Originally posted by mtthwWould you mind explaining why? I was never really comfortable with multiplying both sides by a differential in those sort of differential equations. I know that the solutions worked in the problems given by my professors, but never really understood the formal justification for that method.
'Fraid that doesn't work 🙂. You can't do that with a second derivative.
Roughly speaking, it works with the first derivative because dy/dx = lim[delta y/delta x] as delta x -> 0. Which means that dy/dx = 1/(dx/dy).
Which means that re-arranging in the 'natural' way works. But there's no similarly simple relationship between d2y/dx2 and d2x/dy2.
Just to consider a simple example: y = x^2 (so x = y^1/2)
dy/dx = 2x
dx/dy = 1/(2y^1/2) = 1/(2x) = 1/(dy/dx)
But d2y/dx2 = 2
d2x/dy2 = -1/(4y^3/2)
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Originally posted by mtthwusing trigonometric substitution I believe I have it here is how, and what I arrive at...if someone will check for correctness to this point I would appriciate it.
Using the energy argument, you start from:
mv^2/2 - GMm/r = -GMm/R
Which gives...I think!...
int{r, R} sqrt[r/(1 - r/R)] dr = -t sqrt(2GM)
Not a nice integral to do by hand, though.
Int{r,R}sqrt[r/(1-r/R)]
basic triangle
Hyp: 1
OPP: sqrt( r/R )
ADJ: sqrt ( 1-r/R )
let the angle be represented by "A"
such that the new integral becomes:
sqrt (R) Int{ tanA dr }
for the differential dr:
sinA = sqrt(r/R)
r = R*(sinA)^2
dr = 2R*sinA*cosA dA
plug it in , and simplify.
2*R^(3/2)*Int{ (sinA)^2 dA } = R^(3/2)*Int{ (1 - cos(2A))dA}
= R^(3/2)*[ A - 1/2*sin(2A)]
then changing back to original variable r
= R^(3/2)*[ arcsin(sqrt(r/R)) - 1/2*sin(2 arcsin(sqrt(r/R)))]
just evaluate ( wasn't sure what limit was what) and simplify(which I have not investigated, but have a hunch will be moderately fruitful).
Eric
What the heck I did the evaluation ( assuming "r" was lower limit )
and come to
R^(3/2)*[ arcsin(sqrt(r/R)) - sqrt((r/R)(1- r/R)) - Pi/2 ] =-t sqrt ( 2GM )
Originally posted by mtthwAnd isnt this for position as a function of time? I thought the original poster was asking for velocity as a function of time?
Using the energy argument, you start from:
mv^2/2 - GMm/r = -GMm/R
Which gives...I think!...
int{r, R} sqrt[r/(1 - r/R)] dr = -t sqrt(2GM)
Not a nice integral to do by hand, though.
Originally posted by joe shmoWell, once you've got that, differentiate. Since the force is in terms of distance, you've no real choice but to solve for distance(t) first.
And isnt this for position as a function of time? I thought the original poster was asking for velocity as a function of time?
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Originally posted by mtthwnevermind...when I first looked at it I thought "t" was first degree, now looking at it I realize if I had solved for r(t) explicitly its not. Now Im not even sure its possible to solve for r(t) explicitly. 😕 As a side: Implicit differentiating doesnt look all that inviting either.
No, it just implies r is a function of time. What makes you think it says velocity is constant?
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Originally posted by wittywonkaYou have one independent (time) and one dependent variable (position). It's not multivariable.
I am in AP Calculus in high school, but I have a question about a physics problem I made up that I assume is a multivariable calculus problem.
Force gravitational = GMm / (r^2)
Therefore,
Acceleration gravitational = GM / (r^2)
So, here's the problem I came up with. Let's say you take a basketball and travel out into space, approximate ariable problem.
Any explanations in addition to any answers would be greatly appreciated.
If it were three dimensional motion with each dimension differently dependent on time then I think it would be multidimensional.