Basiccalculus q3 mod12 relatedratesproblems final PDF

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Basic CalculusQuarter 3 – Module 12:Related Rates ProblemsBasic Calculus – Grade 11 Alternative Delivery Mode Quarter 3 – Module 12 : Related Rates Problems First Edition, 2020Republic Act 8293, section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. How...

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Basic Calculus Quarter 3 – Module 12: Related Rates Problems

Basic Calculus – Grade 11 Alternative Delivery Mode Quarter 3 – Module 12: Related Rates Problems First Edition, 2020 Republic Act 8293, section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or office may, among other things, impose as a condition the payment of royalties. Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this module are owned by their respective copyright holders. Every effort has been exerted to locate and seek permission to use these materials from their respective copyright owners. The publisher and authors do not represent nor claim ownership over them. Published by the Department of Education Secretary: Leonor Magtolis Briones Undersecretary: Diosdado M. San Antonio

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Basic Calculus Quarter 3 – Module 12: Related Rates Problems

Introductory Message This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson. Each SLM is composed of different parts. Each part shall guide you step-bystep as you discover and understand the lesson prepared for you. Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self-check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these. In addition to the material in the main text, Notes to the Teacher are also provided to our facilitators and parents for strategies and reminders on how they can best help you on your home-based learning. Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task. If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator. Thank you.

What I Need to Know

One of the main reasons why this module was created is to ensure that it will assist you to understand the concept and know how to solve related rates problems.

When you finish this module, you will be able to: Solve situational problems involving related rates (STEM_BC11D-IIIj-2).

1

What I Know I. For items 1 to 4, answer the following questions taken from the word problem below. Write your answers on a sheet of paper. Two variables 𝑥 and 𝑦 are both differentiable functions of 𝑡 and are related by the equation 𝑦 = 𝑥 2 + 10. Given that 𝑑𝑥⁄ 𝑑𝑡 = 3 when 𝑥 = 2. Find 𝑑𝑦 ⁄𝑑𝑡 when 𝑥 = 2. 1.

What is being asked in the problem? A. derivative of 𝑥 with respect to 𝑡 B. derivative of 𝑦 with respect to 𝑡 C. derivative of 𝑧 with respect to 𝑡 D. derivative of 𝑡 with respect to 𝑡

2.

Which is the set of correct quantities based on the word problem? A. 𝑑𝑦⁄ 𝑑𝑡 = 3 when 𝑥 = 2 and 𝑦 = 𝑥2 + 10 B. 𝑑𝑥⁄ 𝑑𝑡 = 3 when 𝑦 = 2 and 𝑥 = 𝑦 2 + 10 C. 𝑑𝑧⁄𝑑𝑡 = 3 when 𝑦 = 2 and 𝑦 = 𝑦 2 + 10 D. 𝑑𝑥⁄𝑑𝑡 = 3 when 𝑥 = 2 and 𝑦 = 𝑥 2 + 10

3.

Taking the derivative of the related equation, 𝑦 = 𝑥 2 + 10 with respect to time, will result to _____. 𝑑𝑥 𝑑𝑦 𝑑𝑦 𝑑𝑥 C. = 𝑥 A. = 2𝑥 𝑑𝑡

𝑑𝑡

𝑑𝑥

𝑑𝑦 𝑑𝑡

B. 𝑑𝑡 = 2𝑥 4.

𝑑𝑡

D.

𝑑𝑦 𝑑𝑡

𝑑𝑡

= 𝑥2

𝑑𝑥 𝑑𝑡

Once the answers from previous questions, 1 to 3, are complete, solve the word problem above. Choose the correct answer below. A. 10 C. 12 B. 11 D. 13

II. In items 5 to 8, answer the following questions taken from the word problem below. Write the letter of the correct answer on a separate sheet of paper. Assume that a point is moving along the graph 𝑦 = 4𝑥 2 + 3. The horizontal rate of change is 𝑑𝑥⁄𝑑𝑡 = 3 𝑐𝑚 ⁄𝑠 . Find its 𝑑𝑦 ⁄𝑑𝑡 when 𝑥 = −1. 5.

