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Not a test; just a personal challenge

  • Thread starter LareeRudi
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  • #1
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I've been working on this for years; self determination; there IS a time to say "Help"; this is it.

Equations of motion; I'm well acquainted with them, but THIS one is unique, and I can't get to first base.

V is Velocity
X is distance/location
t is time

X = 4*(t^2) - 15 - K*V

so I tried this:

X = 4 * (t^2) - 15 - K * (dx/dt)

Then I shuffle stuff around, and the dx and dt end up in places that I can't handle.

1 Is it possible to get this into a form where X is a function of t?
2 Then a form where V is a function of t?
3 And finally, where V is a function of X?

The first one is the most important to me. And ya MIGHT hafta hold my hand, so don't skip TOOO many steps.

thx,

LarryR : )
 
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Answers and Replies

  • #2
berkeman
Mentor
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I've been working on this for years; self determination; there IS a time to say "Help"; this is it.

Equations of motion; I'm well acquainted with them, but THIS one is unique, and I can't get to first base.

V is Velocity
X is distance/location
t is time

X = 4*(t^2) - 15 - K*V

so I tried this:

X = 4 * (t^2) - 15 - K * (dx/dt)

Then I shuffle stuff around, and the dx and dt end up in places that I can't handle.

1 Is it possible to get this into a form where X is a function of t?
2 Then a form where V is a function of t?
3 And finally, where V is a function of X?

The first one is the most important to me. And ya MIGHT hafta hold my hand, so don't skip TOOO many steps.

thx,

LarryR : )
Welcome to the PF. You don't break up dx/dt, it's just the velocity. If you had x(t), then you could differentiate it to get v(t) = dx(t)/dt.

You have already written x(t) ("x as a function of time"). Does that not give you what you want for 1?
 
  • #3
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Welcome to the PF. You don't break up dx/dt, it's just the velocity. If you had x(t), then you could differentiate it to get v(t) = dx(t)/dt.

You have already written x(t) ("x as a function of time"). Does that not give you what you want for 1?
It's a function of time and velocity. I think he seeks an explicit solution using time to the differential equation.

http://www44.wolframalpha.com/input/?i=y+%3D+4*%28x^2%29+-+15+-+K*y%27

as in "Differential equation solution:
y(x) = c_1 e^(-x/K)+8 K^2-8 K x+4 x^2-15" from the link above

by the way, if you find a way to get #1, you can always get #2 by differentiating the solution from #1. For #3, you can solve for time in the distance equation(it will be time as a function of distance) Then, substitute that solution for t into all the t's present in the function v(t) you derived. I'll leave someone more comfortable with diffyqs to help you with #1(which is the beginning to finding #1 and #2)
 
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  • #4
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Welcome to the PF. You don't break up dx/dt, it's just the velocity. If you had x(t), then you could differentiate it to get v(t) = dx(t)/dt.

You have already written x(t) ("x as a function of time"). Does that not give you what you want for 1?
Yeah, I think the guy on post num 3 explains it better than I can; you see my ORIGINAL equation has X and t and V......................... I don't WANT three variables; that's why I suggested we use dx/dt for the Vel............... leaving just X's and t's to work with, but as I explained, the dx and dt come out in places unfamiliar to me when I shuffle them around.

thx,

LarryR : )
 
  • #5
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It's a function of time and velocity. I think he seeks an explicit solution using time to the differential equation.

http://www44.wolframalpha.com/input/?i=y+%3D+4*%28x^2%29+-+15+-+K*y%27

as in "Differential equation solution:
y(x) = c_1 e^(-x/K)+8 K^2-8 K x+4 x^2-15" from the link above

by the way, if you find a way to get #1, you can always get #2 by differentiating the solution from #1. For #3, you can solve for time in the distance equation(it will be time as a function of distance) Then, substitute that solution for t into all the t's present in the function v(t) you derived. I'll leave someone more comfortable with diffyqs to help you with #1(which is the beginning to finding #1 and #2)
Yes, I see what you MEAN; but that is MY specifice dilemma tooo; to find a solution for num one.

LarryR : )
 
  • #6
258
0
Just simplify the equation.
[tex]
x(t) = 4t^2 - 15 -kx'(t)
[/tex]
Therefore...

[tex]
x'(t) + \frac{x(t)}{k} = \frac{4t^2 - 15}{k}
[/tex]
Can you solve that?

If not, what happens if you multiply both sides by [tex]e^{\frac{t}{k}}[/tex]?
 
  • #7
LCKurtz
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Yes, I see what you MEAN; but that is MY specifice dilemma tooo; to find a solution for num one.

LarryR : )
Didn't he just give you the solution to 1 (changing the variables to yours)?

x(t) = c1 e-t/K+8 K2-8 K t+4 t2-15
 
  • #8
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Didn't he just give you the solution to 1 (changing the variables to yours)?

x(t) = c1 e-t/K+8 K2-8 K t+4 t2-15
I think some of our messages got crossed, that is I was typing while a message was coming in..... however.............

I'm overwhelmed with all your help, and I've gotta leave for a while......... will come back and review all the above.

And thanks again................. if I can get this figured out, I'll have a taste of heaven, ha.

later,

PS; I'm pleasantly surprised at all the help one gets in a VERY SHORT time.

LarryR : )
 
  • #9
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Didn't he just give you the solution to 1 (changing the variables to yours)?

x(t) = c1 e-t/K+8 K2-8 K t+4 t2-15
[I had diff eq way back in 1956-1957 college year, so I've forgotten too much]

Two things;
1 I can't read the exponent in post 6 where they're trying to guide me into answer; it's just a ball of fuzz to my computer and I can't get it go larger; looks like exponent is t/k? but if so, I still wouldn't know WHAT to do with that. So I could still use some help.

