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

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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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Example

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Example

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Example

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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=5c. 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=3c. The instances at which the particle experiences no acceleration (not speeding). When it is speeding up/slowing down?

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1.a. A formula for v(t)v(t) = s'(t) = t2 – 5t + 4

b. 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=3v(t) > 0 if t > 1 or t > 3 →The particle moves in one direction (the positive direction) from before t=1 and after t=3v(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) = 0

s(1) = (1)3 - 6(1)2 + 9(1) = 1- 6 + 9 = 4

s(3) = (3)3 – 6(3)2 + 9(3) = 27 – 54 +27 = 0

s(5) = (5)3 - 6(5)2 + 9(5) = 125 – 150 + 45 = 20

The 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