Meanings of the Derivatives

The Derivative at the Point as the Instantaneous Rate 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

Textbook Examples

Let s(t) = t2 -5t be the position of a moving particle in meters as a function of time t in seconds, 0 ≤ t ≤ 5.Graph S(t), v(t) and a(t)

Example (2) of section 3.5

Solution:v(t) = 2t – 5 = 2 ( t - 5/2 ) ; 0 ≤ t ≤ 5The graph of v is line-segment intersecting the t-axis at (5/2, 0) and the v-axis at (0,-5) Why?V(5)=2(5)-5=5 → the point (5,5) is he furthest point of the graph to the right..

We had v(t)= 2t – 5 = 2 ( t - 5/2 ) ; 0 ≤ t ≤ 5→ a(t) = 2 ; 0 ≤ t ≤ 5The graph of a is a horizontal line-segment intersection the a-axis at (0,2) Why?V(5)=2→ the point (5,2) is he furthest point of the graph to the right.

s(t) = t2 -5t = t(t-5) ; 0 ≤ t ≤ 5The graph of f intersects the axes at (0,0) and (5,0)We had s'(t) = 2( t - 5/2)→ x=5/2 is a critical point for f and we had : s''(t) =2 > 0 → s''(5/2) =2 → s has a local min at t=5/2S(5/2)= (5/2)2 – 5 (5/2) = - 25/4 → The point ( 5/2, -25/4) is a point of local ( and (absolute)min of S(t). The graph of s is a concave upward parabola with vertex (bottom) at the point ( 5/2, -25/4) and with the intersections (0,0) and (5,0) with the t-axis.

A stone is thrown vertically. The position of the stone at any moment t ( measured in seconds) is given (in feet) by the formula:s(t) = -16t2 + 64tWhat is the highest point reached by the stone ( what will be the maximum height it reaches?) and when it will return to the point from which it was thrown (what will be its velocity at that moment?) Graph the velocity , the acceleration and the position as functions of time.

Example (3) of section 3.5

Solution:s(t) = -16t2 + 64t→ v(t) = -32t +64 =- 32 ( t - 2 )The stone reached the highest point at the moment when the v(t)=0 → t = 2The maximum height the stone reaches = S(2) = -16(4) + 64(2) = 64 ft

At the start point S(t) = 0 → -16t2 + 64t = 0 → - 16t (t – 4 ) =0→ t = 0 or = 4 → The stone returns after 4 seconds.

→ The velocity at the moment the stones returns ( at which it hits the earth) = v(4) =- 32 (4) + 64 = -128 + 64 = - 64 ft/sec

v(t) = -32t +64 = -32 ( t - 2 )The graph of v(t) is line-segment through the points (2,0) and (0,64), (4, -64) with the interval [0,4]

We had: v(t) = -32t +64 = -32 ( t - 2 )→ a(t) = -32The graph of a(t) is horizontal line-segment on the interval [0,4]

The graph of s(t) = -16t2 + 64t = -16t (t-4) Is a parabola on the interval [0,4] that intersects the axis at (0,0) and (4,0), concave downward and with vertex ( 2, 64), which is its point of local ( & absolute) max.