and moves along the x-axis in such a way that at time t>0 its velocity v(t) is given by v(t)= t/ (1+t^2).

(A) determine the maximum velocity attained by the particle. Justify your answer.
(B) Determine the position of the particle at t=6.
(C) Find the limiting value of the velocity as t increases without bound.
(D) Does the particle ever pass the point (500,0)? Explain.

ANSWER: Questioner:   Joe Allen
Category:  Calculus
Private:  No
 
Subject:  CaLculus
Question:  A particle starts at the point (5,0) at t=0 and moves along the x-axis in such a way that at time t>0 its velocity v(t) is given by v(t)= t/ (1+t^2).

(A) determine the maximum velocity attained by the particle. Justify your answer.
(B) Determine the position of the particle at t=6.
(C) Find the limiting value of the velocity as t increases without bound.
(D) Does the particle ever pass the point (500,0)? Explain.
..............................
Hi, Joe,
         t
if v(t) = --------
         1 + t^2

Here is what you will do: [You didn't show any of the work you already did, so I assume you just don't know how to approach it.]

(a) Max velocity.  Differentiate v(t), using quotient rule.  Set that v'(t) (just the top) equal to 0 and solve for t.  Put that back into v(t) for your max.

(b) Position. INTEGRATE your v(t) to get x(t).  You get a '+ C', of course.  Put  t=0, x=5 into the equation for x to solve for C.  THEN put t = 6.

(c) Find lim[t->infinity] t/(1+t^2).  I think it will be zero.

(d) Set  x(t) = 500 and see if you get a solution for t.  If you do, say YES.


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

QUESTION: How you do a and b

Answer
Questioner:   Joe Allen
Category:  Calculus
Private:  No
 
Subject:  CaLculus
Question:  A particle starts at the point (5,0) at t=0 and moves along the x-axis in such a way that at time t>0 its velocity v(t) is given by v(t)= t/ (1+t^2).

(A) determine the maximum velocity attained by the particle. Justify your answer.
(B) Determine the position of the particle at t=6.
(C) Find the limiting value of the velocity as t increases without bound.
(D) Does the particle ever pass the point (500,0)? Explain.
..............................
Hi, Joe,
         t
if v(t) = --------
         1 + t^2

Here is what you will do: [You didn't show any of the work you already did, so I assume you just don't know how to approach it.]

(a) Max velocity.  Differentiate v(t), using quotient rule.  Set that v'(t) (just the top) equal to 0 and solve for t.  Put that back into v(t) for your max.

(b) Position. INTEGRATE your v(t) to get x(t).  You get a '+ C', of course.  Put  t=0, x=5 into the equation for x to solve for C.  THEN put t = 6.

(c) Find lim[t->infinity] t/(1+t^2).  I think it will be zero.

(d) Set  x(t) = 500 and see if you get a solution for t.  If you do, say YES.

===================================================

Questioner:   Joe Allen
Category:  Calculus
Private:  No
 
Subject:  Velocity and acceleration.

---------- FOLLOW-UP ----------
Hi, Joe,

I think your question:

QUESTION: How you do a and b

is not what you really mean to ask.  Since I already told you HOW to do a and b, you must mean:

QUESTION: Please do a and b for me.

If that's what you mean, say so.  And, please, read my Instructions to Questioners.  I ask you to tell me (show me) what you already did.  Then I can see what you are doing wrong and help you fix your errors or misunderstandings.  Your questions suggest that you have never heard of the Quotient Rule or how to integrate.  If that is so, you are taking the wrong subject -- it's too hard for you.


         t
if v(t) = --------
         1 + t^2

       (1)(1 + t^2) - (t)(2t)
v'(t) = ------------------------  
         (1 + t^2)^2

       1 + t^2 - 2t^2
v'(t) = -----------------  
        (1 + t^2)^2

        1  - t^2
v'(t) = ------------  
       (1 + t^2)^2

Set the top = 0:

You get  t = 1, t = -1, but t = -1 does not apply.  So t = 1 is your solution, and
        1
v(1) = ----- = 1/2 is your max velocity.
      1 + 1

INTEGRATE:

 t dt
-------
1 + t^2

Use u = 1 + t^2,  du = 2tdt

YOu get  x(t) = (1/2) ln(1 + t^2) + C.

Now use  x(0) = 5, given.

x(0) = 5 =  (1/2) ln(1 + 0) + C.

5 =  (1/2) ln(1) + C.

5 =  (1/2)(0) + C.

C = 5

x(t) = (1/2) ln(1 + t^2) + 5

Now  
x(6) = (1/2) ln(1 + 36) + 5.

x(6) = (1/2) ln(37) + 5.

That's it, unless you want to use your calculator and get a decimal answer.