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From: Gc on 14 Nov 2009 11:27
On 14 marras, 15:13, Peter <Poakfi...(a)msn.com> wrote:
> Hi! Please could someone help me? I would like to know how to solve
> the subject expression in terms of x and t. I gather the correct
> solution is  d^2x/dt^2 = 2x/t^2. If this is correct, how do I arrive
> to it? Could I present it just at it is?

Try to separate the variables.
From: Greg Neill on 14 Nov 2009 11:53
Peter wrote:
> On Nov 14, 10:29 am, "Greg Neill" <gneil...(a)MOVEsympatico.ca> wrote:
>> Peter wrote:
>>> On Nov 14, 8:36 am, KY <wkfkh...(a)yahoo.co.jp> wrote:
>>>> http://www.wolframalpha.com/input/?i=+d%5E2x%2Fdt%5E2+%3D+2x%2Ft%5E2.
>>
>>> Thank you very much. Do you mean that I could put in a paper (which I
>>> am writing) x"(t) = 2x(t)/t^2 without more explanation? Is it
>>> acceptable?
>>
>> No. It is not acceptable. x''(t) = 2x(t)/t^2 only holds
>> under particular conditions; it is not true in general.
>
> I understand it is true when the force is constant. Are there other
> conditions?

Sure:
- Acceleration is constant
- Initial velocity is zero
- Initial position (x) is zero

Example:

A car passes mile marker 3.0 doing 40 mph at time
t0 = 15 seconds, whereupon it accelerates at a contant
rate of 2 mph/second for 10 seconds, after which it
resumes moving with constant velocity. At what time
will it pass mile marker 20? What will be its velocity
at that time?


From: glird on 14 Nov 2009 12:33
On Nov 14, 8:13 am, Peter <Poakfi...(a)msn.com> wrote:
> Hi! Please could someone help me? I would like to know how to solve the subject expression in terms of x and t. >

The subject expression, d^2x/dt^2, is a second differential one. I
think its "solution" would therefore be a first differential
expression, which in this case would be dx/dt.

><I gather the correct solution is
d^2x/dt^2 = 2x/t^2. >

An "expression" is a member of an equation, not the equation itself.
You wrote an expression but not an equation in which it appears. I
think it would, by itself, be equal to 2x/dt. If so, then
dx/dt = 2 units/second = v.
*IF* that is right, then d^2x/dt^2 is the rate of change of v; thus is
an acceleration, so
d^2x/dt^2 = dx/dt^2 = (dx/dt)/dt = a.
(In the latter equation, the "d" symbolizes "delta". In the
expression dx/dt it doesn't denote delta because the parameters of
length and time may be linear; i.e. the space and time they measure
are "homogeneous", i.e. uniform in all directions everywhere
regardless of which system measures them.)

glird

From: glird on 14 Nov 2009 12:45
On Nov 14, 12:33 pm, glird wrote:
>
>< The subject expression, d^2x/dt^2, is a second differential one.  I think its "solution" would therefore be a first differential expression, which in this case would be dx/dt. ... I think it would, by itself, be equal to 2x/dt. >

Please excuse me for writing 2x there. It should have been dx/dt, as
written prior to and after this typo.

glird

From: Ken Pledger on 15 Nov 2009 15:52
In article
<93498fd6-e0c0-44b3-8471-64cc9e8a1f27(a)j9g2000vbp.googlegroups.com>,
Peter <Poakfield(a)msn.com> wrote:

> Hi! Please could someone help me? I would like to know how to solve
> the subject expression in terms of x and t. I gather the correct
> solution is d^2x/dt^2 = 2x/t^2. If this is correct, how do I arrive
> to it? Could I present it just at it is?

Something isn't clear here. You've written a differential
equation for x as a function of t. The next step would be to solve it.
This solution process doesn't appear to be your main concern, i.e. it
doesn't look like homework. :-) So I'll just tell you the answer: x
must have the form a(t^2) + b/t where the constants a and b will
depend upon other conditions in your problem.

HTH

Ken Pledger.
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