Exponential Cell Growth

July 24, 2012

The video shows 300 seconds of purely exponential growth (uninhibited), captured from the exponential growth VAMP scenario. Like the exponential growth function itself, the video starts off slowly then gets a lot more exciting (for a given value of exciting).

The modeled growth is based on the exponential growth function:

N = N_0 e^{rt} (1)

where:

  • N = number of cells (or concentration of biomass);
  • N0 = the starting number of cells;
  • r = the rate constant, which determines how fast growth occurs; and
  • t = time.

Finding the Rate Constant/Doubling Time (r)

You can enter either the rate constant (r) or the doubling time of the particular organism into the model. Determining the doubling time from the exponential growth equation is a nice exercise for pre-calculus students.

Let’s call the doubling time, td. When the organism doubles from it’s initial concentration the growth equation becomes:

2N_0 = N_0 e^{r t_d}

divide through by N0:

2  =  e^{r t_d}

take the natural logs of both sides:

\ln 2  =  \ln (e^{r t_d})

bring the exponent down (that’s one of the rules of logarithms);

\ln 2  =  r t_d \ln (e)

remember that ln(e) = 1:

\ln 2  =  r t_d

and solve for the doubling time:

\frac{\ln 2}{r}  =  t_d

Decay

A nice follow up would be to solve for the half life given the exponential decay function, which differs from the exponential growth function only by the negative in the exponent:

N = N_0 e^{-rt}

The UCSD math website has more details about Exponential Growth and Decay.

Finding the Growth Rate

A useful calculus assignment would be to determine the growth rate at any point in time, because that’s what the model actually uses to calculate the growth in cells from timestep to timestep.

The growth rate would be the change in the number of cells with time:

\frac{dN}{dt}

starting with the exponential growth equation:

N = N_0 e^{rt}

since we have a natural exponent term, we’ll use the rule for differentiating natural exponents:

\frac{d}{dx}(e^u) = e^u \frac{du}{dx}

So to make this work we’ll have to define:

u = rt

which can be differentiated to give:

\frac{du}{dt} = r

and since N0 is a constant:

N = N_0 e^{u}

\frac{dN}{dt} = N_0 e^{u} \frac{du}{dt}

substituting in for u and du/dt gives:

\frac{dN}{dt} = N_0 e^{rt} (r)

rearranging (to make it look prettier (and clearer)):

\frac{dN}{dt} = N_0 r e^{rt} (2)

Numerical Methods: Euler’s method

With this formula, the model could use linear approximations — like in Euler’s method — to simulate the growth of the biomass.

First we can discretize the differential so that the change in N and the change in time (t$) take on discrete values:
\frac{dN}{dt} = \frac{\Delta N}{\Delta t}

Now the change in N is the difference between the current value Nt and the new value Nt+1:

Now using this in our differentiated equation (Eq. 2) gives:

\frac{N^{t+1}-N^t}{\Delta t} = N_0 r e^{r\Delta t}

Which we can solve for the new biomass (N^t+1):

N^{t+1}-N^t = N_0 r e^{r\Delta t} \Delta t

to get:
N^{t+1}     = N_0 r e^{r\Delta t} \Delta t + N^t

This linear approximation, however, does introduce some error.

The approximated exponential growth curve (blue line) deviates from the analytical equation. The deviation compounds itself, getting worse exponentially, as time goes on.

The approximated exponential growth curve (blue line) deviates from the analytical equation. The deviation compounds itself, getting worse exponentially, as time goes on.

Excel file for graphed data: exponential_growth.xls

VAMP

This is the first, basic but useful product of my summer work on the IMPS website, which is centered on the VAMP biochemical model. The VAMP model is, as of this moment, still in it’s alpha stage of development — it’s not terribly user-friendly and is fairly limited in scope — but is improving rapidly.

Citing this post: Urbano, L., 2012. Exponential Cell Growth, Retrieved August 22nd, 2017, from Montessori Muddle: http://MontessoriMuddle.org/ .
Attribution (Curator's Code ): Via: Montessori Muddle; Hat tip: Montessori Muddle.

