Question Video: Finding the Rate of Change of a Polynomial Function Representing the Biomass of a Bacterial Culture at a Certain Time

The biomass of a bacterial culture in milligrams as a function of time in minutes is given by 𝑓(𝑑) = 71𝑑³ + 63. What is the rate of growth of the culture when 𝑑 = 2 minutes?

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Video Transcript

The biomass of a bacterial culture in milligrams as a function of time in minutes is given by 𝑓 of 𝑑 is equal to 71𝑑 cubed plus 63. What is the rate of growth of the culture when 𝑑 is equal to two minutes?

The question gives us a polynomial function, which tells us the biomass of a bacterial culture in milligrams after 𝑑 minutes. And it wants us to calculate the rate of growth of this bacterial culture at the time where 𝑑 is equal to two minutes. There’re many different ways of measuring growth. For example, we could measure the population size. However, in this case, we’re told that we’re measuring the biomass of a bacterial culture in milligrams to represent the growth.

So when we’re asked to find the rate of growth, we’re actually trying to find the rate of change in the biomass of our bacterial culture. And we know how to find the rate of change of a function 𝑓 at π‘Ž. We have the rate of change of the function 𝑓 at π‘Ž given by 𝑓 prime of π‘Ž is equal to the limit as β„Ž approaches zero of 𝑓 evaluated at π‘Ž plus β„Ž minus 𝑓 evaluated at π‘Ž all divided by β„Ž if this limit exists.

The question tells us we want to find the rate of change when π‘Ž is equal to two minutes. Since the question tells us that our biomass function is measured in minutes, we’ll set π‘Ž equal to two. So to find the rate of growth of our culture when 𝑑 is equal to two, we want to find the derivative function 𝑓 prime evaluated at 𝑑 is equal to two. And we know this is equal to the limit as β„Ž approaches zero of 𝑓 evaluated at two plus β„Ž minus 𝑓 evaluated at two all divided by β„Ž if this limit exists.

Evaluating our function 𝑓 at two plus β„Ž and two inside of our limit gives us the limit as β„Ž approaches zero of 71 times two plus β„Ž cubed plus 63 minus 71 times two cubed plus 63 all divided by β„Ž. Distributing the exponent over the first set of parentheses inside of our limit gives us 71 times eight plus 12β„Ž plus six β„Ž squared plus β„Ž cubed. We then add 63 to this. We subtract 71 times two cubed, and two cubed is just equal to eight. Finally, we subtract 63 and then divide all of this by β„Ž.

We see that 63 minus 63 is just equal to zero. Next, we distribute 71 over our parentheses. This gives us the limit as β„Ž approaches zero of 71 times eight plus 71 times 12β„Ž plus 71 times six β„Ž squared plus 71 times β„Ž cubed minus 71 times eight all divided by β„Ž. We see that 71 times eight minus 71 times eight is just equal to zero.

We can’t evaluate this limit by using direct substitution. However, if we cancel the shared factor of β„Ž in our numerator and our denominator, we get the limit as β„Ž approaches zero of 71 times 12 plus 71 times six β„Ž plus 71 times β„Ž squared. And we see this is a polynomial in β„Ž. This means we can attempt to use direct substitution.

Substituting β„Ž is equal to zero gives us 71 times 12 plus 71 times six times zero plus 71 times zero squared. Multiplying a number by zero just gives us zero. So our second and third term are just equal to zero. And we can evaluate 71 times 12 to give us 852. And we could stop there. However, since the output of our function 𝑓 of 𝑑 is measured in milligrams and 𝑑 is measured in minutes, we can add units to our answer since we’re checking the rate of change of growth, which is mass measured in milligrams. And our unit of time is given in minutes. So our units will be milligrams per minute.

Therefore, we’ve shown if the biomass of a bacterial culture measured in milligrams as a function of time in minutes is given by 𝑓 of 𝑑 is equal to 71𝑑 cubed plus 63, then the rate of growth of this culture at the time where 𝑑 is equal to two minutes is 852 milligrams per minute.

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