A particle is moving on the x-axis, where x is in centimeters. Its
velocity, v, in cm/sec, when it is at the point with coordinate x is
given by:

v = x^2 + 3x - 2

Find the acceleration of the particle when it is at the point x = 2.
Give units in your answer. Please write this problem up in a step by
step process, I know the answer I just need it spelled out for those
who don't. Thank you.

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Clarification of Question by wannabeleader-ga on 11 Jul 2005 18:53 PDT

I used the chain rule instead of the power rule, but I don't know when
I'm suppose to use the chain rule and when I'm not suppose to.

This is my inaugural post.  Please let me know your honest opinion.

This comment assumes that you (or the people to whom this explanation
will be given) have a basic understanding of the concepts of velocity
and acceleration and their relationship to each other.  It also
assumes a basic knowledge of the "Power Rule" from Calculus (see
definition here http://mathworld.wolfram.com/PowerRule.html).  Note
that since the units all "match," I will leave them out at first and
save unit analysis for the end of my work.  It looks a bit neater this
way.

Here goes:
Recall that acceleration is defined as the change in velocity over a
given time interval.  That is, acceleration = (change in
velocity)/(change in time).

Another way of writing this is a = dv/dt.  

We are given v = (x^2 + 3x - 2); we must compute dv/dt.

dv/dt = [d/dt](x^2 + 3x - 2).  

This can be broken down into = [d/dt](x^2) + [d/dt](3x) - [d/dt](2).

[d/dt](x^2) = 2x by the power rule (as defined in the above link, with n = 2)
[d/dt](3x) = 3 by the power rule (as defined in the above link, with n = 1)
[d/dt](-2) = 0 by the power rule (as defined in the above link, with n = 0)

Piecing this back together, we have dv/dt = 2x + 3.

Recall that a = dv/dt = 2x + 3.

Now we have our general acceleration equation; the only thing left to
do is to substitute "2" for "x."

So for x = 2, a = 2(2) + 3 = 4 + 3 = 7.

Now to deal with the units.  Velocity is given in cm/sec and x is given in cm.

So the units of dv/dt will be the units of v divided by the units of
t.  That is, (cm/sec)/sec, or cm/(sec^2).

So the acceleration at x = 2 will be a = 7 cm/(sec^2).

I now see that several people have brought your attention to the fact
that problems resembling homework and exam problems are not meant to
be answered.  I have answered this question in good faith, and was not
aware of this stipulation until after posting my earlier comment.  I
don't see a way to remove my prior comment, so please accept this
disclaimer that my response is not intended for use as an answer to a
homework, exam, or quiz problem.  I trust your integrity will steer
you in the right direction.

I'm a little confused on the how you got your derivatives. My end
answer was 56 cm/sec^2

Isn't the chain rule only used when you are multiplying two numbers?

Be careful...the chain rule and power rule are different things.  The
power rule simply states that if you have x raised to some power n,
the derivative of x is: n times x raised to the n-1.  That is:
derivative of x^n = nx^(n-1)  Example: n = 2: derivative of x^2 = 2x^1
= 2x (using n = 2).

I'm not sure exactly where your trouble lies.  Can you offer an
explanation of the method you used to arrive at 56 cm/sec^2?

Sorry, I was confused for a minute. I think I fixed the derivates. Do
you mind checking my work?

DV/DT = D/DX(x^2 + 3x - 2) times DX/DT = (2x + 3)v = (2x +3)(x^2 + 3x - 2)

Substitute x for 2 and:

DV/DT = (2(2) + 3)(2^2 + 3 times 2 - 2) = 56 cm/sec^2 

Does this look right?

I used the chain rule, but I don't know when I am suppose to use it
and when I'm not.

Oh, I misread something in your original problem statement.  Let me
get back to you in a little while.

Ah, I've been brainwashed.  I have done most of these sorts of
problems with velocity as function of time, and here it is
position-dependent.  I didn't even realize that I was mixing my "x's"
and "t's" until just now.

Now, for your long awaited answer:

You can do the problem using the method and steps I stated, but the
part where I mispoke is that it should be dv/dx rather than dv/dt.  So
everywhere you see a "t" in the original answer, make it an "x."  Then
you will see that you don't need to use the chain rule.  I've done so
many of these problems before that were of the form v = at^2 + bt + c
that it didn't even occur to me that you wrote "x's" on the right hand
side and not "t's."

So you have 

a(x) = dv/dx = [d/dx](x^2 + 3x - 2) = 2x + 3
a(2) = 2(2) + 3 = 7 cm/sec^2

Make sense now?

Thank you. Maybe you could help me with my other problems?

v=x^2+3x-2   (1)
a=d/dt(v)    (2)

(2)  ->  a=(dx/dt)*(d/dx)(v)=v*(d/dx)v   :chain rule      (3)
(1)  ->  (d/dx)v=2x+3                                     (4)
(3),(4)  ->   a=(x^2+3x-2)*(2x+3)=2x^3+9x^2+5x-6
              a(x=2)=56 cm/s^2

v(x)=dx/dt=x^2+3x-2
a(x)=d/dt[v(x)]=2x(dx/dt)+3(dx/dt)

if x=2 then dx/dt=4+2*3-2=8
a(2)=2*2*8+3*8=56

as units were not changed, they should be consistent now.

a(x=2)=56 cm/sec^2