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Rate of Change and Derivative Calculus Lesson

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- 1. 3.1 The Limit Definition of the Derivative September 25, 2015
- 2. Objectives O I can find the derivative of a function using the limit definition of a derivative O I can evaluate the slope of a curve (the derivative) at a specific point on the curve O I can write the equation of a line tangent to a curve at a certain point
- 3. Agenda O Discussion of Unit 2 Limits Exam (5 minutes) O Lesson Warm-Up (20 minutes) O Notes on the Limit Definition of a Derivative with built-in Guided Practice (39 minutes) O In-class Practice Time (20 minutes) O Exit Ticket (10 minutes)
- 4. Lesson Warm-Up (10 min.) 1. Find the slope of the line that connects the two points P( 4, 5 ) and Q ( -2, 3 ) 2. Write the equation of the line PQ. 3. For the function f(x) = x2 – 5, evaluate 1. f(4) = ? 2. f(h) = ? 3. f(x+ h) = ?
- 5. Lesson Warm-Up (10 min.) 1. Find the slope of the line that connects the two points P( 4, 5 ) and Q ( -2, 3 ) Slope formula m = y2 - y1 x2 - x1 = 3-5 (-2)- 4 = -2 -6 = 1 3
- 6. Lesson Warm-Up (10 min.) 1. Find the slope of the line that connects the two points P( 4, 5 ) and Q ( -2, 3 ) 2. Write the equation of the line PQ. y- y1 = m(x - x1) y-5= 1 3 (x -3) y-5= 1 3 x -1 y = 1 3 x + 4
- 7. Lesson Warm-Up (10 min.) 1. Find the slope of the line that connects the two points P( 4, 5 ) and Q ( -2, 3 ) 2. Write the equation of the line PQ. 3. For the function f(x) = x2 – 5, evaluate 1. f(4) = ? 2. f(h) = ? 3. f(x+ h) = ? f(4) = 42 – 5 = 16 – 5 = 11 f(h) = h2 – 5 f(x+h) = (x +h)2 – 5 = (x + h) (x + h) – 5 = x2 + xh + xh + h2 – 5 = x2 + 2xh + h2 – 5
- 8. Rate of Change Consider: An object is moving and its position s(t) is measured in meters and depends on t in seconds s(t) = 2t + 1 Where is the object at the 1st second? t = 1 second s(1) = 2(1) + 1 = 3 meters
- 9. Rate of Change Consider: An object is moving and its position s(t) is measured in meters and depends on t in seconds s(t) = 2t + 1 Where is the object at the 2nd second? t = 2 seconds s(1) = 2(2) + 1 =5 meters
- 10. Rate of Change Consider: An object is moving and its position s(t) is measured in meters and depends on t in seconds s(t) = 2t + 1 What is the rate of change ? m = s(t2 )- s(t1) t2 -t1
- 11. Rate of Change Consider: An object is moving and its position s(t) is measured in meters and depends on t in seconds s(t) = 2t + 1 What is the rate of change ? m = 5-3 2-1 = 2
- 12. Consider: An object is moving and its position s(t) is measured in meters and depends on t in seconds s(t) = t2 What is the AVERAGE rate of change between t = 1 and t = 2 seconds? m = s(2)- s(1) 2-1 m = 4-1 2-1 = 3Secant line
- 13. Consider: An object is moving and its position s(t) is measured in meters and depends on t in seconds s(t) = t2 What is the INSTANTANEOUS rate of change between at exactly the FIRST second? tangent line at t = 1
- 14. Consider: An object is moving and its position s(t) is measured in meters and depends on t in seconds s(t) = t2 What is the INSTANTANEOUS rate of change between at exactly the FIRST second? m = s(1.1)- s(1) 1.1-1
