Casio FX-9750G Plus. with the. Activities for the Classroom

PRE-CALCULUS w ith th e Casio FX-9750G Plus Casio, Inc.

PRE-CALCULUS

with the Casio FX-9750G Plus

Activities for the Classroom

9750-PCALC

PRE-CALCULUS

with the

Casio FX-9750G Plus

Activities for the Classroom

Evaluating Trigonometric Functions

Graphing Trigonometric Functions

Curve Fitting with Sine Regression

Amplitude and Period

Evaluating Inverse Trigonometric

Functions

Graphing Inverse Trigonometric

Functions

Trigonometric Applications

Parametric Equations

Polar Equations

Limits

Tangent Lines

Integration

All activities in this resource

are also compatible with the

Casio CFX-9850G Series.

PRE-CALCULUS

with the

Casio FX-9750G Plus

® 2005 by CASIO, Inc. 570 Mt. Pleasant Avenue Dover, NJ 07801

www.casio.com 9750-PCALC

The contents of this book can be used by the classroom teacher to make reproductions for student use. All rights reserved. No part of this publication may be reproduced or utilized in any form by any means, electronic or mechanical, including photocopying, recording, or by any information storage or retrieval system without permission in writing from CASIO. Printed in the United States of America.

Contents

Activity 1: Employment Cycles . . . 1

Teaching Notes

Student Activity

Solutions and Screen Shots

Activity 2: Monthly Temperatures . . . 8

Teaching Notes

Student Activity

Solutions and Screen Shots

Activity 3: Pure Tones . . . 14

Teaching Notes

Student Activity

Solutions and Screen Shots

Activity 4: Total Internal Reflection . . . 22

Teaching Notes

Student Activity

Solutions and Screen Shots

Activity 5: Damped Pendulum . . . 27

Teaching Notes

Student Activity

Solutions and Screen Shots

Activity 6: Area Of A Triangle And Circular Sector . . . 35

Teaching Notes

Student Activity

Solutions and Screen Shots

Activity 7: Projectile Motion . . . 45

Teaching Notes

Student Activity

Activity 8: Rose Petals . . . 53

Teaching Notes

Student Activity

Solutions and Screen Shots

Activity 9: Take It To The Limit . . . 61

Teaching Notes

Student Activity

Solutions and Screen Shots

Activity 10: Slope of Curves . . . 70

Teaching Notes

Student Activity

Solutions and Screen Shots

Activity 11: Up and Down . . . 77

Teaching Notes

Student Activity

Solutions and Screen Shots

Activity 12: Area Under A Curve . . . 85

Teaching Notes

Student Activity

Solutions and Screen Shots

Activity 1

Employment Cycles

Topic Area: Evaluating and Graphing Trigonometric Functions

NCTM Standards:

• Compute fluently by developing fluency in operations with real numbers using technology for more-complicated cases

• Understand functions by interpreting representations of functions

Objective

• To evaluate and graph a trigonometric function

Getting Started

In this activity the students will learn how to evaluate and graph a trigonometric function.

Trigonometric functions are periodic, cyclical or circular functions with the angle measurement (

) in degrees or radians. The sine (sin) and cosine (cos) relation-ships map the angle measurement (

) to the yand xcoordinates of the point on the unit circle that represents the angle.

For example, the sine relationship for the unit circle is defined to be: sin

= y/ r= y/ 1 = y.

The additional trigonometric relationships of tangent (tan), cotangent (cot), secant (sec) and cosecant (csc) are defined in terms of the sine and cosine relationships. For example, tan

= sin

/ cos

= y/ x.

The trigonometric functions are defined by letting the independent variable (x) be the angle measurement and letting the dependent variable (y) be the evaluation of the trigonometric relationship at the angle. Therefore, the six trigonometric func-tions are:

y= sin x, y= cos x, y= tan x, y= cot x, y= sec x, and y= csc x.

The graph of the sine function appears as follows on the Casio fx-9750G PLUS:

Name _____________________________________________ Class ________ Date ________________

Introduction

The sine function is often used to model cyclical or periodic behavior. There are many applications for the sine function where measure data goes up and down. One of these applications is in modeling employment cycles.

When you look at the employment of some industries over a period of years, the number of employees may go up and down. In this activity, you will work with a model that represents the employment cycle for a particular industry.

In the employment model, y= 1.5sin (2x+ 0.3) + 6, yis the number of employees (in hundreds) and xis time in years since the analysis began.

Problems and Questions

1. Using the model, estimate number of employees for each quarter for the three years. Remember time is in terms of years.

Therefore, to calculate the number of employees at the end of the first quarter you will need to enter an xof .25.

Round the number of employees to the nearest whole person. Beginning = __________ End of 1st quarter = __________ End of 2nd quarter = __________ End of 3rd quarter = __________ End of 1st year = __________ End of 5th quarter = __________ End of 6th quarter = __________ End of 7th quarter = __________ End of 2nd year = __________ End of 9th quarter = __________ End of 10th quarter =__________ End of 11th quarter =__________ End of 3rd year = __________

2. Graph the employment model from time = 0 to time = 4 years.

3. What viewing window did you use to see the graph? Xmin = __________ Ymin = __________ Xmax = __________ Ymax = __________

4. Using the trace feature, what appears to be the largest number of employees? _____________________________________

5. What appears to be the least number of employees? _____________________________________

6. What is the baseline or "average" employment? _____________________________________

7. What is the period or length of the employment cycle? _____________________________________

Name _____________________________________________ Class ________ Date ________________

1. To estimate number of employees for each month of the first year you will need to use the Run Menu. Enter the Run Menu from the MAIN MENU by pressing 1

or use the arrow keysto highlight RUN and press EXE. A blank screen should appear. If it does not, press AC/ON to clear the screen.

Make sure the calculator is in radian mode by pressing SHIFT SETUP and

arrow downto highlight Angle. Press F2(Rad) to select radians.

Press EXIT to exit the Set Up screen. Calculate the beginning number of

employees by entering 0 for xand calculating the expression 1.5sin (2(0) + 0.3) + 6. Enter and calculate this by pressing 1 . 5 sin ( 2 ( 0 ) + . 3 ) + 6 EXE.

The beginning number of employees is approximately 644. Find the number of employees for the end of the second quarter by entering .25 for x. Do this by editing the previous calculation. To recall the previous expression press AC/ON

and the up arrowonce. Arrow leftto the 0 and press DEL. Insert the .25 by pressing SHIFT INS . 2 5 . Press EXE to find the number of employees at the end of the first quarter.

The number of employees at the end of the first quarter is approximately 708. Repeat for the additional quarters. The answers will be end of 2nd quarter = 745, end of 3rd quarter = 746, end of 1st year = 712, end of 5th quarter = 650, end of 6th quarter = 576, end of 7th quarter = 508, end of 2nd year = 463, end of 9th quarter = 451, end of 10th quarter = 475, end of 11th quarter = 530, and end of 3rd year = 603.

