1. Activity 5-2: Understanding Rates of Change Click Image To Continue

2. Activity 5-2: Understanding Rates of Change You stand on a cliff that is 30 metres above Lake Superior. You decide to drop a rock to see how long it takes to hit the water. Due to gravity, the rock height in relation to time follows the function: h(t) = 30 – 4.9t 2 Technical Note: Make sure the animations follow the same manner as this slide show. Students check off one of the boxes. If they select QUADRATIC, a green CORRECT appears beside the model. If they get it INCORRECT, the word INCORRECT appears beside the word. Students can have as many selections as required to get the correct answer What graph best describes this model?  LINEAR  QUADRATIC CORRECT!  EXPONENTIAL  SINUSOIDAL BACK HOMENEXT

3. Activity 5-2: Understanding Rates of Change Technical Note: Make sure the animations follow the same manner as this slide show. The 2.47 text will appear on the graph after students answer the first question and the speed animation after the second question. The graph of the function is given below: How long did it take the rock to hit the water to 2 decimal places? 2.47 Students type their answers in the textbox. If students get the answer 2.47 correct, the CORRECT text appears in green beside the answer. If incorrect, the text INCORRECT in red and the answer is replaced with the correct one. Also, the graph will show the answer too. Follow the slide show animation. The solution under the button will appear after CHECK ANSWER is pressed. CHECK ANSWER CORRECT! 2.47 What is the average speed of the rock from the time it was dropped to the time it hits the water? Solution: We can factor h(t) = 30 – 4.9t 2 h(t) = -4.9(t 2 – 6.122) h(t)=0 when the rock hits water 0= -4.9(t 2 – 6.122).: t 2 =6.122 or t = 2.474 (only positive) h(t) = 30 – 4.9t 2 12.13 CHECK ANSWER CORRECT! Solution: Speed is the distance travelled in 2.47 seconds s = d/t = 30m/2.47s s = 12.126m/s Students type their answers in the textbox. If students get the answer 12.13 or 12.126 or 12.14 or 12.146 or 12.15 correct, the CORRECT text appears in green beside the answer. If incorrect, the text INCORRECT in red and the answer is replaced with the correct one. Also, the graph will show the answer too. Follow the slide show animation. The solution under the button will appear after CHECK ANSWER is pressed. Initially, at t=0, the height is 30m When the rock hits the water at 2.47, the height of the rock is 0 m BACK HOMENEXT

4. Activity 5-2: Understanding Rates of Change Technical Note: Make sure the animations follow the same manner as this slide show. The graph of the function is given below: The average speed of the rock is represented by the slope of the line going through the initial point and the final point h(t) = 30 – 4.9t 2 (0, 30) (2.47, 0) VIEW SLOPE Student click the VIEW SLOPE button to SHOW the red dashed line. This line will only appear when the VIEW SLOPE BUTTON is pressed The line joining the two points is called a SECANT. A secant is a line that passes through two points SECANT The average speed is represented by the slope of the secant. SLOPE OF SECANT Student click the SLOPE OF SECANT button to SHOW the calculation of the slope using slope equation below the button. BACK HOMENEXT

5. Activity 5-2: Understanding Rates of Change Technical Note: Make sure the animations follow the same manner as this slide show. The graph of the function is given below: Find the average speed of the rock between 1 and 2.47 seconds. h(t) = 30 – 4.9t 2 (1, 25.1) VIEW ANSWER When student presses the VIEW ANSWER button, the solution below appears in the animation order of this PowerPoint slide show SECANT FIND HEIGHT OF ROCK AT t=1 and t=2 We need to determine the height of the rock at 1 and 2.47 seconds using the equation: h(1) = 30 – 4.9(1) 2 h(1) = 25.1 m h(2.47) = 0 m FIND THE SPEED USING THE SLOPE OF THE SECANT We can draw the secant between the two points. Using the slope equation we get: (2.47, 0) BACK HOMENEXT