- 15. Consider: An object is moving and its position s(t) is measured in meters and depends on t in seconds s(t) = t2 What is the INSTANTANEOUS rate of change between at exactly the FIRST second? m = s(1.1)- s(1) 1.1-1 m = s(1.01)- s(1) 1.01-1
- 16. Consider: An object is moving and its position s(t) is measured in meters and depends on t in seconds s(t) = t2 What is the INSTANTANEOUS rate of change between at exactly the FIRST second? m = s(1.1)- s(1) 1.1-1 m = s(1.01)- s(1) 1.01-1 m = s(1.001)- s(1) 1.001-1
- 17. Consider: An object is moving and its position s(t) is measured in meters and depends on t in seconds s(t) = t2 What is the INSTANTANEOUS rate of change between at exactly the FIRST second? m = s(1.1)- s(1) 1.1-1 m = s(1.01)- s(1) 1.01-1 m = s(1.001)- s(1) 1.001-1
- 18. Consider: An object is moving and its position s(t) is measured in meters and depends on t in seconds s(t) = t2 What is the INSTANTANEOUS rate of change between at exactly the FIRST second? m = s(1.1)- s(1) 1.1-1 m = s(1.01)- s(1) 1.01-1 m = s(1.001)- s(1) 1.001-1
- 19. The Derivative The derivative of f(x) at x = a, f '(a) = limh®0 f (a+h)- f (a) h f '(a) = limx®a f (x)- f (a) x -a
- 20. Finding the Derivative Example 1: Write the equation of the line that is tangent to the curve y = x2 at the point (1, 1).
- 21. Finding the Derivative Example 1: Write the equation of the line that is tangent to the curve y = x2 at the point (1, 1). Step 1: Find the derivative (= slope of the curve) at the point (1, 1) f '(a) = limx®a f (x)- f (a) x -a
- 22. Finding the Derivative Step 1: Find the derivative (= slope of the curve) at the point (1, 1) f '(a) = limx®a f (x)- f (a) x -a f '(1) = limx®1 f (x)- f (1) x -1 f '(1) = limx®1 x2 - f (1) x -1
- 23. Finding the Derivative f '(a) = limx®a f (x)- f (a) x -a f '(1) = limx®1 f (x)- f (1) x -1 f '(1) = limx®1 x2 - f (1) x -1 f '(1) = limx®1 x2 -1 x -1
- 24. Finding the Derivative f '(1) = limx®1 x2 -1 x -1 = limx®1 (x -1)(x +1) x -1 = limx®1(x+1) = limx®1(x+1)=1+1= 2
- 25. Writing the Equation Derivative = slope of the curve = slope of the tangent Slope = 2 m/s Point = (1, 1)
- 26. Writing the Equation Slope = 2 m/s Point = (1, 1) y- y1 = m(x - x1) y-1= 2(x -1) y-1= 2x-2 y = 2x -1
- 27. Finding the Derivative Example 2: Write the equation of the line that is tangent to the curve y = x3 + x when x = 0. f '(a) = limx®a f (x)- f (a) x -a f '(0) = limx®0 f (x)- f (0) x -0 f '(0) = limx®0 x3 + x -0 x -0
- 28. Finding the Derivative f '(0) = limx®0 x3 + x -0 x -0 = limx®0 x3 + x x = x(x2 +1) x = x2 +1= 02 +1=1
- 29. Writing the Equation Slope = 1 Point = (0, 0) y- y1 = m(x - x1) y-0 =1(x -0) y = x
- 30. Guided Practice Problems 1. Write the equation of the line tangent to the curve f(t) = t – 2t2 at a = 3. 2. f(x) = 4 – x2 at a = -1 3. at a = 3 4. at a = -2 f (t)= t2 +1 f (x)= 1 x +3
- 31. Homework Assignment Write the equation of the tangent line of the following curves at the given points. 1. f(x) = 2x2 + 10x , a = 3 2. f(x) = 8x3 , a = 1 3. , a = 1 4. , a = 0 f (x)= x +4 f (x)= 1 x2 +1
- 32. Exit Ticket 1. Compute the derivative and write the equation of the tangent line at a = -1 for the following function: f(x) = 3x2 + 4x + 2 2. In full sentences, explain the relationship how a secant line is different from a tangent line and how average velocity is different from instantaneous velocity.

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