Activity 5-2: Understanding Rates of Change

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Presentation transcript:

Activity 5-2: Understanding Rates of Change

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 – 4.9t2 What graph best describes this model? LINEAR QUADRATIC EXPONENTIAL SINUSOIDAL

Activity 5-2: Understanding Rates of Change How long did it take the rock to hit the water to 2 decimal places? The graph of the function is given below: Solution: We can factor h(t) = 30 – 4.9t2 h(t) = -4.9(t2 – 6.122) h(t)=0 when the rock hits water 0= -4.9(t2 – 6.122) .: t2=6.122 or t = 2.474 (only positive) h(t) = 30 – 4.9t2 What is the average speed of the rock from the time it was dropped to the time it hits the water? When the rock hits the water at 2.47, the height of the rock is 0 m Solution: Speed is the distance travelled in 2.47 seconds s = d/t = 30m/2.47s s = 12.126m/s 2.47

Activity 5-2: Understanding Rates of Change 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 (0, 30) h(t) = 30 – 4.9t2 VIEW SLOPE 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 (2.47, 0)

Activity 5-2: Understanding Rates of Change The graph of the function is given below: h(t) = 30 – 4.9t2 Find the average speed of the rock between 1 and 2.47 seconds. VIEW ANSWER FIND HEIGHT OF ROCK AT t=1 and t=2.47 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 (1, 25.1) SECANT 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)

Activity 5-2: Understanding Rates of Change The graph of the function is given below: h(t) = 30 – 4.9t2 Find the average speed of the rock between 2 and 2.47 seconds. VIEW ANSWER FIND HEIGHT OF ROCK AT t=2 and t=2.47 We need to determine the height of the rock at 2 and 2.47 seconds using the equation: h(2) = 30 – 4.9(2)2 h(2) = 10.4 m h(2.47) = 0 m SECANT (2, 10.4) 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)

Activity 5-2: Understanding Rates of Change Part 1: Review The graph of the function is given below: h(t) = 30 – 4.9t2 Let us view the average speed of the rock from the various times to when it hits ground. VIEW SECANTS -22.1 m/s -17.07m/s -12.15m/s How would you go about estimating the rock’s speed when it hit the water? SHOW ANSWER Bring the point closer and closer to t=2.47 seconds. Find the slope of the secant. (2.47, 0)

Activity 5-2: Understanding Rates of Change Part 1: Review The graph of the function is given below: h(t) = 30 – 4.9t2 Let us view how this occurs VIEW SECANTS How can we find the velocity of the rock the instant it hits the water? VIEW ANSWER Slope of Secant: -25.23 Slope of Secant: -24 Slope of Secant: -23.26 We must find the instantaneous velocity which represents the slope of a point. The slope of a point is called a TANGENT (2.2, 6.28) (2.3, 4.08) (2.4, 1.78) (2.47, 0)