Additional Applications

Bacterium population

Question

Additional Applications

Bacterium population

When a bactericide was added to a nutrient broth in which bacteria were growing, the bacterium population continued to grow for a while, but then stopped growing and began to decline. The size of the population at time t (hours) was b = 10⁶ + 10⁴t − 10³t². Find the growth rates at

a. t = 0 hours.

b. t = 5 hours.

c. t = 10 hours.

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. The concentration of a substance within a chemical solution C in units of milligrams per liter is modeled by the equation C equals 30,000 plus 400 minus 60 squad for 0 is less than. equal to T and T is less than or equal to 20, where T is the time in hours. Find the rate of change of concentration at T equals 2 and T equals 4 hours. Awesome. So it appears for this particular problem, our final answers, because we're trying to solve for two separate answers that we're ultimately trying to figure out, is we're trying to find the rate of change of concentration at T equals 2 as our first answer, and at T equals 4 hours. So it's going to be at T equals 2 hours and T equals 4 hours, and we're trying to figure out the rate of change at these two different time values. So with that said, now that we know what we're ultimately trying to solve for, Our first step is we need to find the derivative of the concentration function. So as we should recall, in order to find the rate of change of concentration, we need to figure out the derivative of the concentration function C with respect to time T. So let us note that we're told that our concentration function is C of T. is equal to 30,000 plus 40T minus 60 square. So now we need to take the derivatives, so we need to take DC by DT. is equal to 0 + 400 minus 120T, which is going to be equal to 400 minus 120 T. Now our second step, so I'll make a note of this, this is our first step. So now our second step is we need to evaluate the derivative at T equals 2 hours. So we need to substitute in T equals 2 in our derivatives, so when you take DC by DT. Evaluated at T equals 2, which will be equal to 400 minus 120 multiplied by 2, which will be equal to when we evaluate that in our calculator, 160. So the rate of change of the concentration at T equals 2 hours will have to be 160 mg per liter per hour. So once again, our first answer that we're trying to solve for has to be 160 mg per liter per hour, and that once again is for T equals 2. So now switching gears to evaluate our derivative at And this is our third step to evaluate our derivative at T equals 4 hours now. So that means we need to substitute T equals 4 into our derivatives. So we need to take DC by DT evaluated at T equals 4, which equals 400 minus 120 multiplied by 4, which when we plug that into our calculator, we'll get 80. So that will mean that the rate of change of concentration at T equals 4 hours is -80 mg per liter per hour, and that is our final and second answer for T equals 4. Hooray, we did it. So therefore, without a doubt, our final two answers have to be, to quickly recap, which is C2 is equal to 160 mg per liter per hour, and our second answer is C 4 is equal to -80 mg per liter per hour. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Key Concepts

Derivative

Polynomial Functions

Evaluating Derivatives at Specific Points

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