The biomass of Cerastium is assumed to follow a logistic growth model with an initial biomass of 0.1 grams and a proportionality factor of 𝑘 is equal to 0.055, using day as a unit of time. At 𝑡 is equal to 75 days, a Cerastium plant’s biomass was 3.0 grams. Find the final biomass of this Cerastium plant when it is fully grown. Give your answer to one decimal place.
We’re told that the biomass of this plant follows a logistic growth model, and we’re given information about this growth model. We’re told the initial biomass of this plant is 0.1 grams. And we’re also told that the proportionality factor in our logistic growth model of 𝑘 is equal to 0.055. And this proportionality factor or growth rate will be using days as its unit of time. And we can see we’re not told the carrying capacity. Instead, we’re told when 𝑡 is equal to 75 days, the plant’s biomass will be 3.0 grams. We need to use this information to find the final biomass of our plant. In other words, what is the mass of the plant when it will be fully grown? And we need to give our answer to one decimal place.
To answer this question, let’s start by recalling what the logistic growth model tells us. The logistic growth model tells us on a population 𝑃 on time 𝑡, d𝑃 by d𝑡 will be equal to 𝑘 times 𝑃 multiplied by one minus 𝑃 divided by 𝐿, where 𝑘 is the growth rate and 𝐿 is the carrying capacity. In our case, we’re not modeling a population. So instead of calling this 𝑃, we’ll call this 𝑀 for biomass. This gives us the following differential equation, and, in fact, we’re told the value of our growth rate 𝑘. It’s equal to 0.055. And it’s also worth pointing out that since our growth rate is given with the units as days, our value of 𝑡 will also be in days.
Now we could start solving this differential equation. In fact, it’s a separable differential equation, and we can solve it by using partial fractions. This would help us find an equation for the mass of our plant at time 𝑡. However, we’ve also solved this differential equation in the general sense, so we can just apply the solution to this problem. We know the general solution will be of this form. 𝑀 of 𝑡 is equal to 𝐿 divided by one plus 𝐴 times 𝑒 to the power of negative 𝑘𝑡, where our constant 𝐴 is equal to 𝐿 minus 𝑀 of zero divided by 𝑀 of zero. And, of course, 𝑀 of zero will be the biomass of our plant when 𝑡 is equal to zero, in other words, the initial biomass.
But we still have a problem. We don’t know the value of 𝐿 or the value of 𝐴. But we do have an equation which links our values of 𝐿 and 𝐴, so we only need to find one of these values. This means we can then find the other. And to do this, we’re going to need to use information given to us in the question. First, we’re told in the question the initial biomass of our plant is 0.1 grams. So we know that 𝑀 of zero must be 0.1. Next, we’re also told in the question that after 75 days, our plant has a biomass of 3.0 grams. This means 𝑀 of 75 must be equal to 3.0. We can then substitute these values into our equation. This would then give us a pair of simultaneous equations in two variables. We can then solve this for either 𝐴 or 𝐿.
We’ll start by substituting 𝑡 is equal to zero. We get 𝑀 of zero, which we know is the initial biomass of our plant which is 0.1, will be equal to 𝐿 divided by one plus 𝐴 times 𝑒 to the power of negative 0.055 times zero. And we can simplify this. We have negative 0.055 multiplied by zero, which simplifies to give us zero. And then 𝑒 raised to the zeroth power is just equal to one. So this entire expression simplifies to give us 0.1 is equal to 𝐿 divided by one plus 𝐴. And in fact, we’ll do one more piece of simplification. We’ll multiply both sides of this equation through by the denominator one plus 𝐴. This gives us 0.1 multiplied by one plus 𝐴 is equal to 𝐿.
We’ll now do the same when 𝑡 is equal to 75. We get 𝑀 of 75, which we’re told in the question is 3.0, is equal to 𝐿 divided by one plus 𝐴 times 𝑒 to the power of negative 0.055 multiplied by 75. And we’ll simplify this in the same way. First, our exponent of 𝑒 can simplify to give us negative 4.125. So this means we can rewrite our equation as 3.0 is equal to 𝐿 divided by one plus 𝐴 times 𝑒 to the power of negative 4.125. And finally, we’ll rearrange this equation to make 𝐿 the subject. We’ll multiply both sides of the equation by the denominator one plus 𝐴 times 𝑒 to the power of negative 4.125. This gives us 3.0 multiplied by one plus 𝐴 times 𝑒 to the power of negative 4.125 is equal to 𝐿.