2. Graph the employment model y= 1.5sin (2x+ 0.3) + 6 by pressing MENU 5

(GRAPH) and entering the expression for Y1. Enter the expression by pressing

1 . 5 sin ( 2 X,

,T + . 3 ) + 6 .

To graph the model from time = 0 to time = 4, press SHIFT F3 (VWIN) to access the View Window screen. Enter 0 for Xmin and 4 for Xmax by pressing

0 EXE 4 EXE

Press EXITto exit the View Window screen and press F6 (DRAW) F2 (ZOOM)

F5 (AUTO) to view the graph.

3. Press SHIFT F3(VWIN) to obtain the View Window screen. The window parameters are Xmin = 0, Xmax = 4, Ymin = 4.5, and Ymax = 7.5.

4. Press EXIT to exit the View Window screen and press F6(DRAW) to view the graph of the model. Press F1 (Trace) to enter the trace mode. A tracer will appear on the curve. Press the left arrow keyto move the tracer to the peak of curve. You will have to watch the y-values at the bottom of the screen to see the maximum y-value. The largest number of employees is estimated to be 750.

5. Press the right arrow keyto move the tracer to the valley of curve. You will have to watch the y-values at the bottom of the screen to see the y-value. The least number of employees is estimated to be 450.

6. The baseline or "average" employment is the horizontal line of symmetry between the peaks and valleys of the graph. This value can be found by averaging the minimum and maximum values for the model. The base line or "average" employment is (750 + 450) / 2 = 600. To calculate this average press

MENU 1 (RUN) ( 7 5 0 + 4 5 0 ) ÷ 2 EXE.

7. The period or length of the employment cycle would be twice the distance from the peak to the valley. The peak or largest employment occurred at an xor time value of .63. The valley or least employment occurred at an xor time value of 2.19. Twice the distance between the peak and valley is 2(2.19 - 0.63) = . Calculate this by pressing 2 ( 2 . 1 9 – . 6 3 ) EXE.

The period or length of the employment cycle is a little over 3 years.

Activity 2

Monthly Temperatures

Topic Area: Curve Fitting with Sine Regression

NCTM Standards:

• Use mathematical models to represent and understand quantitative relationships • Understand functions by interpreting representations of functions

• Compute fluently by developing fluency in operations with real numbers using technology for more-complicated cases

Objective

• To compute a sine curve that best "fits" the given data

Getting Started

In this activity, the students will learn how to compute the best-fitting sine curve for a given set of monthly temperatures. Typically, in the course of a year, the monthly temperatures will rise and fall and lend themselves to curve fitting with a sine regression.

Regression is the process of finding the best-fitting curve through the minimization of error.

The monthly temperatures for a city are provided in the table below.

Teaching Notes

Month 1 2 3 4 5 6 7 8 9 10 11 12 Temp. oF 36 41 50 60 68 77 82 81 73 62 50 40

Name _____________________________________________ Class ________ Date ________________

Introduction

The sine function is often used to model cyclical or periodic behavior. There are many applications for the sine function where measure data goes up and down. One of these applications is in modeling monthly temperatures. Typically, in the course of a year, the monthly temperatures will rise and fall and lend themselves to curve fitting with a sine regression. Regression is the process of finding the best-fitting curve through the minimization of error.

In this activity, you will learn how to compute the best-fitting sine curve for a given set of monthly temperatures.

The monthly temperatures for a city are provided in the table below.

The months are represented by 1 for January through 12 for December.

Problems and Questions

1. Draw the scatter plot for the data in the window [-1,13,1,-1,100,10].

2. Calculate the best-fitting sine curve for the given data.

y= _______________________________

3. Graph the model with the scatter plot in the window [-1,13,1,-1,100,10].

4. Using the graphical solver, find the maximum average temperature according to the model found by the regression.

_______________________________

5. Using the graphical solver, find the minimum average temperature according to the model found by the regression.

_______________________________

6. What is the baseline or "average" temperature for the city? _______________________________

Activity 2 • Monthly Temperatures

Month 1 2 3 4 5 6 7 8 9 10 11 12 Temp. oF 36 41 50 60 68 77 82 81 73 62 50 40

1. To draw the scatter plot for the data you must first enter the data in STAT mode. Enter the STAT mode from the MAIN MENU by pressing 2 or using the arrow keysto highlight STAT and press EXE. If old data exists in List 1 and List 2, then clear them by pressing the up arrow keyto highlight List 1 and press F6 F4(DEL-A) F1(YES). Repeat for List 2.

Enter the month data into List 1 and the temperature data into List 2. Do this by pressing 1 EXE and repeating for all of the month data. Press the right arrow key and enter the temperature data.

Set up the scatter plot by pressing F6 F1(GRPH) F6(SET). Set the Graph Type to Scatter by pressing the down arrow keyto highlight Graph Type and press F1 (Scat).

Set the XList to List1 by pressing the down arrow keyto highlight XList and press F1(List1). Set the YList to List2 by pressing the down arrow keyto highlight Ylist and press F2(List2).

Press EXIT to exit the StatGraph1 set up screen. Press F4(SEL) to make sure StatGraph1 is turned on. If it is off, press F1(On) to set the StatGraph1 to DrawOn.

Press EXIT to exit the draw select screen. Enter the viewing window by pressing SHIFT F3 (V-WIN) and enter the window parameters of

[-1,13,1,-1,100,10]. Do this by pressing (-) 1 EXE 1 3 EXE 1 EXE (-) 1 EXE 1 0 0 EXE 1 0 EXE .

Press EXITto exit the View Window screen. Press F1(GRPH) F1(GPH1) to view the scatter plot in the viewing window.

2. Calculate the best-fitting sine curve by pressing F6 F5(Sin).

The best-fitting sine curve is approximately y= 22.85 sin(.5x– 1.96) + 58.81.

3. Graph the model with the scatter plot in the window [-1,13,1,-1,100,10] by

pressing F5 (COPY) EXEto copy the equation into Y1. If you were to return to the graph entry screen it would look like the screen below. Do not return to this screen.

Press F6(DRAW) to graph the model with the scatter plot.

4. To use the graphical solver, you will need to return to Graph mode. Do this by pressing EXIT MENU 5 (GRAPH). Press F6 (DRAW) to view the graph of the model. Find the maximum average temperature by pressing F5(G-Solv) F2

(MAX).

The average temperature for July is reported with a 7 for the month data. Therefore, 7.0 is considered to be the end of July or July 31st. A value of 7.1 would indicate the maximum average temperature occurs early in August with an average of about 82oF.

5. Use the graphical solver to find the minimum average temperature according to the model by pressing F5 (G-Solv) F3 (MIN).

The minimum average temperature of about 36oF occurs at an xvalue of .78 which is less than 1. A value of 1 would indicate the end of January. So a value of .78 indicates the minimum average temperature occurs in late January.

