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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.

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Presentation on theme: "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."— Presentation transcript:

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) 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

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15 II. The Derivative at the Point as the Instantaneous Rae of Change at the Point

16 The Derivatives as the Instantaneous Rate of Change

17 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? 3. Graph v(t) and a(t) as functions of time t.

18 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 → ( 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

19 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

20 v(t) = t 2 -5t +4=(t-1)(t-4) v(t) = 0 at t=1 and at t=4 v(0) = 4 v(t) = t 2 -5t +4 = (t – 5/2 ) 2 – 9/4

21 a(t) = 2t - 5 = 2(t – 5/2) a(t) = 0 at t = 5/2 = 2.5 a(0) = -5

22 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. Describe the motion of the particle 2. Graph S(t), v(t) and a(t) Solution: v(t) = 3t t + 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 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/3432095/12/slides/slide_21.jpg", "name": "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.", "description": "Describe the motion of the particle 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

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