The problem is about evaluating its horizontal rate of change. True or False? A. True C. Maybe B. False D. Sometimes

6.

Below are the quantities given based on the problem above except for one. Which item is this? 𝑐𝑚 𝑑𝑥 A. 𝑥 = −1 C. 𝑑𝑡 = 3 𝑠 B. 𝑥 = 4𝑦 2 + 3 D. 𝑦 = 4𝑥2 + 3

2

7.

Getting the derivative of the equation, 𝑦 = 4𝑥2 + 3 with respect to time, will result to _____. 𝑑𝑦 𝑑𝑦 𝑑𝑡 𝑑𝑥 A. = 8𝑥 C. = 8𝑥 𝑑𝑡

𝑑𝑡

𝑑𝑥

𝑑𝑦

B. 𝑑𝑡 = 8𝑥 8.

𝑑𝑡

D.

𝑑𝑡

𝑑𝑦 𝑑𝑡

𝑑𝑡

= 8𝑥2

𝑑𝑥 𝑑𝑡

From your answers in questions 5 to 7, solve the answer for the word problem presented above. C. −23𝑐𝑚 ⁄𝑠 A. −21𝑐𝑚 ⁄𝑠 D. −24𝑐𝑚 ⁄𝑠 B. −22𝑐𝑚 ⁄𝑠

III. In items 9 to 12, answer the following questions taken from the word problem below. Write your answer on your paper. For instance, a point is navigating along the graph 4𝑥2 + 3𝑦 2 = 48. When the point is at A(3, 2) its 𝑥-coordinate is decreasing at the rate of 0.5 unit per second. How fast is the y-coordinate changing at that moment? (see figure 1 below for its illustration.)

Graph of 4𝑥 2 + 3𝑦 2 = 48

Point A(3, 2)

Figure 1

9.

What is asked in the problem? A. derivative of 𝑥 with respect to 𝑡 B. derivative of 𝑧 with respect to 𝑡

C. derivative of 𝑦 with respect to 𝑡 D. derivative of 𝑡 with respect to 𝑡

3

10. Below are the quantities presented in the problem except for one item. Which one is this? 𝑢𝑛𝑖𝑡 𝑑𝑥 A. 𝐴(3, 2) C. 𝑑𝑡 = −0.5 𝑠𝑒𝑐 B. 4𝑥2 + 3𝑦 2 = 48

D.

𝑑𝑥 𝑑𝑡

= 0.5

𝑢𝑛𝑖𝑡 𝑠𝑒𝑐

11. The derivative of the graph 4𝑥2 + 3𝑦 2 = 48 with respect to time is what? 𝑑𝑦 𝑑𝑦 𝑑𝑥 𝑑𝑥 C. 8𝑦 𝑑𝑡 + 6𝑥 𝑑𝑡 = 0 A. 6𝑦 𝑑𝑡 + 8𝑥 𝑑𝑡 = 0 B. 8𝑥

𝑑𝑥 𝑑𝑡

𝑑𝑦

𝑑𝑥

𝑑𝑦

D. 6𝑥 𝑑𝑡 + 8𝑦 𝑑𝑡 = 0

+ 6𝑦 𝑑𝑡 = 0

12. From your answers in questions 9 to 11, solve the word problem presented above. Choose the correct answer below. C. 3 𝑢𝑛𝑖𝑡 ⁄𝑠𝑒𝑐 A. 1 𝑢𝑛𝑖𝑡⁄ 𝑠𝑒𝑐 D. 4 𝑢𝑛𝑖𝑡 ⁄𝑠𝑒𝑐 B. 2 𝑢𝑛𝑖𝑡 ⁄𝑠𝑒𝑐 IV. For items 13 to 15, answer each item. Write your answer on a sheet of paper. 13. Tell whether the related rates problem differentiates quantities with respect to time. A. True C. Maybe B. False D. Sometimes 14. Tell whether a mathematical expression that relates other quantities involved in a problem is not a related rate equation. A. True C. Maybe B. False D. Sometimes 15. Tell whether an implicit differentiation helps in solving related rates problems. True or False? A. False B. True