6. The baseline or "average" temperature for the city is the horizontal line of symmetry between the maximum and minimum average temperatures. This value can be found by averaging the minimum and maximum values. The base line or "average" temperature is (82 + 36) / 2 = 59oF. To calculate this average press MENU 1 (RUN) ( 8 2 + 3 6 ) ÷ 2 EXE.

7. To use the graphical solver to find the average temperature for the middle of July, you will need to return to the Graph mode. Do this by pressing MENU 5(GRAPH). Draw the sine curve again by pressing F6(DRAW). The middle of July would be represented by an x-value of 6.5. To find the y-value or average temperature for the x-value of 6.5, press F5(G-Solv) F6 F1(Y-CAL) and then press 6 . 5 EXE.

The average temperature for the middle of July is 81oF.

Activity 3

Pure Tones

Topic Area: Amplitude and Period

NCTM Standards:

• Use mathematical models to represent and understand quantitative relationships • Understand functions by interpreting representations of functions

• Compute fluently by developing fluency in operations with real numbers using technology for more-complicated cases

Objective

• To graph the sine curve representing different pure tones

Getting Started

In this activity, the students will learn how to graph sine curves that represent pure tones. Sounds are vibrations caused by a source and picked up by our ears. Sound waves are usually represented graphically by the sine wave y= Asin(Tx) where A is the amplitude and T is the angular frequency. Frequency is defined to be the num-ber of wave cycles per second and amplitude is the size of the cycles.

According to Fourier theory, all sounds are made up of sine waves at differing fre-quencies and amplitudes. Therefore, sounds can be created by adding different sine waves together. A pure tone is a sound with a single sine wave that has a fixed frequency and amplitude. The frequency of a sound is in terms of Hertz (Hz), for example a piano will play an A above middle C at 440 Hz. This sound is said to take on the form of a 440 Hz sine wave which is represented by the equation

y= sin (2π•440x).

Below is the graph of this sine curve in the window [0, 0.01,1,-1,1,1].

Name _____________________________________________ Class ________ Date ________________

Introduction

In this activity, you will learn how to graph sine curves that represent pure tones. Sounds are vibrations caused by a source and picked up by our ears.

Sound waves are usually represented graphically by the sine wave y= Asin(Tx) where A

is the amplitude and T is the angular frequency. Frequency is defined to be the number of wave cycles per second and amplitude is the size of the cycles.

According to Fourier theory, all sounds are made up of sine waves at differing frequencies and amplitudes. Therefore, sounds can be created by adding different sine waves together. A pure tone is a sound with a single sine wave that has a fixed frequency and amplitude.

The frequency of a sound is in terms of Hertz (Hz), for example a piano will play an A above middle C at 440 Hz. This sound is said to take on the form of a 440 Hz sine wave which is represented by the equation y= sin (2π•440x).

Problems and Questions

1. Draw the 440 Hz sine wave in the window for the data in the window [0, 0.01,1,-1,1,1].

2. What is the amplitude and period of the 440 Hz sine wave? amplitude = ____________________

period = ____________________

3. Graph the 440 Hz sine wave in the window [0, 0.01,1,-2,2,1] using the Dynamic Function and the equation y= Asin (2π•440x).

Vary the amplitude A from 0.5 to 2 with an increment of 0.5. Describe what you see when the amplitude changes.

______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________

Name _____________________________________________ Class ________ Date ________________

4. Graph the 440 Hz sine wave in the window [0, 0.01,1,-1,1,1] using the Dynamic Function and the equation y= sin (2_π•Hx).

Vary the Hertz H from 440 to 470 with an increment of 10. Describe what you see when the Hertz changes.

______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________

5. This sine wave has a very short period. The display of the graph takes on some interesting forms in larger windows. Draw the graph in the following windows: [0,0.1,1,-1,1,1]

[0,0.5,1,-1,1,1] [0,1,1,-1,1,1] [0,5,1,-1,1,1] [0,10,1,-1,1,1]

Describe what is happening in the last window?

______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________

1. To draw the 440 Hz sine wave in the given window, press MENU 5(GRAPH) and enter the sine equation by pressing sin ( 2 SHIFT _π 4 4 0 X,

,T ) EXE.

Enter the viewing window by pressing SHIFT F3 (V-WIN) 0 EXE . 0 1 EXE 1 EXE (-) 1 EXE 1 EXE 1 EXE.

To view the graph press EXIT F6(DRAW).

2. In the general sine function f(x) = A sin (Bx), the amplitude for the sine curve is the coefficient A in front of the sine function. In our function, it an implied 1. Therefore, the amplitude of this sine curve is 1.

The period of a sine curve is determined by the coefficient B of the variable inside the argument of the function. The period of a sine function is 2π/B. In our case, you divide 2_πby the coefficient of 2_π•440. The resulting period is 1/440 or 0.0023. Find this by pressing MENU 1(RUN) AC/ON 1 ÷ 4 4 0 EXE.

3. To enter the 440 Hz sine wave in the Dynamic Function, press MENU 6(DYNA) and press ALPHA A sin ( 2 SHIFT _π 4 4 0 X,

,T ) EXE.

Set the viewing window, by pressing SHIFT F3 (V-WIN) 0 EXE . 0 1 EXE 1 EXE (-) 2 EXE 2 EXE 1 EXE.

Set the variable A from 0.5 to 2 with an increment of 0.5 by pressing EXIT F4 (VAR) F2(RANG) . 5 EXE 2 EXE . 5 EXE.

Set the speed to Stop&Go by pressing EXIT F3(SPEED) F1(SEL).

Draw the graph by pressing EXIT F6(DYNA). The calculator will display "One Moment Please!" while it calculates the graphs. This first graph to appear is when A = 0.5. Press EXE to view the next graph when A = 1. Press EXEto view each additional graph.

As the amplitude increases, the height of the graph also increases.

4. To enter the new Hz sine wave in the Dynamic Function, press EXIT three times and press sin ( 2 SHIFT _π ALPHA H X,

,T ) EXE.

Set the viewing window, by pressing SHIFT F3 (V-WIN) 0 EXE . 0 1 EXE 1 EXE (-) 1 EXE 1 EXE 1 EXE.

Set the variable H from 440 to 470 with an increment of 10 by pressing EXIT F4 (VAR) F2(RANG) 4 4 0 EXE 4 7 0 EXE 1 0 EXE.

Draw the graph by pressing EXIT F6(DYNA). The calculator will display "One Moment Please!" while it calculates the graphs. This first graph to appear is when H = 440 Hz. Press EXEto view the next graph when H = 450 Hz. Press EXE to view each additional graph.

The graphs demonstrate as the Hz value is increased the period of the sine wave decreases. (Watch the movement of the fifth peak from the left.)

5. To graph the sine wave in different windows, press EXIT twice and then

MENU 5(GRAPH). Re-enter the sine equation by pressing sin ( 2 SHIFT π

4 4 0 X,

,T ) EXE.