Could you give me more of a step by step process to show your solution from my original stated problem? That'd be great if you could.

2 One of the things I want to accomplish is that if my "4*t^2" became "2 * t^2" so that I could handle it.

Thanks a million.

LarryR : )
 
  • #11
34
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It's a function of time and velocity. I think he seeks an explicit solution using time to the differential equation.

http://www44.wolframalpha.com/input/?i=y+%3D+4*%28x^2%29+-+15+-+K*y%27

as in "Differential equation solution:
y(x) = c_1 e^(-x/K)+8 K^2-8 K x+4 x^2-15" from the link above

..................................................)
Wow, am I wrestling with this; using all the resources that this forum has suggested, and going back to differentiation of products; integration of products; yes, you reminded me of the [forgot exactly what you called it ? coefficient of integration [the e^(t/k)] etc, and I'm not coming anywhere near close to your answer; Remember, I have LOTS of self doubts so not questioning you, but am wondering;

In the equation above, did you truly intend for the constant to be multiplied to the e^(t/k)? I expected the constant of integration to be "added" someplace.

I don't get all the "K" and K^2 factors you do, BUT I get some x^2 and x factors that you DON'T show..... hmmmmmmmmmmmmmm, I'm going in circles. Would really love if somebody had some time to show me more steps.

Geeeez, after I multiplied both sides by the e^(t/k) and did all the multitudinous steps [tedious to me] , THEN at the end, they ALL CANCELLED OUT; they ended up in EVERY FACTOR that I had come up with, and I HIGHLY suspect that I'm wrong.

Anything you can do to get me further, I'd appreciate. I'm starting to see the "approach", but am getting lost in the tediousness.

thx,

LarryR : )
 
  • #12
258
0
Just simplify the equation.
[tex]
x(t) = 4t^2 - 15 -kx'(t)
[/tex]
Therefore...

[tex]
x'(t) + \frac{x(t)}{k} = \frac{4t^2 - 15}{k}
[/tex]
Can you solve that?

If not, what happens if you multiply both sides by [tex]e^{\frac{t}{k}}[/tex]?
Hint #2:
Notice what happens after you multiply the left side by e^(t/k)
[tex]
e^{t/k}x' + \frac{x}{k} e^{t/k} = (e^{t/k}x)'
[/tex]

Therefore, our equation now becomes...

[tex]
(e^{t/k}x)' = \frac{4t^2 - 15}{k}e^{t/k}
[/tex]

Which shouldn't be too hard to solve.
 
  • #13
34
0
ok ok ok.......... I see what you're trying to do, to make me DO IT; MAKE AN ATTEMPT at it... I just read that THAT is what you people expect of me (smiling), ok, I HAVE been, but I'll post my results, and re-ask some of my puzzling questions:

I'm using the answer in post seven as a comparison to mine:

Post 7: x(t) = c1 e-t/K+8 K2-8 K t+4 t2-15
Mine: x(t) =.......... 8K^2-8Kt +4t^2-15 + C
notice that I don't have an e^(t/k) LEFT; it seems that it operated as a "catalyst", and then when I completed all my work in simplification at the end, it cancelled out ALL of my exponential factors on BOTH sides of the equation???????????

So then I ask, SHOULD they have all dropped out? why does post 7 have the C1e(-t/k) and I don't? Wow, I've done some tedious problems in my life, but this one sure qualifies as that, too. Should I have an exponent factor in my answer? why does THEIR exponent
have a negative value in it? mine was positive as you suggested.

[Edit; ok, maybe I see what post 7 did, they did everything I did while the e^(t/k) was still in all the factors, while the C was "bare"...... THEN they multiplied both sides by the NEGATIVE [ e^(-t/k) ] which then wiped out ALL the e^(t/k) factors, but then required multiplying the constant C by that neg exponent as well.......... right? ok, if that is what post 7 did, is it necessary to do that? does the constant NEED that factor in it for accuracy? ie, is my "bare" C incorrect? ]

Hoping I didn't have any typos.

thx,

LarryR : )


Hint #2:
Notice what happens after you multiply the left side by e^(t/k)
[tex]
e^{t/k}x' + \frac{x}{k} e^{t/k} = (e^{t/k}x)'
[/tex]

Therefore, our equation now becomes...

[tex]
(e^{t/k}x)' = \frac{4t^2 - 15}{k}e^{t/k}
[/tex]

Which shouldn't be too hard to solve.
 
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  • #14
258
0
[Edit; ok, maybe I see what post 7 did, they did everything I did while the e^(t/k) was still in all the factors, while the C was "bare"...... THEN they multiplied both sides by the NEGATIVE [ e^(-t/k) ] which then wiped out ALL the e^(t/k) factors, but then required multiplying the constant C by that neg exponent as well.......... right? ok, if that is what post 7 did, is it necessary to do that? does the constant NEED that factor in it for accuracy? ie, is my "bare" C incorrect? ]
Yes. The C requires the e^(t/k).

Just solve the ODE in my post.

You'll get

[tex]
\int \frac{d((e^{t/k}x))}{dt} dt = \int \frac{4t^2 - 15}{k}e^{t/k} dt
[/tex]

Now, let F(t) be the antiderivative of the right hand side.
[tex]

xe^{t/k} = F(t) + C \rightarrow x = e^{-t/k}F(t) + Ce^{-t/k}
[/tex]

And there's where the C factor comes from.
 

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