1. Meanings of the Derivatives

2. I. The Derivative at the Point as the Slope of the Tangent to the Graph of the Function at the Point

3. The Tangent to a Curve Example: The tangent to the graph of the function f(x) = (x-2) 2 + 3 at the point x = 2 is the line y = 3

4. The Derivatives as the Slope of the Tangent

5. Examples (1) For each of the following functions, find the equation of the tangent and the normal at the given point

10. II. The Derivative at the Point as the Instantaneous Rate of Change at the Point

11. The Derivatives as the Instantaneous Rate of Change

12. Find: 1. a. A formula for v(t) b. The velocity at t=2 and at t=5 c. The instances at which the particle is at rest( stops temporarily before changing direction). When it is moving forward/backward? 2. a. A formula for a(t) b. The acceleration at t=2 and at t=3 c. The instances at which the particle experiences no acceleration (not speeding). When it is speeding up/slowing down?

13. 1. a. A formula for v(t) v(t) = s'(t) = t 2 – 5t + 4 b. The velocity at t=2 and at t=5 v(2) = -2 v(5) = 4 c. The instances at which the particle is at rest( stops temporarily before changing direction). When it is moving forward/backward? c.1. Let: v(t) = 0 = → t 2 – 5t + 4 0 → ( t – 1 )( t – 4 ) = 0 → t = 1 Or t = 4 The particle becomes at rest at t = 1 and at t = 4 c.2. The particle is moving forward when: v(t) > 0 v(t) > 0 → t 2 – 5t + 4 > 0 → ( t – 1 )( t – 4 ) > 0 → t > 4 Or t < 1 c.3. The particle is moving backward when: v(t) < 0 v(t) < 0 → t 2 – 5t + 4 < 0 → ( t – 1 )( t – 4 ) < 0 → 1 < t < 4

14. 2. a. A formula for a(t) a(t) = v'(t) = 2t – 5 b. The acceleration at t=2 and at t=3 a(2) = -1 a(3) = 1 c. The instances at which the experiences no acceleration (not speeding up or slowing down). When it is speeding up/slowing down? c.1. Let: a(t) = 0 = → 2t – 5 = 0 → t = 5/2 The particle experiences no acceleration at t = 5/2 c.2. The particle is speeding up when: a(t) > 0 a(t) > 0 → 2t – 5 > 0 → t > 5/2 c.3. The particle is slowing down when: a(t) < 0 a(t) < 0 → 2t – 5 < 0 → t < 5/2

15. Example (2) Let s(t) = t 3 -6t 2 + 9t be the position of a moving particle in meter as a function of time t in seconds 1. Find the total distance covered by the particle in the first five seconds 2. Graph S(t), v(t) and a(t) Solution: v(t) = 3t 2 - 12t + 9 = 3(t 2 - 4t + 3) = 3(t-1)(t-3) v(t) = 0 if t=1 or t =3 →The particle stops temporarily at t=1 and again at t=3 v(t) > 0 if t > 1 or t > 3 →The particle moves in one direction (the positive direction) from before t=1 and after t=3 v(t) < 0 if 1 <t < 3 →The particle moves in the opposite direction (the negative direction) from between t=1 and t=3

16. s(0) = (0) 3 - 6(0) 2 + 9(0) = 0 s(1) = (1) 3 - 6(1) 2 + 9(1) = 1- 6 + 9 = 4 s(3) = (3) 3 – 6(3) 2 + 9(3) = 27 – 54 +27 = 0 s(5) = (5) 3 - 6(5) 2 + 9(5) = 125 – 150 + 45 = 20 The total distance traveled in 5 seconds = |s(5)-s(3)|+ |s(3)-s(1)|+ |s(1)-s(0)| = |20-0|+ |0-4|+ |4-0| = 20 + 4 + 4 = 28