Enter the first viewing window by pressing SHIFT F3 (V-WIN) 0 EXE . 1 EXE 1 EXE (-) 1 EXE 1 EXE 1 EXE. To view the graph press EXIT F6(DRAW).

Repeat to view in the other windows.

As the window gets larger horizontally, the calculator has a harder time showing all the cycles, so it finally shows a sine curve similar to the one in a smaller viewing window.

Activity 4

Total Internal Reflection

Topic Area: Evaluating and Graphing Inverse Trigonometric Functions

NCTM Standards:

• Compute fluently by developing fluency in operations with real numbers using technology for more-complicated cases

• Understand functions by interpreting representations of functions

Objective

• To evaluate and graph an inverse trigonometric function

Getting Started

In this activity, the students will learn how to evaluate and graph an inverse

trigonometric function. Inverse trigonometric functions are found by swapping the range and domain of the corresponding trigonometric function. For example, the domain of the sine function becomes the range of the inverse sine function and the range of the sine function becomes the domain of the inverse trigonometric func-tion. The inverse sine function is denoted using arcsin or sin-1_.

Total internal reflection (TIR) occurs when all of the light of a beam within a medium strikes a boundary and is reflected inside the medium, instead of exiting the medium. For TIR to occur, the light source must be in a more dense medium and hit a less dense boundary. Examples of a more dense medium might be water or a diamond in comparison to a less dense boundary such as air. Also for TIR to occur, the angle in which the light strikes the boundary must be greater than or equal to a value determined by the two mediums in use. This value is called the critical angle for the two mediums.

The critical angle (

) for the two mediums can be determined by using the formula

= sin-1_(n

r/ ni) where nrand nr are indices of refraction, the ratio of (nr/ ni) is

less than 1.0, and

is in degrees. The value n_ris the index of refraction for the the less dense refractive medium and n_iis the index of refraction for the more dense incident medium.

Several indices of refraction are given in the table below: Medium Index Air 1.00 Ice 1.31 Water 1.33 Ethanol 1.36 Glycerine 1.47 Crown Glass 1.50 Polystyrene 1.59 Flint Glass 1.75 Diamond 2.42

Teaching Notes

Name _____________________________________________ Class ________ Date ________________

Introduction

In this activity, the students will learn how to evaluate and graph an inverse trigonometric function. The inverse sine function is denoted using arcsin or sin-1_.

Total internal reflection (TIR) occurs when all of the light of a beam within a medium strikes a boundary and is reflected inside the medium, instead of exiting the medium. For TIR to occur, the light source must be in a more dense medium and hit a less dense boundary. Examples of a more dense medium might be water or a diamond in comparison to a less dense boundary such as air. Also for TIR to occur, the angle in which the light strikes the boundary must be greater than or equal to a value determined by the two mediums in use. This value is called the critical angle for the two mediums.

The critical angle (

) for the two mediums can be determined by using the formula

= sin-1_(n

r / ni) where nrand nrare indices of refraction, the ratio of (nr/ ni) is

less than 1.0, and

is in degrees. The value n_r is the index of refraction for the the less dense refractive medium and n_i is the index of refraction for the more dense incident medium.

Problems and Questions

1. Using the formula, find the critical angle for the given boundaries:

a. water (n_i = 1.33) - air (n_r = 1.00) boundary

b. diamond (n_i = 2.42) - water (n_r = 1.33) boundary

c. crowned glass (n_i = 1.50) - ice (n_r = 1.31) boundary

2. Graph the model for the fixed refractive medium (air) (n_r = 1) and a variable incident medium (n_i = x) from n_i= x = 1 to x= 3.

3. Using the trace feature, what is the critical angle for:

a. a polystyrene (x= 1.59) - air boundary?

= ____________________

b. a flint glass (x = 1.75) - air boundary?

= ____________________

4. Use the graphical solver feature to find the index of reflection for the incident medium, that would produce a critical angle of:

a. 45o

n_i= ____________________

b. 60o

n_i= ____________________

1. To find the critical angle for the given boundaries you will need to use the Run Menu. Enter the Run Menu from the MAIN MENU by pressing 1(RUN) or use the arrow keysto highlight RUN and press EXE. A blank screen should appear. If it does not, press AC/ON to clear the screen. Make sure the calculator is in degree mode by pressing SHIFT SETUP and arrow downto highlight Angle. Press F1(Deg) to select .

Press EXIT to exit the Set Up screen. Calculate the critical angle for the water (n_i= 1.33) - air (n_r= 1.00) boundary by calculating the expression

sin-1(1/ 1.33). Enter and calculate this by pressing SHIFT sin-1 ( 1 1 . 3 3 ) EXE.

The critical angle for the water - air boundary is 48.75o. Repeat for the other two boundaries.

The critical angle for the diamond–water boundary is 33.34oand the critical angle for the crowned glass–ice boundary is 60.85o.

2. To graph the model for the fixed refractive medium (air) (n_r= 1) and a variable incident medium (n_i = x) press MENU 5 (GRAPH) and enter the expression

y= sin-1_{(1/x) for Y1.}

Enter the expression by pressing SHIFT sin-1 _{( 1 X,}

_{,T ).}

To graph the model fromx= 1 to x = 3, press SHIFT F3(VWIN) to access the View Window screen. Enter 1 for Xmin and 3 for Xmax by pressing 1 EXE 3 EXE.

Press EXITto exit the View Window screen and press F6 (DRAW) F2(ZOOM)

F5 (AUTO) to view the graph.

3. To use the trace feature to find the critical angle for a polystyrene

(x= 1.59) / air boundary, press F1 (Trace) to enter the trace mode. A tracer will appear on the curve. Press the left arrow keyto move the tracer near x= 1.59. The critical angle is approximately 39o.

Find the critical angle for a flint glass (x= 1.75) / air boundary, by pressing the

right arrow keyto move the tracer near 1.75. The critical angle is approximately 35o.

4. To use the graphical solver feature to find the index of reflection for the incident medium, that would produce a critical angle of 45o, press F5(G-Solv) F6 F2 (X-CAL) and press 4 5 EXE. The incident medium would need to have an index of reflection of approximately n_i = 1.41.

Repeat for 60o. The incident medium would need to have an index of reflection of approximately n_i = 1.15.

Activity 5

Damped Pendulum

Topic Area: Trigonometric Applications

NCTM Standards:

• Compute fluently by developing fluency in operations with real numbers using technology for more-complicated cases

• Understand functions by interpreting representations of functions

Objective

• To evaluate and graph a model for a damped pendulum

Getting Started

In this activity, the students will learn how to evaluate and graph a model for a damped pendulum. A pendulum is a weight located on one end of a rod, wire, or string/rope that is attached on the other end to a fixed position. When pushed, the pendulum’s weight will swing back and forth, under the influence of gravity, over the lowest or center point.

In a perfect setting, the pendulum would not be met with air resistance or friction and it’s motion would continue indefinitely and if the motion was tracked it would form a regular periodic curve.

