1. Lesson 2-4 Tangent, Velocity and Rates of Change Revisited

2. Objectives Identify the average and instantaneous rates of change

3. Vocabulary Average rate of change – ∆y/∆x (the slope of the secant line between two points on the curve) Instantaneous rate of change – lim ∆y/∆x (∆x→0) (the slope of the tangent line at a point on the curve)

4. Average Rate of Change Average Rate of Change – slope of the secant line P Q a f(a) f(a+h) hb ∆y f(a+h) – f(a) m s = ----- = ----------------- ∆x h secant h = b - a

5. Instantaneous Rate of Change Instantaneous Rate of Change – slope of the tangent line (derivative of the function evaluated at the point) ∆y f(a+h) – f(a) m t = lim ----- = lim ----------------- ∆x h ∆x→0 h→0 P Q a f(a) f(a+h) hb mtmt As Q gets closer and closer to P, h (∆x) gets closer to 0; and the slope of the secant line, m s, approaches the slope of the tangent line, m t msms

6. Average vs Instantaneous Rate of Change Average Rate of Change – slope of the secant line Instantaneous Rate of Change – slope of the tangent line (derivative of the function evaluated at the point) P Q a f(a) f(a+h) hb P Q a f(a) f(a+h) hb ∆y f(a+h) – f(a) m s = ----- = ----------------- ∆x h ∆y f(a+h) – f(a) m t = lim ----- = lim ----------------- ∆x h ∆x→0 h→0 mtmt As Q gets closer and closer to P, h (∆x) gets closer to 0; and the slope of the secant line, m s, approaches the slope of the tangent line, m t

7. Example 1 Find the slope of the tangent line to the curve f(x) = 3 – 2x – 2x 2 where x = 1. f(a+h) – f(a) lim --------------------- = h  0 h f(1+h) – f(1) lim --------------------- = h  0 h f(1) = 3 - 2 - 2 = -1 f(1+h) = 3 - 2(1+h) - 2(1+h)² = 3 - 2 - 2h - 2 - 4h -2h² = -1 - 6h - 2h² (-1 - 6h - 2h²) – (-1) lim --------------------------- = h  0 h -6h -2h² lim ------------- = h  0 h lim -6 – 2h = -6 h  0

8. Example 2 Find the slope of the tangent line (using the definition) where f(x) = 1/(x-1) and a = 2. f(a+h) – f(a) lim --------------------- = h  0 h f(2+h) – f(2) lim --------------------- = h  0 h f(2) = 1/(2-1) = 1 f(2+h) = 1/(2+h – 1) = 1/(h + 1) (1/(h+1)) – (1) lim --------------------- = h  0 h - h / (h + 1) lim ---------------- = h  0 h - 1 lim ---------- = - 1 h  0 h + 1 1 h + 1 - h ------ - --------- = ----------- h + 1 h + 1 h + 1

9. Example 3 Find the instantaneous velocity at time t = 3 seconds if the particle’s position at time is given by f(t) = t 2 + 2t ft. f(a+h) – f(a) lim --------------------- = h  0 h f(3+h) – f(3) lim --------------------- = h  0 h f(3) = 9 + 6 = 15 f(3+h) = (3+h)² + 2(3+h) = h² + 6h + 9 + 6 + 2h = h² + 8h + 15 (h² + 8h + 15) – (15) lim --------------------------- = h  0 h h² + 8h lim ------------- = h  0 h lim h + 8 = 8 h  0

10. Example 4 For a particle whose position at time t is f(t) = 6t 2 - 4t +1 ft.: a. Find the average velocity over the following intervals: i.[1, 4] ii.[1, 2] iii.[1, 1.2] iv.[1, 1.01] ∆y Ave vel = m s = --------- ∆x f(4) – f(1) 81 – 3 78 ---------------- = ------------- = -------- = 26 4 – 1 3 3 f(2) – f(1) 17 - 3 ---------------- = ------------ = 14 2 – 1 1 f(1.2) – f(1) 4.84 - 3 1.84 ------------------ = -------------- = -------------- = 9.2 1.2 – 1 0.2 0.2 f(1.01) – f(1) 3.0806 - 3.0806 -------------------- = --------------- = ------------- = 8.06 1.01 – 1.01.01

11. Example 4 continued For a particle whose position at time t is f(t) = 6t 2 - 4t +1 ft.: b. Find the instantaneous velocity of the particle at t = 1 sec. f(a+h) – f(a) lim --------------------- = h  0 h f(1+h) – f(1) lim --------------------- = h  0 h f(1) = 6 - 4 + 1 = 3 f(1+h) = 6(1+h)² - 4(1+h) + 1 = 6h² + 12h + 6 – 4h – 4 + 1 = 6h² + 8h + 3 (6h² + 8h + 3) – (3) lim --------------------------- = h  0 h 6h² + 8h lim ------------- = h  0 h lim 6h + 8 = 8 h  0

12. Summary Summary: –Average rate of change is the slope of the secant line –Instantaneous rate of change is the slope of the tangent line –Limit of m secant = m tangent as ∆x →0