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# Meanings of the Derivatives

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. Télécharger la présentation ## Meanings of the Derivatives

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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 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

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

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

6. The Derivatives as the Instantaneous Rate of Change

7. 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?

8. 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

9. 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

10. 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

11. 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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