However, we do not live in a perfect setting, so the motion of the pendulum is damped or slowed by friction. The motion of a damped pendulum is modeled by a sine function whose amplitude is decreasing as time increases. This amplitude decreases in proportion to an exponential function with a base less than 1. The model for a particular damped pendulum is:

y= (0.85)X_{1.03 sin (3.5 X - 1.8)}

where yis the distance (in feet) and direction (negative is left, positive is right) away from the center point 0, and xis time passed in seconds. The graph of the damped pendulum’s motion is shown below:

Name _____________________________________________ Class ________ Date ________________

Introduction

A pendulum is a weight located on one end of a rod, wire, or string/rope that is attached on the other end to a fixed position. When pushed, the pendulum’s weight will swing back and forth, under the influence of gravity, over the lowest or center point In a perfect setting, the pendulum would not be met with air resist-ance or friction and it’s motion would continue indefinitely and if the motion was tracked it would form a regular periodic curve.

However, we do not live in a perfect setting, so the motion of the pendulum is damped or slowed by friction. The motion of a damped pendulum is modeled by a sine function whose amplitude is decreasing as time increases. This amplitude decreases in proportion to an exponential function with a base less than 1. The model for a particular damped pendulum is:

y= (0.85)X1.03 sin (3.5 X - 1.8)

where yis the distance (in feet) and direction (negative is left, positive is right) away from the center point 0, and xis time passed in seconds.

Problems and Questions

1. Using the formula, find the position of the damped pendulum for the given time:

a. x= 2 seconds y= __________

b. x= 4 seconds y= __________

c. x= 6 seconds y= __________

d. x= 8 seconds y= __________

2. Graph the model for the motion of the damped pendulum from x= -0.1 seconds to 10 seconds, and from y= -2 feet to 2 feet. Draw your graph in the box

below.

3. Using the solver feature, find the initial position of the pendulum. This is represented by the y-intercept of the curve. The initial position is _____ feet to the left/right (choose one) of the center point.

4. Using the trace feature, find the first five peaks of the graph which represent the first five right-hand swing positions of the damped pendulum. Then, find the ratio for each successive peak change by dividing the latter peak height by the previous peak height. What did you find?

Name _____________________________________________ Class ________ Date ________________

a. peak 1, x= _____ y= _____

b. peak 2, x= _____ y= _____ "yratio 2" = "ypeak 2" "ypeak 1" = _____

c. peak 3, x= _____ y= _____ "yratio 3" = "ypeak 3" "ypeak 2" = _____

d. peak 4, x= _____ y= _____ "yratio 4" = "ypeak 4" "ypeak 3" = _____

e. peak 5, x= _____ y= _____ "yratio 5" = "ypeak 5" "ypeak 4" = _____

5. Find the ratio for each successive peak change by dividing the previous peak time by the latter peak time. What did you find?

"xratio 2" = "xpeak 1" "xpeak 2" = _____ "xratio 3" = "xpeak 2" "xpeak 3" = _____ "xratio 4" = "xpeak 3" "xpeak 4" = _____ "xratio 5" = "xpeak 4" "xpeak 5" = _____

6. Increase Xmax in the viewing window, and determine at what elapsed time has the pendulum virtually stopped (peak is less than .05).

x= __________

1. To find the position for the damped pendulum for the given time you will need to use the Run Menu. Enter the Run Menu from the MAIN MENU by pressing

1 (RUN)or use the arrow keysto highlight RUN and press EXE. A blank screen should appear. If it does not, press AC/ON to clear the screen. Make sure the calculator is in radian mode by pressing SHIFT SETUP and arrow downto highlight Angle. Press F2(Rad) to select .

Press EXIT to exit the Set Up screen. Calculate the position for the damped pendulum at 2 seconds by calculating the expression:

0.852(1.03) sin (3.5 (2) - 1.8)

Enter and calculate this by pressing . 8 5 ^ 2 1 . 0 3 sin ( 3 . 5 2 – 1 . 8 ) EXE.

The position of the pendulum at x= 2 seconds is -0.657 feet or 0.657 feet to the left of center. Repeat for the other three times. Do this by pressing the right arrow keyto replay the expression and then press the right arrow keyto highlight the 2 and press 4 . Do this for both xor 2 values.

Press EXE to see the answer. Repeat for x= 6 and x= 8.

The position of the pendulum at x= 4 seconds is -0.19 feet or 0.19 feet to the left of center. The position of the pendulum at x= 6 seconds is 0.13 feet or .13 feet to the right of center.

The position of the pendulum at x= 8 seconds is 0.246 feet or 0.246 feet to the right of center.

2. To graph the model for the motion of the damped pendulum press MENU 5 (GRAPH) and enter the expression y= 0.85X 1.03 sin (3.5 X - 1.8) for Y1. Enter the expression by pressing . 8 5 ^ X,

,T 1 . 0 3 sin ( 3 . 5 X,

,T – 1 . 8 ) EXE .

To graph the model from x= -0.1 seconds to 10 seconds, and from y= -2 feet to 2 feet, press SHIFT F3 (VWIN) to access the View Window screen. Enter -0.1 for Xmin and 10 for Xmax by pressing (-) . 1 EXE 1 0 EXE 1 EXE. Enter -2 for Ymin and 2 for Ymax by pressing (-) 2 EXE 2 EXE 1 EXE.

Press EXITto exit the View Window screen and press F6 (DRAW) to view the graph.

3. To use the graphical solver feature to find the initial position of the pendulum, represented by the y-intercept of the curve, press F5(G-Solv) F4 (Y-ICPT) .

The initial position was 1 foot to the left of the center point.

4. To use the trace feature to find the first five peaks of the graph which represent the first five right-hand swing positions of the damped pendulum, press

F1 (Trace) to enter the trace mode. A tracer will appear on the curve. Press the

left arrow key to move the tracer to the first peak at x= 0.942 seconds. The peak occurs at .881 feet.

Press the right arrow keyto move the tracer to the second and third peaks. The peaks occur at x= 2.706 seconds and y= 0.652 feet and x= 4.549 seconds and

y= 0.492 feet respectively.

Press the right arrow key to move the tracer to the fourth and fifth peaks. The peaks occur at x= 6.313 seconds and y= 0.366 feet and x= 8.156 seconds and

y= 0.273 feet respectively.

To find the ratio for each successive peak change, return to the Run menu by pressing MENU 1 (RUN). Divide the "ypeak 2" = .652 by "ypeak 1" = .881 by pressing

.

6 5 2 . 8 8 1 EXE. The "yratio 2" = .74. Repeat for the other ratios.

The resulting ratios are virtually the same, therefore the friction is damping the pendulum in a proportional manner. In other words, the pendulum is slowing down in a constant manner.