So now we have a pair of simultaneous equations in two variables, and both of these are written with 𝐿 as the subject. So in actual fact, we must have 0.1 times one plus 𝐴 is equal to three times one plus 𝐴 times 𝑒 to the power of negative 4.125. So by equating these two parts and rewriting 3.0 as three and keeping in mind we only have one decimal place of accuracy, we get the following equation. And we want to solve this for 𝐴. We’ll do this by distributing both of our parentheses. This gives us 0.1 plus 0.1𝐴 is equal to three plus three 𝐴 times 𝑒 to the power of negative 4.125.
Now we can just solve this for 𝐴. We’ll subtract 0.1𝐴 from both sides of the equation and subtract three from both sides of the equation. This gives us 0.1 minus three is equal to three 𝐴 times 𝑒 to the power of negative 4.125 minus 0.1 times 𝐴. Now, we’ll take out the common factor of 𝐴 on the right-hand side of our equation, giving us the following equation. And now we can solve for 𝐴 by dividing through both sides of our equation by three 𝑒 to the power of negative 4.125 minus 0.1. This gives us the following expression for 𝐴. And if we calculate this expression, to one decimal place, we get 𝐴 is approximately equal to 56.3.
And now that we found our value of 𝐴, we can use this to find our value of 𝐿. We’ll do this by using the equation 𝐴 is equal to 𝐿 minus 𝑀 of zero all divided by 𝑀 of zero. Substituting in our values, we get that 56.3 is approximately equal to 𝐿 minus 0.1 divided by 0.1. And we can then solve this for 𝐿. We multiply through by 0.1 and then add 0.1 to both sides. We get 𝐿 is approximately equal to 5.73. And it’s worth pointing out we have the same number of significant figures of accuracy.
But remember, the question is not asking us to find the logistic growth model in this case. It wants us to use this to determine the final biomass of our plant when it’s fully grown. And one way to consider this is to ask the question, what happens to our function 𝑀 of 𝑡 as 𝑡 approaches ∞? So let’s clear some space and ask the question of what happens. We remember our value of 𝐿 was approximately 5.73 and 𝐴 was approximately 56.3. We can then substitute these values into our growth model 𝑀 of 𝑡. This gives us that 𝑀 of 𝑡 will be approximately equal to 5.73 divided by one plus 56.3 multiplied by 𝑒 to the power of negative 0.055𝑡. And we want to know what happens to the biomass of our plant as we take 𝑡 goes to ∞.
There’s a few different ways of doing this. We’ll do this formally from the definition of a limit. We see 𝑡 is approaching ∞. We can see that our numerator remains constant. Of course, one and 56.3 are also constant with respect to 𝑡. We can see as 𝑡 approaches ∞, negative 0.055𝑡 is approaching negative ∞. So we’re raising 𝑒 to the power of a number which is getting more and more negative. And this is in fact growing without bound. This means this is approaching zero, and everything else in this expression is a constant. So this limit evaluates to give us 5.73 divided by one plus 56.3 times zero.
But of course, 56.3 times zero is equal to zero and then we’re dividing by one, which doesn’t change anything. So this entire limit just simplified to give us our numerator of 𝐿, which was 5.73. And remember, the question told us to give this answer to one decimal place. So we’ll write this as 5.7. We could in fact add a unit of grams; however, it’s not necessary in this case.
In this question, we were given a real-world problem involving the biomass of Cerastium. And we were told this follows a logistic growth model. We were given the initial biomass and proportionality factor, and we were told that days we used as the unit of time. And then we were also told the biomass of a Cerastium plant after 75 days. We were able to use all of this information to estimate the final biomass of this Cerastium plant when it’s fully grown. We got this was approximately equal to, to one decimal place, 5.7 grams.