6. Activity 5-2: Understanding Rates of Change Technical Note: Make sure the animations follow the same manner as this slide show. The graph of the function is given below: Find the average speed of the rock between 2 and 2.47 seconds. h(t) = 30 – 4.9t 2 (2, 25.1) VIEW ANSWER When student presses the VIEW ANSWER button, the solution below appears in the animation order of this PowerPoint slide show SECANT FIND HEIGHT OF ROCK AT t=1 and t=2 We need to determine the height of the rock at 2 and 2.47 seconds using the equation: h(2) = 30 – 4.9(1) 2 h(2) = 10.4 m h(2.47) = 0 m FIND THE SPEED USING THE SLOPE OF THE SECANT We can draw the secant between the two points. Using the slope equation we get: (2.47, 0) BACK HOMENEXT

7. Activity 5-2: Understanding Rates of Change Part 1: Review Technical Note: Make sure the animations follow the same manner as this slide show. The graph of the function is given below: Let us view the average speed of the rock from the various times to when it hits ground. h(t) = 30 – 4.9t 2 VIEW SECANTS When student presses the VIEW SECANTS, the secants display one at a time as shown through the powerpoint animation. Bring the point closer and closer to t=2.47 seconds. Find the slope of the secant. -53.4m/s BACK HOMENEXT -17.07m/s (2.47, 0) 12.15m/s How would you go about estimating the rock’s speed when it hit the water? SHOW ANSWER When student presses the SHOW ANSWER, the text appears below

8. Activity 5-2: Understanding Rates of Change Part 1: Review Technical Note: Make sure the animations follow the same manner as this slide show. The graph of the function is given below: Let us view how this occurs h(t) = 30 – 4.9t 2 VIEW SECANTS When student presses the VIEW SECANTS, the secants display one at a time as shown through the powerpoint animation. BACK HOMENEXT (2.47, 0) (2.2, 6.28) Slope of Secant: -23.26 (2.3, 4.08) Slope of Secant: -24 (2.4, 1.78) Slope of Secant: -25.23 How can we find the velocity of the rock the instant it hits the water? We must find the instantaneous velocity which represents the slope of a point. The slope of a point is called a TANGENT VIEW ANSWER When student presses the VIEW ANSWER, the text shows below

9. Activity 5-2: Understanding Rates of Change Technical Note: Make sure the animations follow the same manner as this slide show. The graph of the function is given below: To approximate the instantaeous speed of the rock we will zoom in the graph. You can press zoom multiple times h(t) = 30 – 4.9t 2 ZOOM When students press the zoom button, a red dashed box appears where the graph will zoom and a new zoomed graph will appear. Students can press zoom three times and it will zoom three times. BACK HOMENEXT (2.47, 0) (2.474,0) (2.45,0.59) Slope of Secant=-29.5 (2.474,0) (2.4705,0.093) Slope of Secant=-26.57 Notice that function seems looks like a line. Now we have formed a TANGENT! The slope of the tangent represents the instantaneous rate. Slope of Tangent = -24.25

10. Activity 5-2: Understanding Rates of Change Estimate the instantaneous rate of change for each function by reducing the interval over which the average rate of change occurs at x. Technical Note: Make sure the animations follow the same manner as this slide show. Students type their answers in the textbox. If they are correct, CORRECT appears next to the text box, if they are incorrect, INCORRECT appears beside the textbox with the answer beside the INCORRECT statement as shown above. Allow students an accuracy of 1 decimal place. BACK HOMENEXT CHECK ANSWER f(x) = 2x + 3 at x = 3 2 CORRECT! Students type their answers in the textbox. If students get the answer 2.47 correct, the CORRECT text appears in green beside the answer. If incorrect, the text INCORRECT in red and the answer is replaced with the correct one. Also, the graph will show the answer too. Follow the slide show animation. The solution under the button will appear after CHECK ANSWER is pressed. CHECK ANSWER f(x) = 3 x at x = 3 31 INCORRECT! 29.7 CHECK ANSWER f(x) = 3sinx at x = 4 -1.96 CORRECT!