5. To find the ratio for each successive peak change, divide the previous peak time by the latter peak time. The first ratio (xratio 2) is found by dividing "xpeak 1" = 0.942 by "xpeak 2" = 2.706. Do this by pressing . 9 4 2 2 . 7 0 6

EXE . The "xratio 2" = .. Repeat for the other ratios.

The resulting ratios are increasing, which means the time interval between peaks is decreasing.

6. To find when the pendulum has virtually stopped, you must return to the graph. Do this by pressing MENU 5 (GRAPH). Increase the Xmax to 20 by pressing

SHIFT F3 (V-Window), pressing the down arrow keyand pressing 2 0 EXE.

Press EXE F6(DRAW) to view the graph.

Press F1(Trace) and a tracer will appear on the curve. Press the right arrow and find the peak values. The eleventh peak is y= 0.048.

The elapsed time in which the the pendulum has virtually stopped (a peak less than 0.05) is x= 18.883 seconds.

Activity 6

Area Of A Triangle And Circular Sector

Topic Area: Trigonometric Applications

NCTM Standards:

• Compute fluently by developing fluency in operations with real numbers using technology for more-complicated cases

• Understand functions by interpreting representations of functions

Objective

• To calculate the area of a triangle and a circular sector using trigonometry

Getting Started

In this activity, the students will learn how to calculate the area of a triangle and a circular sector using trigonometry. The area of a triangle is defined to be one-half of the product of the lengths of two sides (a, b) and the sine of the angle (C ) included between those two sides. The formula looks like:

area = 0.5•a•b•sin C

Another useful formula for finding the area of a triangle is Heron’s formula. In this formula, all you need to know is the length of the three sides (a, b, c) to find the area of the triangle. Heron’s formula looks like:

area = [ s(s–a)(s–b)(s–c) where s = 0.5 (a+ b + c)

The area of a circular sector is defined to be one-half of the product of the radius (r) squared and the central angle

. The formula looks like:

area = 0.5•r2•

or

Teaching Notes

a c b r

_r a c b

Introduction

In this activity, you will learn how to calculate the area of a triangle and a circular sector using trigonometry. The area of a triangle is defined to be one-half of the product of the lengths of two sides (a, b) and the sine of the angle (C ) included between those two sides. The formula looks like:

area = 0.5•a•b•sin C

Another useful formula for finding the area of a triangle is Heron’s formula. In this formula, all you need to know is the length of the three sides (a, b, c) to find the area of the triangle. Heron’s formula looks like:

area = [ s(s–a)(s–b)(s–c) where s = 0.5 (a+ b+ c)

The area of a circular sector is defined to be one-half of the product of the radius (r) squared and the central angle

. The formula looks like:

area = 0.5•r2•

or

Problems and Questions

1. Find the area of the following triangles for the given values using the formula in the Run Menu.

a. a= 5, b= 7, C= 35o area = _______________

b. b= 10, c= 8, A= 28o area = _______________

c. a = 6, c= 4, B= 40o area = _______________

Name _____________________________________________ Class ________ Date ________________

Activity 6 • Area Of A Triangle And Circular Sector

a c b r

r

a c b

2. Find the area of the following triangles for the given values using Heron’s formula in the Solver feature.

a. a= 5, b= 7, c= 10 s = _______________ area = _______________ b. a= 10, b = 8, c= 6 s = _______________ area = _______________ c. a= 6, b= 4, c= 8 s = _______________ area = _______________

3. Find the area of the following circular sectors for the given values using the formula in the Solve feature.

a. r= 8,

= _π/5 area = _______________

b. r= 10,

= 2_π/3 area = _______________

c. r = 7,

= 4_π/9 area = _______________

Name _____________________________________________ Class ________ Date ________________

1. To find the area of a triangle for the given values, you will need to use the Run Menu. Enter the Run Menu from the MAIN MENU by pressing 1(RUN) or use the arrow keysto highlight RUN and press EXE. A blank screen should

appear. If it does not, press AC/ON to clear the screen. Make sure the calculator is in degree mode by pressing SHIFT SETUP and arrow downto highlight Angle. Press F1(Deg) to select .

Press EXIT to exit the Set Up screen. Calculate the area of a triangle given

a= 5, b= 7, and C= 35o. Do this by calculating the expression 0.5•5•7•sin 35. Enter and calculate this by pressing . 5 5 7 sin 3 5 EXE. The area is approximately 10.04.

Calculate the area of a triangle given b= 10, c= 8, A= 28o. Do this by calculating the expression 0.5•10•8•sin 28. Enter and calculate this by pressing . 5 1 0

8 sin 2 8 EXE. The area is approximately 18.78.

Calculate the area of a triangle given a= 6, c= 4, B = 40o. Do this by calculating the expression 0.5•6•4•sin 40. Enter and calculate this by pressing . 5 6 4 sin 4 0 EXE. The area is approximately 7.71.

2. To use the Solver feature to find the area of a triangle for the given values, you will need to use the Equation Menu. Enter the Equation Menu from the MAIN MENU by pressing A (EQUA) or use the arrow keysto highlight EQUA and press EXE.

The Equation menu will appear.

Press F3(Solv) to access the Solver feature. If an equation is already present, press F2 (DEL) F1(YES) to delete it.

Enter Heron’s formula into the calculator by pressing ALPHA H SHIFT = SHIFT _√ ( ALPHA S ( ALPHA S – ALPHA A ) ( ALPHA S – ALPHA B ) ( ALPHA S – ALPHA C ) ) . We used a H for the value of Heron’s formula for the area.

Press EXEto store the equation.

Calculate the area of a triangle given a= 5, b = 7, and c= 10. First, find the value of s, which is s= .5(a+ b+ c) = .5(5 + 7 + 10) = .5(22) = 11. Enter the values for s, a, b, and cby pressing the down arrow keyfollowed by 1 1 EXE 5 EXE 7 EXE 1 0 EXE .

To find the area, press the up arrow keyto highlight the H and then press

F6 (SOLV). The area is approximately 16.25.

Calculate the area of a triangle given a= 10, b= 8, c= 6. Press F1(REPT) to return to the variable entry screen. Find the value of s, which is s= .5(a+ b+ c) = .5(10 + 8 + 6) = .5(24) = 12. Enter the values for s, a, b, and cby pressing the

down arrow key followed by 1 2 EXE 1 0 EXE 8 EXE 6 EXE.

To find the area, press the up arrow keyto highlight the H and then press

F6(SOLV). The area is 24.

Calculate the area of a triangle given a= 6, b = 4, c= 8. Press F1(REPT) to return to the variable entry screen. Find the value of s, which is s= .5(a+ b+ c) = .5(6 + 4 + 8) = .5(18) = 9. Enter the values for s, a, b, and c by pressing the

down arrow keyfollowed by 9 EXE 6 EXE 4 EXE 8 EXE.

To find the area, press the up arrow keyto highlight the H and then press

F6(SOLV). The area is 11.62.

