Expert: Paul Klarreich - 4/23/2006

Question
Hi Paul,
A particle moves along the x-axis so that its velocity at time t is given by v(t)=-(t+1)sin(t^2/2)
At time t=0, the particle is at position x=1.
During the time interval 0<=t<=3, what is the greatest distance between the particle and the origin.
I don't understand how they get their answer([they did the integral of v(t) from 0 to sqrt(2pi) then xsqrt(2pi)=x(0)+ integral from 0 to sqrt(2pi)of v(t)=-2.265. Since total distance from t=0 to 3 is 4.334[I do understand how they got 4.334, but not anything beyonf that.Please help me.], the particle is still to the left of the origin at t=3so answer is 2.265) and also don't understand why they don't use derivative analysis for finding the maximum.
This comes from free response 2003formA#2d at apcentral.com(collegeboard)

Answer
Hi, Jeff,

I found the reference.  Of course, they made me register and log in, but I was able to fake it.

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OK.  This is a rather nasty function to try to integrate directly.  If you could do that, you can find the position at any time t.  But I think you are expected to use a graphing calculator to find your definite integrals.  Alas, I don't have one of them, so I have to guess what you (they) are doing.

HOWEVER, if  v = -(t + 1) sin(t^2/2)  then the object changes direction at a value of t for which v(t) = 0:

Can we solve  -(t + 1) sin(t^2/2) = 0?

Obviously t = -1 is a solution, but this is not in your [0,3] interval, so forget that.

So we solve  sin(t^2/2) = 0.  Now sin A = 0 when A is a multiple of pi; just remember the graph of  y = sin x.  It crosses at pi, 2pi, etc.

If  t^2/2 = pi,  then  t = sqrt(2 PI) = 2.506, which is within our [0,3] interval, so we can use it.

Which way was it going before and after it changed direction?  If  t < sqrt(2pi), then  t^2/2 < pi, and sin(t^2/2) is positive.  Then we have:

v(t) = -(t  +   1)(sin(t^2/2)
v(t) = -(positive)(positive ) = negative.  

In other words, from t=0 to 2.506, the object is moving to the left.  When t > 2.506, it will be moving to the right.

Now the total distance traveled in either direction (I suppose this is to compute your gasoline consumption) was  4.334.  They got that by (numerically) integrating the absolute value of the v(t) from 0 to 3.

But we can split the interval into before and after the 'reversing'.  Before it, we integrate v(t) from 0 to 2.506, again numerically, and we get -3.265.

Subtract:   3.265 from 4.334, and you get 1.069.  So we know that the object, during this period, moved  3.265 to the left, then turned and moved 1.069 to the right.  Where was it at the start?

Oh, yes, it says:  at t = 0,  x = 1.  So now we can trace the movement of the particle.

At t = 0, it was at x = 1.
From t = 0 to t = 2.506, it moved 3.265 to the left, reaching -2.265.  
Then it moved 1.069 units to the right, reaching about -1.2.

So the farthest it ever was from the origin was at its turning point, -2.265.

Do you see that the initial condition, x(0) = 1,  is important?  Suppose we had  x(0) = 1000.  Then the object started at 1000, moved a few spaces to the left, went back about one space, and so never got back to 1000, which would have been the farthest it ever got.