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

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## Meanings of the Derivatives

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**I. The Derivative at the Point as the Slope of the Tangent**to the Graph of the Function at the Point**The Tangent to a CurveExample: The tangent to the graph of**the function f(x) = (x-2)2 + 3 at the point x = 2 is the line y = 3**Examples (1)For each of the following functions, find the**equation of the tangent and the normal at the given point**II. The Derivative at the Point as the Instantaneous Rae of**Change at the Point**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?**1.**a. A formula for v(t)v(t) = s'(t) = t2 – 5t + 4b. The velocity at t=2 and at t=5v(2) = -2v(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 = → t2 – 5t + 4 0 → ( t – 1 )( t – 4 ) = 0 → t = 1 Or t = 4The particle becomes at rest at t = 1 and at t = 4 c.2.The particle is moving forward when: v(t) > 0v(t) > 0 → t2 – 5t + 4 > 0 → ( t – 1 )( t – 4 ) > 0 → t > 4 Or t < 1 c.3.The particle is moving backward when: v(t) < 0v(t) < 0 → t2 – 5t + 4 < 0 → ( t – 1 )( t – 4 ) < 0 → 1 < t < 4**2.**a. A formula for a(t)a(t) = v'(t) = 2t – 5 b. The acceleration at t=2 and at t=3a(2) = -1a(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) > 0a(t) > 0 → 2t – 5 > 0 → t > 5/2 c.3.The particle is slowing down when: a(t) < 0a(t) < 0 → 2t – 5 < 0 → t < 5/2**Example (2)Let s(t) = t3 -6t2 + 9t be the position of a**moving particle in meter as a function of time t in seconds1. Find the total distance covered by the particle in the first five seconds2. Graph S(t), v(t) and a(t) Solution: v(t) = 3t2 - 12t + 9 = 3(t2- 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**s(0) = (0)3 - 6(0)2+ 9(0) = 0s(1) = (1)3 - 6(1)2+ 9(1) = 1-**6 + 9 = 4 s(3) = (3)3 – 6(3)2+ 9(3) = 27 – 54 +27 = 0s(5) = (5)3 - 6(5)2+ 9(5) = 125 – 150 + 45 = 20The 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

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