3. To use the Solver feature to find the area of a circular sector for the given values, you will need to use the Equation Menu. Enter the Equation Menu from the MAIN MENU by pressing A (EQUA) or use the arrow keys to highlight EQUA and press EXE.

The Equation menu will appear.

Press F3(Solv) to access the Solver feature. If an equation is already present, press F2(DEL) F1(YES) to delete it.

Enter formula to find the area of a circular sector by pressing ALPHA A SHIFT = . 5 ALPHA R x2 _{ALPHA T . We use variable Rfor r, and T}

for

.

Press EXEto store the equation.

If necessary, you may change the angle mode of the calculator by pressing

SHIFT SETUP F2(Rad).

Press EXITto return to the variable entry screen. Calculate the area of a triangle given r = 8 and

= π/5. Enter the values for r and

by pressing the

down arrow key followed by 8 EXE SHIFT _π 5 EXE.

To find the area, press the up arrow keyto highlight the A and then press

F6 (SOLV). The area is approximately 20.11.

Press F1(REPT) to return to the variable entry screen. Calculate the area of a triangle given r = 10, 2 = 2_π/3. Enter the values for rand

by pressing the down arrow keyfollowed by 1 0 EXE 2 SHIFT π 3 EXE.

To find the area, press the up arrow keyto highlight the A and then press

F6(SOLV). The area is approximately 104.72.

Press F1(REPT) to return to the variable entry screen. Calculate the area of a triangle given r = 7, 2 = 4π/9. Enter the values for rand

by pressing the down arrow keyfollowed by 7 EXE 4 SHIFT _π 9 EXE.

To find the area, press the up arrow keyto highlight the A and then press

F6 (SOLV). The area is approximately 34.21.

Activity 7

Projectile Motion

Topic Area: Parametric Equations

NCTM Standards:

• Compute fluently by developing fluency in operations with real numbers using technology for more-complicated cases

• Understand functions by interpreting representations of functions

Objective

• To evaluate and graph a set of parametric equations that model the motion of a projectile

Getting Started

In this activity, the students will learn how to evaluate and graph a set of parametric equations that model the motion of a projectile. In parametric equations, the xand

ycoordinates are generated by separate functions of a third variable t (usually time). Parametric equations are represented as x= f(t) and y= g(t).

A projectile is an object moving on a path, like a ball that is thrown or a missile thatis fired. The model for projectile motion assumes no air resistance and the object is traveling on a parabolic path. The x-axis represents the ground and the x -coordinate represents the distance traveled horizontally at any time t. The y -coor-dinate represents the height of the object at time t. The parametric equations for projectile motion are:

x= v•cos(

)•t y= v•sin(

)•t –g•t2/2

where vis the initial velocity (ft/sec) of the object,

is the initial angle (radians) of trajectory from the ground, t is elapsed time in seconds, and gis the acceleration due to gravity (32 ft/sec2).

The graph of the activity’s projectile is shown below:

Introduction

In this activity, you will learn how to evaluate and graph a set of parametric equations that model the motion of a projectile. In parametric equations, the xand

y-coordinates are generated by separate functions of a third variable t (usually time). Parametric equations are represented as x= f(t) and y= g(t).

A projectile is an object moving on a path, like a ball that is thrown or a missile thatis fired. The model for projectile motion assumes no air resistance and the object is traveling on a parabolic path. The x-axis represents the ground and the x -coordinate represents the distance traveled horizontally at any time t. The y -coor-dinate represents the height of the object at time t. The parametric equations for projectile motion are:

x= v•cos(

)•t y= v•sin(

)•t–g•t2/2

where vis the initial velocity (ft/sec) of the object,

is the initial angle (radians) of trajectory from the ground, tis elapsed time in seconds, and gis the acceleration due to gravity (32 ft/sec2).

Problems and Questions

Investigate a projectile that is fired at an angle of 60o(= _π/3) to the ground with an initial velocity of 200 ft/sec.

1. Using the information provided, find the set of parametric equations.

x= _______________

y= _______________

2. Using the equations, find the coordinates of the projectile after 2 seconds.

x= _______________

y= _______________

3. Graph the parametric equations from x= -100 to 1500 feet, scale 100; from

y= -100 to 500 feet, scale 100; and from t= 0 to 15 seconds, pitch 0.1. Draw your graph in the window below.

Name _____________________________________________ Class ________ Date ________________

4. Using the trace feature, find the maximum height of the projectile. At what time is the maximum height reached.

Maximum height = ____________________ Time it was achieved = ____________________

5. Using the trace feature, find the point of impact for the projectile. What is the range of the projectile (xvalue at impact)? What is the time of flight (tvalue at impact)?

Range of projectile = ____________________ Time of flight = ____________________

6. Repeat the investigation for a projectile that is fired at an angle of 45o(= _π/4) to the ground with an initial velocity of 100 ft/sec.

a. Parametric Equations are:

x= _______________

y= _______________

b. Coordinates of the projectile after 2 seconds:

x= _______________

y= _______________

c. Graph from x= -10 to 500 feet, scale 10; from y= -10 to 100 feet, scale 10; and from t = 0 to 10 seconds, pitch 0.1. Draw your graph in the window below.

Maximum height = ____________________ Time it was achieved = ____________________ Range of projectile = ____________________ Time of flight = ____________________

Name _____________________________________________ Class ________ Date ________________

1. Plug in the given information to find parametric equations: x= 200•cos(π/3)•t,

y= 200•sin(_π/3)•t –32•t2/2. Simplify the equations by calculating the coefficients.

Enter the Run Menu from the MAIN MENU by pressing 1(RUN) or use the

arrow keysto highlight RUN and press EXE. A blank screen should appear. If it does not, press AC/ON to clear the screen. Make sure the calculator is in radian mode by pressing SHIFT SETUP and arrow downto highlight Angle. Press F2(Rad) to select.

Press EXIT to exit the Set Up screen. Find the coefficients by pressing 2 0 0 cos ( SHIFT _π ÷ 3 ) EXE 2 0 0 sin ( SHIFT _π ) 3 ÷ EXE.

The simplified equations are x= 100t, y= 173.2t –16t2.

2. Using the equations, find the coordinates of the projectile after 2 seconds. Do this by plugging in 2 for tin the equations and using the Run menu to calculate the coordinates:

x= 100 (2), y= 173.2 (2) –16 (2)2. Enter the Run menu by pressingMENU 1(RUN). Calculate the coordinates by pressing AC/ON 1 0 0 2 EXE 1 7 3 . 2 2 – 1 6 2 x2 EXE.

The coordinates of the projectile after 2 seconds is x= 200 feet and y= 282.4 feet.

3. To graph the parametric equations, press MENU 5 (GRAPH) and delete equations by highlighting them and pressing F2(DEL) F1 (YES). Change the type of graph to parametric by pressing F3(TYPE) F3(Parm). Enter the

parametric equations by pressing 1 0 0 X,

,T EXE 1 7 3 . 2 X,

,T – 1 6 X,

,T x2 EXE.

