Bacteria Population The total number of bacteria (in millions) present in a culture is given by

Bacteria Population The total number of bacteria (in...

Key Concepts

Rate of Change

This concept measures how one quantity changes in response to a change in another quantity. In the context of a bacteria population, it indicates how quickly the population size varies with respect to time, representing the derivative of the population function.

Derivative

A derivative is a fundamental concept in calculus that captures the instantaneous rate of change of a function with respect to one of its variables. In this problem, the derivative of the bacteria population function with respect to time reveals how fast the bacteria count is increasing at any given moment.

Product Rule

The product rule is a technique used to differentiate expressions where two functions are multiplied together. Since the bacteria population function is given as a product of time and a square root function, applying the product rule is essential to correctly compute its derivative.

Chain Rule

The chain rule is applied when differentiating a composite function—a function made up of an inner and an outer function. In this case, the square root function, which has a linear function inside it, requires the use of the chain rule to find its derivative accurately.

Evaluation of Derivatives

This concept involves substituting specific values into the derivative to determine the instantaneous rate of change at those points. Evaluating the derivative at given time values helps to quantify how briskly the bacteria population is growing at different stages of the experiment.

Transcript

00:01 So if we're given that the bacteria population in millions is given by the equation here, and i forgot to put the one -half power.

00:16 So if we have that the bacteria population is given by n of t, and this is equal to 2t times t 5 t plus 9 to the one -half power, plus 12 where our time is an hours.

00:37 And we want to find the rate of change of the bacteria of population with respect to time after 0 hours, 7 1⁄2 hours, and 8 hours.

00:47 Remember any time we talk about rate of change of some formula, this should tell us that we're looking for a derivative.

00:56 So we need to find what n prime of t is, and then we'll plug in, zero hours, seven and a half hours, and eight hours into this equation here.

01:09 So on the side, let's go ahead and find what this is.

01:18 So n prime of t will equal.

01:24 So remember, i can distribute the derivative across plus signs, as well as i can pull out any constant since the derivative is a linear operator.

01:37 So i can rewrite this like, this here, so 2 times the derivative with respect to t of 5t plus 9 to the 1 half power plus the derivative with respect to time of 12.

02:03 So first recall that the derivative of a constant is 0, and then to take this derivative here, oh, and i dropped my t.

02:17 That should be right there being multiplied by it.

02:21 Now, if i want to take the derivative of this, i'll have to use the product rule.

02:25 Since i have two functions being multiplied together, so i have t here, and i have 5t plus 9 to the 1⁄2 power.

02:34 So i can go ahead and rewrite this as two times.

02:39 So first, i'll write down t.

02:42 Then i take the derivative of 5t plus 9 to the 1 half power.

02:59 So 5t plus 9 to the 1 half.

03:05 Then i'll switch the places of these two functions.

03:10 So i'll have 5t plus 9, the 1⁄2 times the derivative of t.

03:35 So 2.

03:36 So the t gets dropped down, and then recall to take the derivative of 5t plus 9 to the 1 half.

03:43 It is the generalized power rule, so i'll take this 1 half, move it out front, and then subtract 1 off of it.

03:52 So i get t times 1⁄ 1⁄2 times, so 5t plus 9.

04:01 But since i subtracted 1 off from my power, this will now be a negative.

04:08 One half power and then remember since this is essentially chain rule i have to take the derivative of the inside function and multiply by it so the derivative of 5t plus 9 is 5 and then i'm done with this first expression and then to take the derivative over here i should first just go ahead and write down 5t plus 9 to the 1 half and then the derivative of t is one and so at this point all we need to do is clean this up a little bit make it look a little bit nicer since we're wanting to plug in three terms into this so let's simplify this as much as we can so we won't have to do as much work when we plug in those numbers so two so actually before i do that notice that we have this five t plus nine common term in each of these.

05:21 So what we can do is pull out the smallest power of this.

05:26 So the smallest power is negative half.

05:29 So i'll go ahead and pull out five plus nine raised to the negative one half.

05:40 And so this first expression that will leave me with.

05:45 So again combined on my constant, so i end up with five half and then the only variable i'll be left with is t and then if i pull out a negative one -half power from 5 t plus 9 to the one -half that will be like i'm adding a half to its power since i have to divide it by 5t plus 9 to the negative half so i'd be left with 5 t plus 9 plus 9 then i can go ahead and clean this up a little bit more.

06:28 I'll go ahead and combine my like terms by adding them.

06:33 And so i'll have two times.

06:40 So 5 .5 plus 5 will give me 15 over 2 t plus 9.

06:49 And then changing this negative half power into a positive.

06:53 It will be 5t plus 9.

06:57 But remember, a half power is actually a square root, so i'll write it like this here.

07:04 And for the last step, i can go ahead and distribute this 2 -in.

07:08 And in doing so, i end up with 15t plus 18 all over.

07:19 And i can't write any further down, so let me actually move this up a little bit...