4

Lesson

1

Related Rates Problems

One of the challenging lessons in Calculus is related rates problems. Simply because it is the application of derivatives in real-life situations and each one requires multiple approach in solving the problem. In this lesson, there are guides that will help learners understand and know how to start solving such problems.

What’s In Match items in Column A with its derivative in Column B. Write your answer on a sheet of paper. Column A

Column B

1. 𝑥 + 𝑦 = 6

A. 𝑑𝑦 ⁄𝑑𝑥 = 1

2. 𝑦 2 + 3𝑥 = 3𝑥2 − 5

B. 𝑑𝑦 ⁄𝑑𝑥 = −1 C. 𝑑𝑦 ⁄𝑑𝑥 = 3/2

3. 5𝑦 − 4𝑥 = 𝑥 + 7 4. 2𝑦 − 3𝑥 = 10

D. 𝑑𝑦 ⁄𝑑𝑥 =

6𝑥−3 2𝑦

What’s New

Do you recall the steps in solving mathematical word problems? Enumerate them and write your answers on your paper.

5

What is It

Related rates refer to the ratio of two or more quantities which are closely related or affiliated to each other. Often these related quantities vary with respect to time. Conventionally, related rates problem features finding the rate of change of one quantity that has relationship with the other. The application of implicit differentiation is very helpful in solving such word problem. Below are some helpful guides in solving related rates problems. 1. Read and understand the problem carefully. 2. Identify and write down all given quantities and quantities to be determined. Make a sketch or draw a diagram based on the given word problem whenever necessary. 3. Write a mathematical relation or equation that relates all variables concerned including its rate of change. 4. Apply implicit differentiation on the chosen equation. 5. Substitute all given quantities in the equation. 6. Solve for the required rate of change. Use algebraic manipulation whenever necessary. Example 1 Given two variables 𝑎 and 𝑏 which are both differentiable functions of 𝑡. They are related by the equation 𝑏 = 𝑎 2 − 7. Given that 𝑑𝑎⁄ 𝑑𝑡 = 2, find 𝑑𝑏 ⁄𝑑𝑡 when 𝑎 = 4. Explanation

Solution

Identification of quantities Rate of change of b with respect to time t, 𝑑𝑏⁄ 𝑑𝑡 =? that are given and needs to Related rates equation, 𝑏 = 𝑎 2 − 7 be determined in the problem. Rate of change of a with respect to time t, 𝑑𝑎⁄ 𝑑𝑡 = 2 Coordinate, 𝑎 = 4 𝑏 = 𝑎2 − 7

Use the given related equation and take its derivatives on both terms 6

𝑑(𝑏)

with respect to 𝑡 or time. (Note, apply implicit differentiation and other derivative rules depending on the given equation.)

𝑑𝑡

=

𝑑(𝑎 2 − 7) 𝑑𝑡

𝑑𝑏 𝑑(𝑎 2 ) 𝑑(7) = − 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑏 𝑑𝑎 −0 = 2𝑎 𝑑𝑡 𝑑𝑡

Substitute the given quantities on the resulting equation to get the answer.

𝑑𝑎 𝑑𝑏 = 2𝑎 𝑑𝑡 𝑑𝑡 𝑑𝑏 = 2(4)(2) 𝑑𝑡 𝑑𝑏 = 16 𝑑𝑡

Example 2 Assume that a point is moving along the graph 𝑥 2 + 2𝑦 2 = 12. When the point is at (−2, 2), its 𝑥-coordinate is increasing at the rate of 0.4 unit per second. How fast is the y-coordinate changing at that moment? (note: see figure 2 below.) Graph of 𝑥 2 + 2𝑦 2 = 12

Point (-2, 2)

Figure 2 Note: All graphing and plotting of coordinates was created through GeoGebra and Desmos Graphing Calculators. Visit www.geogebra.org and www.desmos.com to learn how to use their free online graph calculator application.