To graph the model from x= -100 to 1500 feet, scale 100; from y= -100 to 500 feet, scale 100; and from t = 0 to 15 seconds, scale 0.1; press SHIFT F3(VWIN) to access the View Window screen. Enter -100 for Xmin, 1500 for Xmax, and 100 for Xscale by pressing (-) 1 0 0 EXE 1 5 0 0 EXE 1 0 0 EXE. Enter -100 for Ymin, 500 for Ymax, and 100 for Yscale by pressing (-) 1 0 0 EXE 5 0 0 EXE 1 0 0 EXE .

Enter 0 for Tmin, 15 for Tmax, and 0.1 for Tpitch by pressing the down arrow

followed by 0 EXE 1 5 EXE . 1 EXE.

Press EXITto exit the View Window screen and press F6 (DRAW) to view the graph.

4. To use the trace feature to find the maximum height of the projectile, press

F1 (Trace) and a tracer will appear on the screen at t= 0. Press the right arrow keyto move the tracer to find the maximum height (y) reached.

The maximum height reached is approximately 468.72 feet at a time of 5.4 seconds.

5. Continue tracing to find the approximate point of impact (near y= 0).

The range of the projectile is approximately 1080 feet with a time of flight of around 10.8 seconds.

6. To repeat the investigation for a projectile that is fired at an angle of 45o(= π/4) to the ground with an initial velocity of 100 ft/sec, first find the simplified parametric equations:

x= 100•cos(π/4)•t, y= 100•sin(π/4)•t–32•t2/2.

Using the Run Menu, find the coefficients by pressing 1 0 0 cos ( SHIFT π

4 ) EXE 1 0 0 sin ( SHIFT _π ÷ 4 ) EXE.

The simplified equations are x= 70.7 t, y= 70.7 t– 16 t2. Use the equations to find the coordinates of the projectile after 2 seconds. Do this by plugging in 2 for tin the equations and using the Run menu to calculate the coordinates:

x= 70.7 (2), y= 70.7 (2) –16 (2)2. Calculate the coordinates from the RUN menu by pressing 7 0 . 7 2 EXE 7 0 . 7 2 – 1 6 2 x2 EXE .

The coordinates of the projectile after 2 seconds is x= 141.4 feet and y= 77.4 feet. To graph the parametric equations, press MENU 5(GRAPH) and and enter the parametric equations by pressing 7 0 . 7 X,

,T EXE 7 0 . 7 X,

,T – 1 6 X,

,T x2 _EXE.

To graph the model from x= -100 to 1500 feet, scale 100; from y= -100 to 500 feet, scale 100; and from t = 0 to 15 seconds, scale 0.1; press SHIFT F3 (VWIN) to access the View Window screen.

Enter -100 for Xmin, 1500 for Xmax, and 100 for Xscale by pressing (-) 1 0 0 EXE 1 5 0 0 EXE 1 0 0 EXE. Enter -100 for Ymin, 500 for Ymax, and 100 for Yscale by pressing (-) 1 0 0 EXE 5 0 0 EXE 1 0 0 EXE. Enter 0 for Tmin, 10 for Tmax, and 0.1 for Tpitch by pressing the down arrowfollowed by 0 EXE 1 0 EXE . 1 EXE.

Press EXITto exit the View Window screen and press F6 (DRAW) to view the graph.

Use the trace feature to find the maximum height of the projectile by pressing

F1 (Trace) and a tracer will appear on the screen at t= 0. Press the right arrow keyto move the tracer to find the maximum height (y) reached.

The maximum height reached is approximately 78.1 feet at a time of 2.2 seconds. Continue tracing to find the approximate point of impact (near y= 0).

The range of the projectile is approximately 311 feet with a time of flight of around 4.4 seconds.

Change the type of graph back to Y= by pressing MENU 5(GRAPH) F3 (TYPE)

F1 (Y=).

Activity 8

Rose Petals

Topic Area: Polar Equations

NCTM Standards:

• Compute fluently by developing fluency in operations with real numbers using technology for more-complicated cases

• Understand functions by interpreting representations of functions

Objective

• To convert between polar and rectangular coordinates, graph polar equations • To determine characteristics of the general model for rose petal graphs

Getting Started

In this activity, the students will learn how to convert between polar and rectangu-lar coordinates, graph porectangu-lar equations, and determine characteristics of the general model for rose petal graphs.

In polar equations, the location of a point in a plane is determined by the polar coordinates (r,

) and not rectangular coordinates (x,y). In polar coordinates, r

represents the distance from the origin to the point, and

represents the angle from the positive horizontal axis to the ray connecting the origin and the point in the plane.

The relationship between rectangular coordinates and polar coordinates can be expressed in the following equations: x= r•cos

and y= r•sin

, or r= √(x2+ y2) and

= tan-1(y/x).

Polar equations can generate some beautiful curves. One family of these curves is rose petals. The general model for rose petal curves is r = a+ b•cos (k

), where a,

band kare constants.

The graph of the basic rose petal, r = 1 + 1• cos(1•

), is shown below

Introduction

In this activity, you will learn how to convert between polar and rectangular coordinates, graph polar equations, and determine characteristics of the general model for rose petal graphs.

In polar equations, the location of a point in a plane is determined by the polar coordinates (r,

) and not rectangular coordinates (x,y). In polar coordinates, r

represents the distance from the origin to the point, and

represents the angle from the positive horizontal axis to the ray connecting the origin and the point in the plane.

The relationship between rectangular coordinates and polar coordinates can be expressed in the following equations: x= r•cos

and y= r•sin

, or r = √(x2+ y2) and

= tan-1(y/x).

Polar equations can generate some beautiful curves. One family of these curves is rose petals. The general model for rose petal curves is r= a+ b•cos (k

), where a,

band kare constants.

Problems and Questions

1. Convert the point in the plane represented by the rectangular coordinates (2, 1) to polar coordinates.

r= _______________

= _______________

2. Convert the point in the plane represented by the polar coordinates (1, _π/3) to rectangular coordinates.

x= _______________

y= _______________

3. Graph the basic rose petal (one petal), r = 1 + 1• cos(1•

) in the initial window. Draw the graph in the window below.

Name _____________________________________________ Class ________ Date ________________

4. Using the trace feature, where does the graph begin (

=0)? From this initial point, which direction does the graph generate?

______________________________________________________________________________ ______________________________________________________________________________

5. When the leading constant is increased from 1 to 2, and 3 respectively, what happens to the graph?

______________________________________________________________________________ ______________________________________________________________________________

6. When the coefficient of the cosine is increased from 1 to 2, and 3 respectively, what happens to the graph?

______________________________________________________________________________ ______________________________________________________________________________ 7. When the coefficient of the

is increased from 1 to 2, 3, and 4 respectively, what

happens to the graph?

______________________________________________________________________________ ______________________________________________________________________________

Name _____________________________________________ Class ________ Date ________________