Course:MATH110/Archive/2010-2011/003/Teams/Schaffhausen/Homework 12

Homework 12

Starting with the function:

P(t)={\frac {1}{1+e^{-t}}}

-Change the height of the horizontal asymptote on the right and denote it by K.

-Change the y-intercept to any number between 0 and K. BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.

Solution 1

We saw that there are two possible ways to change the horizontal asymptote in this equation we can either replace the numerator by K which will then become the horizontal asymptote so for example like this.

P(t)={\frac {K}{1+e^{-t}}}

So any what ever value K becomes we will gain a new horizontal asymptote.

We also saw that if we place a K value under the 1 in the denominator that also changed the value of the horizontal asymptote so it would look like this.

P(t)={\frac {1}{1/k+e^{-t}}}

As we moved around the K value we noticed that the y-intercept was constantly moving depending what we did with our K values, and we observed a pattern that when we used the first equation to change the horizontal asymptote that the y-intercept was usually about half of the value of K. So if we had imputed 6 into our K value we would thus get a y-intercept of 3. So if we wanted the y-intercept to be between 0 and K it we would just have to plug a value for K that is greater than 0.

Bonus

Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical. (If you graph it, it should be quite clear)

Well firstly we noticed that no matter how big we made our K value the equation would always reach the horizontal asymptote about the same t value which was around 12-13. Which means that as we increase the K value the slope normally also increases with it in order to keep up with a constant encroachment of the horizontal asymptote. Another way we noticed to change the slope of the curved part was to add an additional value before the T, by doing so the rate at which it increased was also increased so for example.

P(t)={\frac {1}{1+e^{-4t}}}

by adding this 4 in front of the T, we increased the rate at which the slope reaches the horizontal asymptote, normally this slope would reach its asymptote of 1 at around T= 13 but when we add the 4 in font the value at which it reaches the asymptote is T= 4. Thus by inserting this new value we can increase the slope of the curved part.

Once you've played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it's carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity). When doing this last part, explain well where you're taking your data from (real data or imagined data), what it is that you're modelling and how you are doing the math to answer a predictive question.

So an example of this would be that of exponential growth of a colony of bacteria that is growing on a rock. Lets say that initially we have 1,000 bacteria growing on this rock but their is only enough space and nutrients to support 5,000 bacteria on the rock. To calculate the time it would take for the bacteria to get to 95% would look like this.

5,000 x .95 = 4750