Solving for a variable, transposing equation


by Pharrahnox
Tags: equation, friction, solving, transposing, variable, velocity
Pharrahnox
Pharrahnox is offline
#1
Dec30-13, 08:28 PM
P: 86
I have an equation, where acceleration is affected by a driving power and air resistance. The acceleration is given by:

a = (2P / m + u^2)^0.5 - (k*p*A*u^2 / 2m) - u

I'm trying to make "u" the subject, which is previous velocity, to find at what velocity does acceleration become 0, the maximum speed. However, this has proven to be very difficult for me, and the closest I have gotten is:

8P / k*m*p = k*p*u^4 + 4*u^3

However, this does not have "u" by itself. I have gotten "u" by itself, but it requires cube rooting or even 4th rooting (I don't know the term) the other side of the equation, and still has "u" in that...

I'm hoping there's a nice easy way of fixing this, but I can't seem to find it. I have tried online calculators, and my ClassPad, but they give very large and complex answers.

Thanks for any help.
Phys.Org News Partner Mathematics news on Phys.org
Researchers help Boston Marathon organizers plan for 2014 race
'Math detective' analyzes odds for suspicious lottery wins
Pseudo-mathematics and financial charlatanism
SteamKing
SteamKing is offline
#2
Dec30-13, 09:19 PM
HW Helper
Thanks
P: 5,552
Your second equation is a quartic polynomial in u. There is a rather complicated formula to find all of the roots such that f(u) = 0.

http://en.wikipedia.org/wiki/Quartic_function
Pharrahnox
Pharrahnox is offline
#3
Dec30-13, 11:17 PM
P: 86
I went to this website (http://www.1728.org/quartic2.htm) to figure out the quartic equation, I haven't done them before, and I came up with a cubic equation of y^3-3y^2-3y-1=0, which then gave:

x1,2 and 3 = 1-1/kp
x4 = -3-1/kp

But I don't think this is correct somehow, or is not the end result, because it gives values of about -2000 for specific values of k and p, when it should be about 32.69.

What should I do with the quartic equation? I put it in the form: k*p*u^4 + 4*u^3 - 8*P / k*m*p

I tried to graph the equation and this gives me the correct result at x = 0 (y being the maximum speed): x = (8*P / k^2*p^2*m - 4*y^3 / k*p)^0.25

But it is not as straightforward as I had hoped. Will the quartic equation solver give me the answer I'm looking for?


EDIT: I have managed to figure out the formula for it: u = cuberoot(2*P / k*m*p), but it took a fair few steps, and one of them still doesn't properly make sense to me.
Even though I have the answer, I would still like to have a better understanding of how to work through it, so that I can replicate it for other equations.

NascentOxygen
NascentOxygen is offline
#4
Dec31-13, 06:32 AM
HW Helper
P: 4,715

Solving for a variable, transposing equation


If you want the solutions of a quartic to be in algebraic form, you will find it disappointingly complicated and unwieldy.

A quartic will have one, two, or four real solutions. If you are able to express the polynomial with numerical coefficients so that the only algebraic term is the unknown, x, then you can use a plotting facility to reveal the graph, as well as provide that equation's numeric solutions.

e.g., an example using wolframalpha http://m.wolframalpha.com/input/?i=y...-1%3D0&x=0&y=0
nmf77
nmf77 is offline
#5
Dec31-13, 10:24 AM
P: 13
something worries me slightly about your original equation .... I'm not sure it looks sensible in terms of dimensions/units. You have one term on the left (acceleration), and three terms on the right. The third term on the right is velocity I think, which obviously is dimensionally different to acceleration. Perhaps I'm being a bit dim, but I can't see how that be correct. If you could explain what the various terms are that might be helpful in understanding the overall problem
sjb-2812
sjb-2812 is offline
#6
Dec31-13, 11:02 AM
P: 418
Quote Quote by NascentOxygen View Post
If you want the solutions of a quartic to be in algebraic form, you will find it disappointingly complicated and unwieldy.

A quartic will have one, two, or four real solutions. If you are able to express the polynomial with numerical coefficients so that the only algebraic term is the unknown, x, then you can use a plotting facility to reveal the graph, as well as provide that equation's numeric solutions.
This should be zero, two or four, surely?
Pharrahnox
Pharrahnox is offline
#7
Dec31-13, 08:27 PM
P: 86
a = (2P / m + u^2)^0.5 - (k*p*A*u^2 / 2m) - u

^ 1 ^ 2 ^ 3

1 - The velocity due to power being supplied, driving it forward, using power to directly change kinetic energy.
3 - Is the velocity before that power was supplied, so velocity - previous velocity is the change in velocity.
2 - Is the negative acceleration due to air resistance. In my equation, I replace the reference area "A" with mass "m", as it is proportional for my purposes anyway.

I guess, after looking at this, it is more finding the change in velocity rather than acceleration, because time isn't important. So I am trying to find the point at which the force from the driving power and the force from friction cancel out, to find the maximum speed.

The formula I put in my second post isn't actually correct, and it is actually:

u = cuberoot( 2*P / k*m*p - k*p*u^4 / 4)

But the problem is that there is "u" on both sides of the equation, so it hardly solves anything.

EDIT: The wolfram site gave me the correct answer, so the equation to graph it is:

k*p*u^4 + 4*u^3 - (8*P / k*m*p) = 0

How can this be converted to make "u" the subject? If there is an easy way...
NascentOxygen
NascentOxygen is offline
#8
Dec31-13, 10:25 PM
HW Helper
P: 4,715
Quote Quote by sjb-2812 View Post
This should be zero, two or four, surely?
That's a good point, but I was focussed on emphasising in the vernacular of the poster that it may be that there is only one real number that satisfies the equality condition.

It can be left for another time to explain how one can be counted as two, etc.


Register to reply

Related Discussions
Transposing formulas/Solving for variables General Math 9
Solving an equation with variable as denominator? Calculus & Beyond Homework 3
Solving a two variable equation Calculus & Beyond Homework 4
Solving non-homogeneous differential equation with variable coefficients Differential Equations 3
Exponential equation, solving for one variable Precalculus Mathematics Homework 11