7

Explanation

Solution

Identification of quantities Rate of change of y with respect to time t, 𝑑𝑦⁄ 𝑑𝑡 =? that are given and needs to 2 2 be determined in the Related rates equation, 𝑥 + 2𝑦 = 12 problem. Rate of change of x with respect to time t, 𝑑𝑥⁄ 𝑑𝑡 = 0.4 unit/sec. Coordinate, (−2, 2) 𝑥 2 + 2𝑦 2 = 12

Use the given related equation and take its derivatives on both terms with respect to 𝑡 or time. (Apply implicit differentiation and other derivative rules depending on the given equation.)

Substitute the given quantities on the resulting equation and apply algebraic manipulation to get the value of 𝑑𝑦⁄ 𝑑𝑡.

𝑑(𝑥 2 ) 𝑑(2𝑦 2 ) 𝑑(12) = + 𝑑𝑡 𝑑𝑡 𝑑𝑡 2𝑥

𝑑𝑥 𝑑𝑦 + 4𝑦 =0 𝑑𝑡 𝑑𝑡

(2)(−2)(0.4 𝑢𝑛𝑖𝑡⁄𝑠𝑒𝑐 ) + 4(2) −1.6 𝑢𝑛𝑖𝑡⁄𝑠𝑒𝑐 + 8

8

𝑑𝑦 𝑑𝑡

𝑑𝑦 =0 𝑑𝑡

𝑑𝑦 =0 𝑑𝑡

= 1.6 𝑢𝑛𝑖𝑡⁄ 𝑠𝑒𝑐

𝑑𝑦 𝑑𝑡 = 1.6 𝑢𝑛𝑖𝑡 ⁄𝑠𝑒𝑐 8 8

8

𝑑𝑦 = 0.2 𝑢𝑛𝑖𝑡⁄𝑠𝑒𝑐 𝑑𝑡 Note: always include the unit of the resulting value.

Example 3 A wooden plank stands on the ground and leans against a brick wall. The plank is 7 feet long. The base of the plank is slipping away from the wall at the rate of 2.5 feet per second. How fast is the top of the wooden plank slipping down when it is 5 feet above the horizontal ground?

8

𝑧 = 7 𝑓𝑡.

Figure 3

Explanation

Solution

Identification of quantities that are given and needs to Rate of change of y with respect to time t, 𝑑𝑦⁄ 𝑑𝑡 =? be determined in the problem. Related rates equation, 𝑥 2 + 𝑦 2 = 𝑧2 (Note 1: Since the plank that leans on the wall forms a right triangle, the Pythagorean equation suites the related rates equation.) (Note 2: the horizontal rate is increasing therefore

Wooden plank measurement, 𝑧 = 7 ft. (Movement of plank at its bottom part) Rate of change of x with respect to time, 𝑑𝑥⁄ 𝑑𝑡 = 2.5 ft./sec. The vertical distance from the ground, 𝑦 = 5𝑓𝑡.

𝑑𝑥⁄ 𝑑𝑡 is a positive.) 𝑥 2 + 𝑦 2 = 𝑧2

Use the related equation and take its derivatives on both terms with respect to 𝑡 or time. (Apply implicit differentiation and other derivative rules depending on the given equation.)

𝑑(𝑥 2 ) 𝑑(𝑦 2 ) 𝑑(7) + = 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑧 = 7ft which is a constant,

Therefore, 9

(Note on this situation, the variable 𝑧 is a constant because it has a fixed measurement.) Substitute the given quantities on the resulting equation and apply algebraic manipulation to get the value of 𝑑𝑦⁄ 𝑑𝑡. (Note 1: since the value of 𝑥 is not directly given, we can get it by using the Pythagorean theorem equation.)

