Basic Cal Quarter 4 Week 4 Problems Involving Exponential Growth and Decay PDF

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Republic of the PhilippinesDepartment of EducationREGION IV-A CALABARZONSCHOOLS DIVISION OF BATANGASTUY SENIOR HIGH SCHOOLQuarter 4 Week 4SIMPLIFIED ACTIVITY SHEET IN BASIC CALCULUS Reference: Basic Calculus by: Carlene Perpetua P. Arceo, Ph.I. Learning Competency/iesa. Solve situational problems in...

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Republic of the Philippines

Department of Education REGION IV-A CALABARZON SCHOOLS DIVISION OF BATANGAS TUY SENIOR HIGH SCHOOL

Quarter 4 Week 4 SIMPLIFIED ACTIVITY SHEET IN BASIC CALCULUS Reference: Basic Calculus by: Carlene Perpetua P. Arceo, Ph.D. I.

Learning Competency/ies a.

II.

Solve situational problems involving exponential growth and decay

Directions/ Instructions The following are some reminders in using this module: 1. Read the instruction carefully before doing each task. 2. Observe honesty and integrity in doing the tasks. 3. Finish the task at hand before proceeding to the next. 4. If you encounter any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator. 5. Always bear in mind that you are not alone. We hope that through this material, you will experience meaningful learning and gain deep understanding of the relevant competencies. You can do it!

III.

Objectives a. Solve situational problems involving exponential growth and decay 1. 2. 3.

IV.

Cognitive: Apply the concept of antidifferentiation in solving problems involving growth and decay Affective: Appreciate the application of the lesson in real life situation. Psychomotor: Apply anti-differentiation in solving problems involving growth and decay.

Discussion Lesson 1:Situational Problems Involving Growth and Decay Problems In the previous module, we have discussed about solving separable differential equations using antidifferentiation. It is important to note the following important terms: • Differential Equation (DE) is an equation that involves x, y, and the derivatives of y. The following are examples of differential equations:

• •

• •

The order of a differential equation per tains to the highest order of the derivative that appears in the differential equation. The first two examples above are first-order DEs because they involve only the first derivative, while the last example is a second-order DE because y” appears in the equation. A solution to a differential equation is a function 𝑦 = (𝑥) or a relation (𝑥, 𝑦) = 0 that satisfies the equation. Solving a differential equation means finding all possible solutions to the DE. If f(x) and g(y) are functions in terms of x and y, respectively, then

First of which is a first-order differential equation that can be solved using a technique called separation of variables. Thus, it is called separable differential equation.

• •

If there are initial conditions, or if we know that the solution passes through a point, we can solve this constant and get a particular solution to the differential equation. When a problem involves finding a particular solution to the differential equation, i.e., a function y of x given its derivative and its initial value 𝑦0 at a point 𝑥0, then we have an initial value problem.

Let 𝑦 = 𝑓(𝑡) be the size of a certain population at time t, and let the birth and death rates be the positive 𝑑𝑦 constants b and d, respectively. The rate of change in the population y with respect to the time t is given by 𝑑𝑡

If k is positive, that is when 𝑏 > 𝑑, then there are more births than deaths and

negative, that is when 𝑏 < 𝑑, then there are more deaths than births and Let us try solving

𝑑𝑦

𝑑𝑡

= 𝑘𝑦. By separating the variables, we have

𝑑𝑦 𝑑𝑡

𝑑𝑦

𝑑𝑡

denotes growth. If k is

denotes decay.

Integrating both sides will yield to

Taking the exponentials of both sides, we get

But since we are operating with respect to time, we can have the initial value of y when 𝑡 = 0. Thus, Substituting 𝑦0 to 𝑒 𝑐 in the equation 𝑦𝑡 = (𝑒 𝑐 )(𝑒)𝑘𝑡 , we have 𝑦𝑡 = 𝑦0𝑒 𝑘𝑡 We now have the Exponential Growth Law and Exponential Decay Law which states that some quantities grow or decay at a rate proportional to their size. This relationship can be written using the differential 𝑑𝑦 equation = 𝑘𝑦 which can be solved using the general function 𝑦𝑡 = 𝑦0𝑒 𝑘𝑡 . 𝑑𝑡 Example 1: Consider a population of a certain country that is observed to grow exponentially. Based from the available record, there were 8000 people at the start. After 10 years, it increased to 14,000. If the increase in number is constant, what is the estimated number of people in that country after 100 years? Solution: At the start, we have 𝑦𝑡 = 𝑦0 = 800. The rate of change of a number of people is actually be written as

𝑑𝑦

𝑑𝑡

= 8000𝑦 which can be solved using the equation 𝑦𝑡 = 8000

𝑒𝑘𝑡.

𝑑𝑦 . 𝑑𝑡

This can

The value of the growth constant k is obtained by substituting 𝑡 = 10 and 𝑦10 = 14,000 because there are 14,000 people after 10 years. Thus, we have

Solving for k, we have

Having found the constant rate 𝑘 = ln 1.75 10 , the solution to the differential equation that satisfies the initial value problem is

If 𝑡 = 100, then

Therefore, it is estimated that there will be 2,155,115 people in that country after 100 years.

