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Understanding Tangent, Velocity, and Rates of Change in Calculus

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This lesson revisits the concepts of average and instantaneous rates of change, crucial for understanding calculus. We define average rate of change as the slope of the secant line between two points on a curve, calculated as Δy/Δx. Conversely, the instantaneous rate of change, denoted as the limit of Δy/Δx as Δx approaches zero, represents the slope of the tangent line at a specific point. Through examples, we illustrate how to compute these rates, particularly in the context of particle motion and function behavior.

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Understanding Tangent, Velocity, and Rates of Change in Calculus

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  1. Lesson 2-7 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 Q f(a+h) P f(a) secant h = b - a a h b ∆y f(a+h) – f(a) ms = ----- = ----------------- ∆x h

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

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

  7. Example 1 Find the slope of the tangent line to the curve f(x) = 3 – 2x – 2x2 where x = 1. f(1+h) – f(1) lim --------------------- = h0 h f(a+h) – f(a) lim --------------------- = h0 h (-1 - 6h - 2h²) – (-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 -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(2+h) – f(2) lim --------------------- = h0 h f(a+h) – f(a) lim --------------------- = h0 h (1/(h+1)) – (1) lim --------------------- = h0 h f(2) = 1/(2-1) = 1 f(2+h) = 1/(2+h – 1) = 1/(h + 1) - h / (h + 1) lim ---------------- = h0 h 1 h + 1 - h ------ - --------- = ----------- h + 1 h + 1 h + 1 - 1 lim ---------- = - 1 h0 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) = t2 + 2t ft. f(3+h) – f(3) lim --------------------- = h0 h f(a+h) – f(a) lim --------------------- = h0 h (h² + 8h + 15) – (15) 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 lim ------------- = h0 h lim h + 8 = 8 h0

  10. Example 4 For a particle whose position at time t is f(t) = 6t2 - 4t +1ft.: a. Find the average velocity over the following intervals: • [1, 4] • [1, 2] • [1, 1.2] • [1, 1.01] ∆y Ave vel = ms = --------- ∆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) = 6t2 - 4t +1 ft.: b. Find the instantaneous velocity of the particle at t = 1 sec.   f(1+h) – f(1) lim --------------------- = h0 h f(a+h) – f(a) lim --------------------- = h0 h (6h² + 8h + 3) – (3) 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 lim ------------- = h0 h lim 6h + 8 = 8 h0

  12. Summary & Homework • 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 • Homework: pg 155 - 157: 2, 7, 15, 22, 27

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