2𝑥

𝑑𝑥 𝑑𝑡

+ 2𝑦

𝑑𝑦 =0 𝑑𝑡

(2)(𝑥)(2.5 𝑓𝑡 ⁄𝑠𝑒𝑐 ) + (2)(5𝑓𝑡)

𝑑𝑦 =0 𝑑𝑡

Solving for the value of 𝑥, 𝑧 2 = 𝑥2 + 𝑦 2 𝑥 = √𝑧 2 − 𝑦 2 𝑥 = √72 − 52 𝑥 = √74 or ≈ 8.6 ft 𝑑𝑦

Solving for the value of 𝑑𝑡 , (2)(8.6𝑓𝑡)(2.5 𝑓𝑡 ⁄𝑠𝑒𝑐 ) + (2)(5𝑓𝑡) 𝑑𝑦 =0 𝑑𝑡 𝑑𝑦 = −43 𝑓𝑡 2⁄ 𝑠𝑒𝑐 10 𝑓𝑡 𝑑𝑡

43 𝑓𝑡 2⁄ 𝑠𝑒𝑐 + 10 𝑓𝑡

𝑑𝑦 2 𝑑𝑡 = −43 𝑓𝑡 ⁄𝑠𝑒𝑐 10 𝑓𝑡 10 𝑓𝑡

10 𝑓𝑡 Note 2: the negative sign denotes the downward direction of the motion of the plank. Always include the unit of the resulting value.

𝑑𝑦 𝑑𝑡

10

= −4.3 𝑓𝑡 ⁄𝑠𝑒𝑐

𝑑𝑦 =0 𝑑𝑡

What’s More

Using related rates problem-solving method, solve for each item. Write your answers on a sheet of paper. Table 1 Find 1.

2.

3.

Given = 4, when 𝑥 = 3

𝑦 = 3𝑥 2 − 2

= 2, when 𝑥 = 2

𝑦 = 𝑥3 + 4

= 2, when 𝑥 = −3 and 𝑦=3

𝑥 2 + 𝑦 2 = 18

𝑑𝑦 𝑑𝑡

𝑑𝑥

𝑑𝑥

𝑑𝑦 𝑑𝑡

𝑑𝑡

𝑑𝑡

𝑑𝑦 𝑑𝑡

𝑑𝑥 𝑑𝑡

Related Rates Equation

What I Have Learned Apply what you have learned in this lesson by answering the questions below. Write your answers on a sheet of paper. (5pts. each) 1. Describe related rates problem.

2. How can you solve related rates problems?

11

What I Can Do

Solve the related rates problem below. Write your answer on a sheet of paper. A surfboard 9 ft. long stands on the ground and leans against a wall. Due to a strong external force, the bottom part of the board moves horizontally against the wall at a rate of 0.8 ft./sec. How fast is the top of the surfboard moving up the wall when its bottom part is 4 ft. from the wall? (See Figure 4)

Figure 4

12

Assessment I.

For items 1 to 4, answer each item taken from the word problem. Write your answer on your paper.

Two variables 𝑎 and 𝑏 are both differentiable functions of 𝑡 and are related by the equation 𝑏 = 2𝑎2 − 5. Given that 𝑑𝑎 ⁄𝑑𝑡 = 5. Find 𝑑𝑏 ⁄𝑑𝑡 when 𝑎 = 3. 1. What is being asked on the problem? A. derivative of 𝑥 with respect to 𝑡 B. derivative of 𝑦 with respect to 𝑡

