Derivatives as Rates of Change 3.6

3.6: Derivatives as Rates of Change

Introduction to Derivatives as Rates of Change

In calculus, derivatives are used to describe how a quantity changes with respect to another. When an object moves along a straight line, its position, velocity, and acceleration can all be described using derivatives of its position function.

Velocity, Speed, and Acceleration

• Position Function: The position of an object at time t is given by s(t).

• Velocity: The velocity is the first derivative of position with respect to time: $v(t) = s'(t)$

• Speed: The speed is the absolute value of velocity: $|v(t)|$

• Direction of Motion:

  • If $v(t) > 0$, the object moves in the positive direction (right/up).

  • If $v(t) < 0$, the object moves in the negative direction (left/down).

• Acceleration: The acceleration is the derivative of velocity (second derivative of position): $a(t) = v'(t) = s''(t)$

• Relationship between Velocity and Acceleration:

  • If $v(t)$ and $a(t)$ have the same sign, the object is accelerating.

  • If $v(t)$ and $a(t)$ have opposite signs, the object is decelerating.

Worked Examples

Example 1: Basic Velocity and Acceleration

Given $s(t) = 4t^2 - 9t$, find $v(t)$, $a(t)$, and the velocity and acceleration when $t = 3$.

• $v(t) = s'(t) = 8t - 9$

• $a(t) = v'(t) = 8$

• $v(3) = 8 \times 3 - 9 = 24 - 9 = 15$

• $a(3) = 8$

Example 2: Analyzing Motion and Rest

Given $s(t) = t^3 - 6t^2 + 9t$ (in feet, $t$ in seconds):

• Velocity at time t: $v(t) = s'(t) = 3t^2 - 12t + 9$ ft/s

• When is the particle at rest? Set $v(t) = 0$: $3t^2 - 12t + 9 = 0 \implies t = 1$ or $t = 3$

• When is the particle moving forward? $v(t) > 0 \implies 3(t-1)(t-3) > 0 \implies t < 1$ or $t > 3$

• Change in position over first 5 seconds: $s(5) - s(0) = 5^3 - 6(5^2) + 9(5) - 0 = 125 - 150 + 45 = 20$ ft

Example 3: Speed Increasing Intervals

Given $s(t) = 12t - 3t^2$ (in feet, $t$ in seconds):

• Velocity: $v(t) = s'(t) = 12 - 6t$

• Acceleration: $a(t) = v'(t) = -6$

• At $t = 1$: $v(1) = 12 - 6 \times 1 = 6$ ft/s, $a(1) = -6$ ft/s2

• Speed is increasing when: $a(t) < 0$ and $v(t) < 0$ $12 - 6t < 0 \implies t > 2$

Example 4: Object Dropped from a Tower

An object is dropped from a tower. Its height (in feet) after $t$ seconds is $s(t) = 1024 - 16t^2$.

• Velocity: $v(t) = s'(t) = -32t$ ft/s

• Acceleration: $a(t) = v'(t) = -32$ ft/s2

• When does it hit the ground? $s(t) = 0 \implies 1024 - 16t^2 = 0 \implies t^2 = 64 \implies t = 8$ s

• Velocity at impact: $v(8) = -32 \times 8 = -256$ ft/s

• Acceleration at impact: $a(8) = -32$ ft/s2

Example 5: Object Thrown Vertically Upward

An object is thrown upward from the ground at 42 m/s. Its height after $t$ seconds is $s(t) = -0.7t^2 + 42t$ (in meters).

• Velocity: $v(t) = s'(t) = -1.4t + 42$ m/s

• Highest point: $v(t) = 0 \implies -1.4t + 42 = 0 \implies t = 30$ s

• Height at highest point: $s(30) = -0.7(30)^2 + 42 \times 30 = -0.7 \times 900 + 1260 = -630 + 1260 = 630$ m

• When does it strike the ground? $s(t) = 0 \implies -0.7t^2 + 42t = 0 \implies t = 0$ or $t = 60$ s

• Velocity at impact: $v(60) = -1.4 \times 60 + 42 = -84 + 42 = -42$ m/s

Summary Table: Key Relationships

Quantity	Definition	Formula
Position	Location of object at time t	$s(t)$
Velocity	Rate of change of position	$v(t) = s'(t)$
Speed	Magnitude of velocity	$|v(t)|$
Acceleration	Rate of change of velocity	$a(t) = v'(t) = s''(t)$

Additional info: The examples provided illustrate how to apply derivatives to real-world motion problems, including finding when an object is at rest, moving forward, or reaching its highest or lowest point. These concepts are foundational for understanding motion in physics and engineering.