Mastering the Tricky Equation: How to Solve for T Without a Calculator

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Homework Help Overview

The discussion revolves around solving a complex equation derived from a differential equation, specifically focusing on finding the variable t. The equation involves cubic terms and is set within the context of a physical scenario involving the amount of an element in a container over time.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore rewriting the equation to standard form and discuss the implications of the cubic nature of the equation. There are attempts to clarify the equation's structure and the significance of certain terms, as well as suggestions for rescaling variables to simplify the problem.

Discussion Status

The discussion is ongoing, with various participants providing insights and suggestions for approaching the problem. Some participants have offered methods for rewriting the equation and checking assumptions, while others are still grappling with the initial setup and seeking clarity on the problem's requirements.

Contextual Notes

Participants note that calculators are not permitted, which adds a layer of complexity to the problem-solving process. There are also mentions of initial conditions and constants that may influence the solution.

tedwillis
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Hi,

I need to solve this equation that stems from a differential equation, though that isn't really important. Here's the equation that needs to be solved for t:

3t+100=4((9t^3+900t^2+30000t)/(3t+100)…

I've tried myself and always seem to hit a dead end. Using a calculator, I know the answer is 100/3(2^2/3-1) or ~19.58, but can't find a way to get to this answer.

I should mention, I would like this in a fully worked solution if possible. The subject does not allow calculators.
 
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3t+100=4((9t^3+900t^2+30000t)/(3t+100)…
The three dots at the end of an equation usually means "and so on" - meaning to continue some sort of pattern. Is that what you meant here?
(there's also an extra bracket in there.)

Guessing "no" ... rewriting - you need t to satisfy the relation: $$(3t+100)^2 = 4(9t^3+900t^2+30000t)$$your first task (brute force) is to put the equation into standard form.

From the look of it, it is a cubic, so as many as three values of t may make this equation true. You should already know how to solve for the roots of a cubic.
 
Last edited:
Sorry, wrote the question wrong:
3t+100=4((9t^3+900t^2+30000t)/(3t+100)^2)
 
Last edited:
Here's some more info on the problem that may help:

x(t) is the amount of an element in a container

The original d.e:
$$dx/dt+6x/(100+3t)=3$$

100+3t=v, the total volume of liquid in a container over time. The max of this is 300.

Solving the d.e gives:
$$x(t)=(9t^3+900t^2+30000t)/(3t+100)^2)$$

Now, and this is where I am stuck, I have to find when the percentage of the element within the total liquid is ≥ 25%. This occurs when $$x/v=1/4$$, so $$3t+100=4x$$, hence:
$$3t+100=4((9t^3+900t^2+30000t)/(3t+100)^2)$$
 
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Hmmm I get ##x(t)= c(3t-100)^2-3t+100## where c is the constant of integration.

But if you are still happy with yours - you still have a cubic.
Put it in standard form.
 
Simon Bridge said:
Hmmm I get ##x(t)= c(3t-100)^2-3t+100## where c is the constant of integration.

But if you are still happy with yours - you still have a cubic.
Put it in standard form.

Oh sorry, I forgot to mention the intial vlaue is x(0)=0
 
So I guess this is what you mean by standard form:

$$9t^3+900t^2+30000t-1000000=0$$
 
tedwillis said:
Sorry, wrote the question wrong:
3t+100=4((9t^3+900t^2+30000t)/(3t+100)^2)

It is easier (and more insightful) to re-scale this: set t = 100s in the equation, to get
[tex](3s+1)^3 = 12(3s^3 + 3s^2 + s).[/tex] It is easier to get high accuracy on a calculator when you do this, plus it is easier to manipulate, etc.

RGV
 
Ray Vickson said:
It is easier (and more insightful) to re-scale this: set t = 100s in the equation, to get
[tex](3s+1)^3 = 12(3s^3 + 3s^2 + s).[/tex] It is easier to get high accuracy on a calculator when you do this, plus it is easier to manipulate, etc.

RGV
Following Ray's suggestion, let u = 3s+1. so that [itex]\displaystyle s=\frac{u-1}{3}\ .[/itex]

That should make the algebra even cleaner.
 
  • #10
tedwillis said:
Oh sorry, I forgot to mention the intial vlaue is x(0)=0
In which case, c=-1/100 ... have you checked your DE solution?

Sammy and Ray have good suggestions for making things clearer.
You can look up how to solve a cubic equation... and that was "standard form" yes.
 
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