Expert: Josh - 4/29/2006

Question
Hi John,
I think I'm doing pretty well thanks to your great explanations. I do have one more question which I'm hoping you can answer for me. My AP exam is this Wednesday at 8AM.
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.
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
This comes from free response 2003formA#2d at apcentral.com(collegeboard)

Why does the following problem follow a different procedure?
Find the point on the curve y=sqrt(x) that is a minimum distance from the point (4,0). You do D=(x-4)^2+(y-0)^2 using the distance formula then take the derivative and set it = to 0. Why don't you use this procedure for the previous problem? What's the difference between the 2 problems that makes it that you can't you use the same method for both problems?
Thank you once again.
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Followup To
Question -

Answer -
Hi Jeff,

How's your preparation going for your AP exams?

Answer
Hey Jeff,

How did you go in the AP exam?
I guess you must be taking everything in your stride.

Cheers
..
.
Jeff,

The integral is non-trivial, I don't think you can use the methods you've learnt to find a closed-form solution (exact expression for x(t)). Nonetheless, an understanding of simple harmonic motion from physics would impart some meaning on this problem.

Given v(t)=-(t+1)sin(t^2/2), we should find its anti-derivative to determine the maximum displacement x(t) in 0<=t<=3.

In physics, if you consider an ideal object which undergoes simple harmonic motion -- such as a pendulum, it reaches its maximum displacement (furtherest point during the swing) when it changes direction (as v(t)=0). [This principle is carefully described in standard physics textbooks]. The basic form for the displacement and velocity functions may be described simply as
x(t)=A*sin(wt) and v(t)=x'(t)=Aw*cos(wt), where w is the angular velocity.

The idea is the same with v(t)=-(t+1)sin(t^2/2), we observed that the time axis is warped non-linearly and scaled by a constant, because we have "half t-square" instead of t, as our time-variable. Furthermore, the envelope (amplitude) of oscillation is modulated by a linear slope "t", which grows with time. If the observation period becomes infinite, the system will become unstable, in the sense that the output becomes unbounded. (If you want a graph of this, I can send you one via email)

The basic ideas behind harmonic motion still hold. There is still a sense of periodicity, albeit, in a strangely warped time-scale (proportional to t^2, not t). We expect the maximum displacement to be reached when v(t)=0. Clearly, in v(t), only the sin(t^2/2) part is periodic. Its zeros (nulls) appear at t^2/2=pi*n, for any integer n. This implies that x(t) has local maxima at t=sqrt(2*pi*n). Since the velocity (potential energy) increases with t, the amplitude of oscillation must also increase with time. Thus, we favor the largest value of t=sqrt(2*pi*n) closest to t=3. Called this to=sqrt(2*pi).

Now, x(t)= integral -(t+1)sin(t^2/2) dt = p(t)+r(t)+constant,
evaluated from t=0 to t=to=sqrt(2*pi) ......[#1]

where p(t)= integral -t*sin(t^2/2) dt;
     r(t)= integral - sin(t^2/2) dt;
     constant = x(0) = 1.

Let u=t^2/2, du=t dt,
p(t)= - integral sin(u) du, with limits u=0, u=pi.
   = [cos(u)]|u=pi - [cos(u)]|u=0
   = cos(pi) - 1
   = -2

r(t) is very tricky :P). Here, we will approximate this by taking the power series expansion of sin(u), where sin(u) ~= u - (u^3)/3! + (u^5)/5! - (u^7)/7! + (u^9)/9! - ... and n! represents n factorial (i.e., n*(n-1)*(n-2)*...*1). This is usually taught in first year calculus.
-r(t)~= integral u - (u^3)/3! + (u^5)/5! - (u^7)/7! + (u^9)/9! dt,
    [Note 1: note carefully that we are still integrating w.r.t. "t", not w.r.t. "u"]
    [Note 2: for convenience, I compute -r(t), not r(t)]
    [Note 3: the number of terms included in this approximation must be sufficiently large to ensure convergence (for an accurate numerical estimate)]
    = integral (t^2/2) - (t^6)/(8*3!) + (t^10)/(32*5!) - (u^14)/(128*7!) + (u^18)/(512*9!) dt,
    = [t^3/(3*2) -t^7/(7*8*3!) + t^11/(11*32*5!) - t^15/(15*128*7!) + t^19/(19*512*9!)] evaluated at t=sqrt(2*pi)
    = 1.2662

Finally, combining p(t),r(t) and the initial condition x(0), from [#1] we get
x(to) = -2-1.2662+1 = -2.2662 (to 4 decimal points)

Yes, quite a bit of work. A sketch solution should be enough if you are unlucky enough to be given such a question in the exam. Don't bother using the calculator unless specifically asked to do so. You don't earn much credit for knowing how to push the buttons on a calculator. It's your method (the ability to find a way through a problem) that is going to make you stand out. Your outline solution should make such an impression on the examiner, demonstrating what you've learnt.

In an exam, you cannot possibly replicate everything you're reading here. At the same time, I find the solution you referred to at "apcentral.com" rather unsatisfactory. You've got to find the right balance I guess.

Re: second question

Actually both are minimization problems that require you to set the first derivative to zero.
In the first, it involves a tricky integral (the sin(t^2) part) which cannot be solved analytically. [Aside: It is actually related to the normal distribution through the imaginary part of integral exp(it^2) dt, where i=sqrt(-1). Don't worry too much, you don't need to know this] Important thing is that we are evaluating the line integral from t=0 to t=to=sqrt(2*pi); and how did we determine the value "to"? We considered the solutions to x'(t)=v(t)=0 and restrict "to" to the maximum value in the interval [0,3] based on physical principles (refer to paragraph 5).
In the second, through substitution, we get an expression containing one variable which involves no integration. This is purely an algebraic problem.

Cheers.