An Unsolvable Differential Calculus Problem: What is the Limit?”

In summary: The rate at which substance is eliminated from the body is proportional to the amount of substance present, and satisfies the differentail equation dx/dt=a-bx, where a and b are positive constants. What is the limiting concentration of substance(limit as t goes to infinity of x(t))?Can't integrate for this because it's a course on differential calculus only. Any pointers?Well, what is the rate of change of the amount once the amount has reached a constant level?Rate of change is zero... So, x=a/b when the limiting concentration is reached. What do I do with this?Thank you for replying so fastWell, so what is then the limiting
  • #1
laminar
15
0
This is biological application question, so the limit can't be negative infinity, which is what I would use as my answer for a regular question.

The rate at which substance is eliminated from the body is proportional to the amount of substance present, and satisfies the differentail equation dx/dt=a-bx, where a and b are positive constants. What is the limiting concentration of substance(limit as t goes to infinity of x(t))?

Can't integrate for this because it's a course on differential calculus only. Any pointers?
 
Physics news on Phys.org
  • #2
Well, what is the rate of change of the amount once the amount has reached a constant level?
 
  • #3
Rate of change is zero... So, x=a/b when the limiting concentration is reached. What do I do with this?

Thank you for replying so fast
 
  • #4
Well, so what is then the limiting concentration you were asked about?
 
  • #5
laminar, the rate of change is zero as t goes to infinity (this should be intuitive).
 
  • #6
But then another part of the question asks the time it takes to reach half of the limiting concentration... I'm lost.
 
  • #7
You DO know that the solution of the diff.eq f'=af for some constant f is an expontential function, right?
 
  • #8
Yes, I know that I would find the time by x=x0e^kt.
 
  • #9
laminar said:
Yes, I know that I would find the time by x=x0e^kt.
No, you don't, because your diff.eq is slightly different!
First, do the following rewriting of your diff.eq:
[tex]\frac{dx}{dt}=-b(x-\frac{a}{b})[/tex]
with initial condition: [tex]x(0)=x_{0}[/tex]

Then, introduce the new variable [itex]y(t)=x(t)-\frac{a}{b}\to\frac{dy}{dt}=\frac{dx}{dt}[/tex]

Thus, your diff.eq can be written as:
[tex]y(t)=-by(t)\to{y}(t)=Ae^{-bt}\to{x}(t)=\frac{a}{b}+Ae^{-bt}[/tex]
where A is a constant to be determined by the initial condition:
[tex]x_{0}=x(0)=\frac{a}{b}+A\to{A}=x_{0}-\frac{a}{b}[/tex]

Therefore, your solution is:
[tex]x(t)=\frac{a}{b}+(x_{0}-\frac{a}{b})e^{-bt}[/tex]

Now, you are to find the time T, so that [tex]x(T)=\frac{x_{0}}{2}[/tex]

We therefore have to solve the following equation for T in terms of the other parameters:
[tex]\frac{x_{0}}{2}=\frac{a}{b}+(x_{0}-\frac{a}{b})e^{-bT}[/tex]
Can you do that?
 
  • #10
I'm sorry. I thought the question was to find the time T so that you have half the INITIAL concentration.
In order to find the time t* to half the limiting concentration, adjust the last step accordingly.
 
Last edited:
  • #11
Ah, much thanks. Forgive me, I'm an arts student taking math courses.
 
  • #12
Arts or no arts, have you found your answer now?
 
  • #13
Yeah I get how to find t, but I still don't get how to find the limiting concentration.
 

Related to An Unsolvable Differential Calculus Problem: What is the Limit?”

1. What is a differential calculus problem?

A differential calculus problem is a mathematical problem that involves the application of the principles and techniques of differential calculus. This branch of mathematics deals with the study of rates of change and how they impact the behavior of functions.

2. What is the limit in differential calculus?

The limit is a fundamental concept in differential calculus that represents the value that a function approaches as its input approaches a certain value. It can also be thought of as the value that a function "approaches" as its input gets closer and closer to a specific point on the graph.

3. Why is the limit considered an unsolvable problem?

The limit is not necessarily an unsolvable problem. In many cases, it can be easily calculated using algebraic techniques. However, there are certain functions for which the limit cannot be determined using traditional methods. These functions are known as "unsolvable" or "indeterminate" and require more advanced techniques to determine their limit.

4. What makes the "An Unsolvable Differential Calculus Problem" unique?

The "An Unsolvable Differential Calculus Problem" is unique because it refers to a specific function, known as the "continuously differentiable nowhere function," which is an example of an unsolvable differential calculus problem. This function is continuous everywhere but does not have a derivative at any point, making it particularly difficult to determine its limit.

5. How do scientists approach solving this unsolvable problem?

Scientists approach solving this unsolvable problem by using advanced techniques and concepts from differential calculus, such as the concept of "differentiability" and the "mean value theorem." These techniques allow them to determine the limit of the continuously differentiable nowhere function and other similar unsolvable problems.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
337
  • Calculus and Beyond Homework Help
Replies
28
Views
1K
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
32
Views
2K
Replies
1
Views
100
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Differential Equations
Replies
25
Views
2K
  • Calculus and Beyond Homework Help
Replies
12
Views
4K
  • Differential Equations
Replies
1
Views
1K
Back
Top