# 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 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:

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

The Derivatives as the Slope of the Tangent

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

The Derivatives as the Instantaneous Rate of Change

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.

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

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

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

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

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 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 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/12/3432095/slides/slide_22.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

Download ppt "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."

Similar presentations