C. derivative of 𝑏 with respect to 𝑡 D. derivative of 𝑎 with respect to 𝑡

2. Which are the correct quantities based on the word problem? A. 𝑑𝑦⁄ 𝑑𝑡 = 5 when 𝑥 = 3 and 𝑦 = 2𝑥2 − 5 B. 𝑑𝑎⁄ 𝑑𝑡 = 5, 𝑤ℎ𝑒𝑛 𝑎 = 3 and 𝑏 = 2𝑎 2 − 5 C. 𝑑𝑥⁄ 𝑑𝑡 = 5, 𝑤ℎ𝑒𝑛 𝑦 = 3 and 𝑥 = 2𝑦 2 − 5 D. 𝑑𝑏 ⁄𝑑𝑡 = 5 when 𝑏 = 3 and 𝑎 = 2𝑏 2 − 5 3. Taking the derivative of the related equation 𝑏 = 2𝑎2 − 5 with respect to time results to ___. A. B.

𝑑𝑎 𝑑𝑡 𝑑𝑥 𝑑𝑡

= 4𝑏

𝑑𝑏

= 4𝑥

𝑑𝑦 𝑑𝑡

C.

𝑑𝑡

D.

𝑑𝑦 𝑑𝑡 𝑑𝑏 𝑑𝑡

= 4𝑦

𝑑𝑥 𝑑𝑡 𝑑𝑎

= 4𝑎 𝑑𝑡

4. Once done from questions 1 to 3, solve the word problem given above.

II.

A. 40

C. 60

B. 50

D. 70

In items 5 to 9, answer each item taken from the word problem below. Write your answers on your paper.

Imagine that a point is moving along the graph 𝑦 = 5𝑥 2 + 3𝑥. The horizontal rate of change is 𝑑𝑥⁄𝑑𝑡 = 3 𝑐𝑚 ⁄𝑠 . Find its 𝑑𝑦⁄𝑑𝑡 when 𝑥 = −2. 5. The problem is about evaluating its vertical rate of change. True or False? A. True B. False

C. Maybe D. Sometimes

13

6. Below are the quantities found in the problem except for one. Which item is it? A. 𝑥 = −2

C.

B. 𝑦 = 5𝑥 2 + 3𝑥

𝑑𝑥 𝑑𝑡

=3

D. 𝑥 =

𝑐𝑚 𝑠

5𝑦 2

+3

7. Getting the derivative of the equation, 𝑦 = 5𝑥2 + 3𝑥 with respect to time, will result to _____. 𝑑𝑦

A. 𝑑𝑡 = 10𝑥 𝑑𝑦

B. 𝑑𝑡 = 3𝑥

+3

𝑑𝑥

+ 10

𝑑𝑥

𝑑𝑥 𝑑𝑡

𝑑𝑥 𝑑𝑡

𝑑𝑡

𝑑𝑡

C.

𝑑𝑥 𝑑𝑡

= 10𝑥

D.

𝑑𝑥 𝑑𝑡

= 3𝑥

+3

𝑑𝑦

+ 10

𝑑𝑦

𝑑𝑦 𝑑𝑡

𝑑𝑦 𝑑𝑡

𝑑𝑡

𝑑𝑡

8. From questions 5 to 7, solve the word problem above. Choose the correct answer below. A. −48𝑐𝑚⁄𝑠

C. −50𝑐𝑚 ⁄𝑠

B. −49𝑐𝑚⁄𝑠

D. −51𝑐𝑚 ⁄𝑠

9. The negative sign at the answer in question number 8 denotes an increasing rate. A. True B. False

III.

C. Maybe D. Sometimes

In items 10 to 13, answer each of the items taken from the word problem below. Write your answer on a sheet of paper. A point is moving along the graph, 𝑥 2 + 𝑦 2 = 5. At point (−2, 1 ), its 𝑥-coordinate is increasing at the rate of 0.6 unit per minute. How fast is the y-coordinate changing at that moment? (See Figure 5)

Figure 5.

14

10.What is asked in the problem? A. derivative of 𝑎 with respect to 𝑡 B. derivative of 𝑧 with respect to 𝑡 C. derivative of 𝑥 with respect to 𝑡 D. derivative of 𝑦 with respect to 𝑡

11. Which one refers to the quantities presented in the ...