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In Example 6 we considered a bacteria population that doubles every hour. Suppose that another population of bacteria triples every hour and starts with 400 bacteria. Find an expression for the number $ n $ of bacteria after $ t $ hours and use it to estimate the rate of growth of the bacteria population after 2.5 hours.

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00:27

Amrita Bhasin

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 7

Rates of Change in the Natural and Social Sciences

Derivatives

Differentiation

Campbell University

Baylor University

University of Michigan - Ann Arbor

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

44:57

In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.

02:01

A bacteria population trip…

02:17

Suppose that a population …

02:16

A population of 500 bacter…

03:23

Population Growth A popula…

03:20

here we have a bacterial growth problems so we can use the equation and equals and not eat the Katie. That's our basic exponential growth equation. We're told that the bacteria colony starts with 400 and it takes them one hour to triple, so we can use those pieces of information to find the value of K, the growth constant. So if you triple 400 you get 1200. So we have 1200 for end. We have 400 for and not and we have one for T one hour. So this is going to be to the K Times one or just k, just all that for K. Let's divide both sides of the equation by 400 we get three equals e to the K, and then we'll take the natural log of both sides. So K is the natural log of three so we can substitute that and get our model. We have n equals 400 times e to the natural log of three times T. Now we know that e to the natural log of three can be simplified and it's just three so we can write this model as an equals 400 times, three to the T, and what we want to do with that is use it to find the rate of growth and the rate of growth would be the derivative. So let's find the derivative of this in prime of tea would be 400 times three to the tee times natural log of three. And then we're going to find end prime of 2.5 to find the rate of growth after 2.5 hours and prime of 2.5, we would substitute the 2.5 in there and use a calculator, and you're going to get approximately 6850. That's the rate of growth. So we're saying that at that moment the bacteria is growing at that many bacteria per hour.

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