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Example 2: Consider the constant rate of decrease of a certain microorganism. At the start, there are 10,000 microorganisms in a certain place. After 1 minute, the population decreased to 6,000. If the microorganisms is directly proportional to the constant rate of decay and such decrease in number continues at the same rate, then how many microorganisms will there be after 1.5 minutes? Solution: 𝑦0 = 10000. After 1 minute, the population decreased in number to 6000. So, we have

Solving for k, we have

The value of k is negative, which means a decay or decrease in number of microorganisms. The resulting differential solution would be

To get the number of organisms after 1.5 minutes, let 𝑡 = 1.5. Thus,

Therefore, it is estimated that there will be 4648 microorganisms after 1.5 minutes. Example 3: Suppose that a colony of ants grows exponentially. After 1 day, 50 ants are counted. After 3 days, 200 were counted. How many ants are there originally? What is the exponential growth equation for the colony? Solution: Recall the exponential growth equation 𝑦𝑡 = 𝑦0𝑒 𝑘 • •

𝑦1 = 50 𝑚𝑒𝑎𝑛𝑠 50 = 𝑦0 𝑒1𝑘 𝑦3 = 200 𝑚𝑒𝑎𝑛𝑠 200 = 𝑦0 𝑒3𝑘

Note that these two equations 50 = 𝑦0 𝑒𝑘 and 200 = 𝑦0 𝑒3𝑘 will give us the values for the two unknowns, 𝑦0 and 𝑒𝑘. From the first equation, 𝑦0 = 50 𝑒−k . Using this in the second equation, we have

Substituting 𝑒 𝑘 = 2 in the first equation,

Therefore, there were originally 25 ants in the colony. The exponential growth equation for the given problem is 𝑦𝑡 = 25 (2 ). EXAMPLE 2: The rate of decay of radium is said to be proportional to the amount of radium present. If the halflife of radium is 1690 years and there are 200 grams on hand now, how much radium will be present in 845 years? Solution. The exponential decay equation again starts o f as 𝑌 = 𝐶𝑒 𝑘𝑡 Since there are 200 grams present at the start, the equation immediately evolves to 𝑦 = 200𝑒 𝑘𝑡 .

3

A half-life of 1690 years means that the initial amount of 200 grams of radium will reduce to half, or just 100 grams, in 100 years. Thus, 100 = 200𝑒 𝑘∙1690. This gives 1 1/1690 , 𝑒𝑘 = ( ) 2 and consequently, 1 𝑡/1690 . 𝑦 = 200 ( ) 2 To answer the problem, 1 845/1690 𝑦 = 200 ( ) 2 1

1 2 𝑦 = 200 ( ) 2 1 𝑦 = 200 ( ) √2 1 𝑦 = 200 ( ) 0.707 𝑦 = 141.4

Therefore, after 845 years, there will be approximately 141.4 grams of radium left.

To wrap-up, answer the following questions. Write your answers on a separate sheet of papers. 1.

What is differential equation used in solving situational problems involving exponential growth and decay?

2.

What is the general function that is used in solving the differential equation involving exponential growth and decay?

3.

When do we say that the differential equation 𝑑𝑦 𝑑𝑡 denote growth or decay?

4. Can you give a real-life situation that model exponential growth or decay? How can we solve this problem? Discuss it.

V.

Activities/Instructions Directions: Solve the following problems involving Growth and Decay. Please refer to the rubrics on page 5. Write your answer on a separate sheet of papers. 1.

Initially, there are 1 million bacteria present in a Petri dish. After 2 minutes, there are already 5 million of them. If the bacterial population follows a law of natural growth, how many bacteria should be present in the Petri dish after 5 minutes?

2.

Consider the constant rate of depreciation value of a certain car. You bought a car amounting to 1.5 million pesos. After 2 years, the value decreased to 1.2 million pesos. If the amount of the car is directly proportional to the constant rate of decay in amount and such decrease in amount continues at the same rate, then how much is the estimated value of the car after 10 years?

3.

The state of decay of radium is proportional to the amount present at any time. Suppose that 400 mg of radium are currently present, and the amount decreases after every hour. After 1 hour, it decreases to 295. How much radium will be left after 2.5 hours if the decrease in number is constant?

4.

Bacteria grown in a certain culture increase at a rate proportional to the amount present. If there are 1500 bacteria present initially and the population tripled in 2 hours, how many bacteria will there be after 3 hours?

5.

The population of Barangay Siksikan is increasing at a rate proportional to its current population. In the year 2000, the population was 10,000. In 2003, it became 15,000. What was its population in 2009? In approximately what year will its population be 100,000?

6.

Certain bacteria cells are being observed in an experiment. The population triples in 1 hour. If at the end of 3 hours, the population is 27,000, how many bacteria cells were present at the start of the experiment? After how many hours, approximately, will the number of cells reach 1 million?

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VI.

Reflection Write your insights about the lesson on the spaces provided. I understand that_______________________________________________________________________________ ________________________________________________________________________________________________________ I realized that __________________________________________________________________________________ ________________________________________________________________________________________________________ ________________________________________________________________________________________________________

VII.

Mathematical Problem Solving Rubric

Conceptual Understanding

Computation and Execution

Exemplary(5) 1. Hidden or implied information not readily apparent was uncovered and used in problem solving. 2. Mathematical procedures were selected that would lead to an accurate solution. All aspects of the solution were complete and accurate

Proficient (4) 1. Relevant information from the problem was used in the solution 2. Mathematical procedures were selected would lead to a correct solution.

Developing (3) 1. Some but not all of the relevant information from the problem was used. 2. Mathematical procedures used would lead to partially correct solution.

Emerging (2) 1. The wrong information was used in trying to solve the problem 2. Mathematical procedures used would not lead to a correct solution.

Computations were essentially accurate

Minor computational errors were present

Errors in computation were serious enough to flaw the solution.

PREPARED: ELSIE M. DE LOS REYES Math Teacher

NOTED: MA. VERLA AFRICA ALVARAN Principal II

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