Expert: Scotto - 9/3/2008

Question
QUESTION: I do not know if you will understand.

I am retired and amusing myself by studying relativity. I think that I am quite well advanced in my understanding, but I keep banging my head against simple problems. The queries are so simple that I feel ashamed, but I simply cannot find an answer. I hope that you can help me.

1. Momentum is mv. Force is mv^2.
I recall 'acceleration' as being d/t^2, not d^2/t^2.
What is the significance of the squaring of the 'distance' in 'v^2'?

2. In the expression 'd^2x/dt^2',the 't' is squared in the denominator but the 'd' is squared in the nominator (not the 'x'). Why?

I hope that you will understand that I can get my head around 'time dilation' and length contraction' and stuff. It is the nitty-gritty which stumps me!

Thank you for your help.

James Watson.




ANSWER: 1. Acceleration is really d^2f/dt^2 where f(t) is the position function.

2. The number refers to how many times the derivative is taken. d^2f/dt^2 means the second derivative of f with respect to the variable t.

More notes:

For derivatives, d is always used.  Note that d^n in these problems refers to the nth derivative.  The power in the top and bottom is always the same on derivatives and refers to which derivative is being looked at.  Whenever you have a d^n on top, the sum of 'exponets' on the bottom must be the same.

Suppose that we have f(x)=2x^5, then
df/dx=the derivative of f(x)=10x^4,
d^2f/dx^2=the second derivative of f(x)=40x^3,
d^3f/dx^3=the third derivative of f(x)=120x^2, and
d^4f/dx^4=the fourth derivative of f(x)=240x.

Another way to write derivatives is df/dx=f'(x), d^2f/dx^2=f''(x),
d^3f/dx^3=f'''(x).

The reason n is put after the d on top is so we know that it is the nth derivative of whatevers next.  The reason it is put after the variable on the bottom is so that we know which variable is the function is being differentiated against.

For example, if we had f=x^3y^4z^5=(x^3)(y^4)(z^5),
df/dx=3x^2y^4z^5, d^2f/dx^2=6xy^4z^5, d^2f/dxdy=12x^2y^3z^5,
d^3f/dxdz^2=(3)(5)(4)x^2y^4z^3=60x^2y^4z^3.


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

QUESTION: Thank you for your reply.
I think that, once I have studied your reply to the second part of my question, I will 'get it'. Yes, I think so.

I do not think that I explained the first part of my question clearly. It was meant to be very simple but I think that I confused the issue by using the letter 'd'.

What I meant to say was that I always thought that acceleration is simply 'distance per second per second', so that a cyclist, for example, by pedaling harder,  might increase his speed by 10 meters per second per second. What I do not understand is why, in the expression 'v^2', the distance is squared as well as the time. Can you tell me what the significance of the squaring of the distance is? Can you give me an imaginary example (similar to my cyclist example)?

Thanks for your time.

Answer
Let the space travelled by this bicyclist be s.  His velocity is v=ds/dt.  His acceleration is a=dv/dt=d²s/dt².  The ² in this expression does not refer to squaring the values, but since it has the d's signifying a derivative is to be taken, ² means taking the second derivative.
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That is why s is used for distance (space travelled).  In mathematics, c²b/ce² is known to be (c squared times b) over (c squared times e squared).  This is reduced to cb/e² before anything else is done, since mathematicians usually reduce problems to lowest terms defore proceeding.

If it were d²b/de², that implies that we are looking at the second derivative of b with respect to e.  There b must really be b(e), a function of e.  It's dependence on e is not known at this time since the derivative wasn't given.
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Again, it does not refer to a square, but to the second derivative when there is a d in the numerator with the same exponet and a d in the denominator.  The d must be the first letter and there must be one function in the numerator and one variable in the denominator for each derivative taken.
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For example, suppose f(x,y) = sin(x)cos(y).
We know that d(sin(x))/dx=cos(x) and that d(cos(x))/dx=-sin(x).

Given this, it can be seen that df/dx = cos(x)cos(y),
df/dy = -sin(x)sin(y), df²/dx² = -sin(x)cos(y),  
d²f/(dxdy) = -cos(x)sin(y), and d²f/dy² = -sin(x)cos(y).
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Once again, the ² does not refer to d squared, x squared, or y squared, but to the second derivative.