Euler’s Method for Approximating a Solution to a Differential Equation

March 3, 2012

Based on Euler's Method, this interactive graph illustrates a numerical method for solving differential equations. This approach is at the core of many sophisticated computer models of physical phenomena (like climate and weather).
  • Starting point: (x,y) = (   ,   )
  • Step size:
  • Direction:
Your browser does not support the canvas element.
  • Slope equation: dy/dx =   x +   
  • Show analytical solution:
If you know the equation for the slope of a curve (the red line for example),
and a point that the curve passes through, such as , you can integrate to find the equation of the curve:

If you don't have a starting point (initial condition), you can draw a slope field to see what the general pattern of all the possible solutions.

Even with a starting point, however, there are just times when you can't integrate the slope equation -- it's either too difficult or even impossible.

Then, what you can do is come up with an approximation of what the curve looks like by projecting along the slope from the starting point.

The program above demonstrates how it's done. This approach is called Euler's Method, and is gives a numerical approximation rather than finding the exact, analytical solution using calculus (integration).

So why use an approximation when you can find the exact solution? Because, there are quite a number of problems that are impossible or extremely difficult to solve analytically, things like: the diffusion of pollution in a lake; how changing temperature in the atmosphere gives you weather and climate; the flow of groundwater in aquifers; stresses on structural members of buildings; and the list goes on and on.

As with most types of numerical approximations, you get better results if you can reduce the step size between projections of the slope. Try changing the numbers and see.

A more detailed version, with solutions, is here: Euler's Method.

A good reference: Euler's Method by Paul Dawkins.

Citing this post: Urbano, L., 2012. Euler's Method for Approximating a Solution to a Differential Equation, Retrieved August 22nd, 2017, from Montessori Muddle: http://MontessoriMuddle.org/ .
Attribution (Curator's Code ): Via: Montessori Muddle; Hat tip: Montessori Muddle.

Numerical Integration

February 23, 2012

Your browser does not support the canvas element.

This is an attempt to illustrate numerical integration by animating an HTML5's canvas.

We're trying to find the area between x = 1 and x = 5, beneath the parabola:

y = -\frac{1}{4} x^2 + x + 4

By integrating, the area under the curve can be calculated as being 17 ⅔ (see below for the analytical solution). For numerical integration, however, the area under the curve is filled with trapezoids and the total area is calculated from the sum of all the areas. As you increase the number of trapezoids, the approximation becomes more accurate. The reduction in the error can be seen on the graph: with 1 trapezoid there is a large gap between the shaded area and the curve; more trapezoids fill in the gap better and better.

The table below show how the error (defined as the difference between the calculation using trapezoids and the analytic solution) gets smaller with increasing numbers of trapezoids (n).

Number of trapezoids Area (units2) Error (difference from 17.66)
1 15.00 2.66
2 17.00 0.66
3 17.37 0.29
4 17.50 0.16
5 17.56 0.10
6 17.59 0.07
7 17.61 0.05
8 17.63 0.03
9 17.63 0.03
10 17.64 0.02

Analytic solution

The area under the curve, between x = 1 and x = 5 can be figured out analytically by integrating between these limits.

Area = \int_{_1}^{^5} \left(-\frac{x^2}{4}  + x + 4 \right) \,dx

Area = \left[-\frac{x^3}{12}  + \frac{x^2}{2} + 4x \right]_1^5

Area = \left[-\frac{(5)^3}{12}  + \frac{(5)^2}{2} + 4(5) \right] - \left[-\frac{(1)^3}{12}  + \frac{(1)^2}{2} + 4(1) \right]

Area = \left[-\frac{125}{12}  + \frac{25}{2} + 16 \right] - \left[-\frac{1}{12}  + \frac{1}{2} + 4 \right]

Area = \left[ 22 \frac{1}{12} \right]  - \left[4 \frac{5}{12} \right]

Area = 17 \frac{2}{3}  = 17.6\bar{6}

(See also WolframAlpha's solution).

Update

Demonstration of this numerical integration using four trapezoids in embeddable graphs.

Citing this post: Urbano, L., 2012. Numerical Integration, Retrieved August 22nd, 2017, from Montessori Muddle: http://MontessoriMuddle.org/ .
Attribution (Curator's Code ): Via: Montessori Muddle; Hat tip: Montessori Muddle.

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