Using exponential functions to find half capacity

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Joy Odigbo
Joy Odigbo el 25 de Feb. de 2021
Editada: Joy Odigbo el 25 de Feb. de 2021
How long does it take for the population to reach one half of the carrying capacity? Compare with theoretical results.
This is what I have so far

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Walter Roberson
Walter Roberson el 25 de Feb. de 2021
you need to find the first place that P is 150 or higher. Once you have the location then use it to index the time

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Johnny B
Johnny B el 25 de Feb. de 2021
First of all, "how can I answer this question" is a really bad name for the topic.
The syntax 150(t) doesn't make any sense; the value in parentheses is the index of the variable whose name precedes it, so you're asking for the t-th element of 150.
Your code is creating a vector of values of t from 0 to 20 and then calculating a P value for each value of t. So what you need is to find the P value closest to 150 and return the t value that corresponds to it. There's very little chance that you have a value of P in your array that is exactly 150, so testing for P==150 isn't going to work. One way to find the closest value is to take advantage of the second, optional, output of the min function:
[ ~, ix ] = min(abs(P - 150));
t_half = t(ix)
Then minimum of the absolute value of P-150 will, of course, occur where P is closest to 150. The min function will return this minimum value and optionally the index where it occurs. The tilde (~) tells MATLAB that we don't care about the actual minimum, just the index. Since t and P are the same size and have corresponding values (e.g., the 3rd value of P corresponds to the 3rd value of t), you just need t(ix).
Another approach is to use the interp1 function to find the value of t when P is 150. Although the function is written so that P is a function of t, you can think of them the other way around -- pretend that t is a function of P -- and interpolate:
t_half = interp1(P, t, 150)

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