# Help creating a dif. eq. for a real life model?

• sutupidmath
In summary, the woman estimates that her retirement account will have an amount of $40000 after 5 years. She will have made$30000 in salary during this time, along with $6000 in interest. sutupidmath Help creating a dif. eq. for a real life model.help? A 30 year old woman accepts an engeineeering position with a starting salary of$30000 per year. Her salary S(t) increases exponentially, with $$S(t)=30e^{\frac{t}{20}$$ thousand dollars after t years. Meanwhile, 12% of her salary is deposited continuously in a retirement account, which accumulates interest at a continuous annual rate of 6%.
a) Estimate dA in terms of dt to derive the differential equation satisfied by the amount A(t) in her retirement account after t years.

Well, i am not any good at differential equations, but i am just trying to give a shot on my own, although i currently am in calculus I, so i am facing some difficulties from time to time absorbing these kind of problems, especially when they are wording ones that require us to derive a differential equation to describe the situation.

I have tried many things on this problem, first i thought that the differential equation derived for measuring the concentration on a tank- as one pipe brings water in while the other pipes water out of tank- would work but i think it does not quite fit in this situation.

so any hints on how to begin to derive a diff. eq for describing this situation??

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sutupidmath said:
A 30 year old woman accepts an engeineeering position with a starting salary of $30000 per year. Her salary S(t) increases exponentially, with $$S(t)=30e^{\frac{t}{20}$$ thousand dollars after t years. Meanwhile, 12% of her salary is deposited continuously in a retirement account, which accumulates interest at a continuous annual rate of 6%. a) Estimate dA in terms of dt to derive the differential equation satisfied by the amount A(t) in her retirement account after t years. A(t) increases in two ways: 1) She deposits 12% of her salary: 0.12 S= 0.12(30 et/20[/itex]= 3.6et/20. 2) interest, compounded continuously at 6% is added: A(t)e0.06t. Those are the two ways in which the account changes and the rate of change is dA/dt. Looks to me like dA/dt= e0.06tA(t)+ 3.6 et/20. Well, i am not any good at differential equations, but i am just trying to give a shot on my own, although i currently am in calculus I, so i am facing some difficulties from time to time absorbing these kind of problems, especially when they are wording ones that require us to derive a differential equation to describe the situation. I have tried many things on this problem, first i thought that the differential equation derived for measuring the concentration on a tank- as one pipe brings water in while the other pipes water out of tank- would work but i think it does not quite fit in this situation. so any hints on how to begin to derive a diff. eq for describing this situation?? thnx in advance Shouldn't we take into consideration somewhere her starting salary, maybe you incorporated it somewhere but i just cannot se where? If not why would it be ok, even if we do not take it into consideration? Also how did you derive A(t)e^0.06t ? In other words, how did you incorporate the 6% interest here? sutupidmath said: A 30 year old woman accepts an engeineeering position with a starting salary of$30000 per year.

Geeze, I certainly hope not! That's an obscenely low salary for any 30-year-old, especially an engineer!

sutupidmath said:
Shouldn't we take into consideration somewhere her starting salary, maybe you incorporated it somewhere but i just cannot se where? If not why would it be ok, even if we do not take it into consideration?
Also how did you derive A(t)e^0.06t ? In other words, how did you incorporate the 6% interest here?

The initial salary is already taken into account in the S(t)= 30et/20 although we should write it as 30000et/20.

The 6% is the 0.06 in the exponential. Amount A, compounded continuously, at interest rate r, for t years yields Aert.

Ben Niehoff said:
Geeze, I certainly hope not! That's an obscenely low salary for any 30-year-old, especially an engineer!

First college teaching job I got, after I got my Ph.D, paid $12000 a year. And that was more than my father had ever earned in one year in his life. Yeah, but what decade was that? Hell, even my first internship was$18/hr (though I realize I'm a bit lucky there). :P

HallsofIvy said:
The 6% is the 0.06 in the exponential. Amount A, compounded continuously, at interest rate r, for t years yields Aert.

Where can i find some things to read on my own about this interest rate, because i do not actually know why amount A, compounded continuously, at interest rate r, for t years yeilds Aert??

Well, thank you HallsofIvy, i really appreciate it. I will come back from time to time with these kind of problems, until i get used to them.

Thnx once more!

## 1. What is a differential equation (dif. eq.) and why is it important in real life modeling?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It is important in real life modeling because many natural phenomena can be expressed mathematically using differential equations, allowing scientists to make predictions and understand the behavior of complex systems.

## 2. How do I create a differential equation for a real life model?

To create a differential equation for a real life model, you first need to identify the variables and their relationships in the system. Then, you can use known physical laws and principles to formulate the equation, making sure to include all relevant terms and parameters.

## 3. Can I use a pre-existing differential equation for my real life model?

Yes, you can use pre-existing differential equations as long as they accurately represent the system you are trying to model. However, it is important to understand the assumptions and limitations of the equation and make any necessary adjustments for your specific model.

## 4. What are some common techniques for solving differential equations in real life modeling?

Some common techniques for solving differential equations in real life modeling include analytical methods, such as separation of variables and variation of parameters, and numerical methods, such as Euler's method and Runge-Kutta methods.

## 5. How can I validate my differential equation for a real life model?

To validate your differential equation for a real life model, you can compare the results of the equation with real-world data or observations. You can also perform sensitivity analyses to determine the impact of changing parameters on the behavior of the system and see if it aligns with your expectations.

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