Velocity At A Moment And Instantaneous Rate Of Change

We are often interested in finding the rate of change—how quickly phenomena occur or in which direction a change is taking place. In this section, we examine the concept of velocity for rectilinear motion and generalize to rates of change for arbitrary functions.

Quick Reference

Concept	Formula	Description
Average Velocity	$v_{avg} = \frac{Δ s}{Δ t} = \frac{f (t_{0} + Δ t) - f (t_{0})}{Δ t}$	Ratio of displacement $Δ s$ to elapsed time $Δ t$ over interval $[t_{0}, t_{0} + Δ t]$ .
Instantaneous Velocity	$v (t_{0}) = lim_{Δ t \to 0} \frac{f (t_{0} + Δ t) - f (t_{0})}{Δ t}$	Limit of average velocity as time interval $Δ t \to 0$ (derivative of position).
Instantaneous Speed	$Speed = \| v (t_{0}) \|$	Absolute value (magnitude) of instantaneous velocity.
Instantaneous Rate of Change	$lim_{Δ x \to 0} \frac{f (x_{0} + Δ x) - f (x_{0})}{Δ x}$	Rate of change of $y = f (x)$ with respect to $x$ at $x_{0}$ .

Average Velocity

If a car travels a distance of 240 kilometers in 3 hours, we say it has traveled at the rate of 80 kilometers per hour. But this does not mean the speedometer registers 80 km/h at all times; 80 km/h is its average speed (or average velocity).

In general, suppose an object moves along a straight line with position $s = f (t)$ at time $t$ .

Position of a particle on a line at times t0 and t1 — Position of an object on a line at times $t_{0}$ and $t_{1}$ .

The object's average velocity from time $t_{0}$ to time $t_{1}$ is found by dividing displacement $Δ s = f (t_{1}) - f (t_{0})$ by $Δ t = t_{1} - t_{0}$ :

v_{avg} = \frac{f (t_{1}) - f (t_{0})}{t_{1} - t_{0}} = \frac{Δ s}{Δ t} .

Because $t_{1} = t_{0} + Δ t$ , we can rewrite this as:

v_{avg} = \frac{f (t_{0} + Δ t) - f (t_{0})}{Δ t} .

What is the Difference Between Velocity and Speed?

Average velocity is calculated by dividing displacement $Δ s$ by elapsed time $Δ t$ :

v_{avg} = \frac{displacement}{travel time} = \frac{final position - initial position}{travel time} .

Velocity has a direction (sign), so it can be positive or negative.

Average speed is calculated by dividing total distance traveled by elapsed time:

average speed = \frac{distance traveled}{travel time} .

Speed is always non-negative.

For example, if you travel to a city 150 km away and return in 4 hours, your total distance is $300$ km, so your average speed is $300 / 4 = 75$ km/h. However, your average velocity is zero because your final position equals your initial position.

Assume you drop a stone from rest and air resistance is negligible. The stone's position fallen after $t$ seconds is $s = \frac{1}{2} g t^{2}$ , where $g \approx 32 {ft/s}^{2}$ . Find the average velocity of the stone:

(a) during the first 3 seconds of fall;

(b) during the 1-second interval between second 2 and second 3.

Solution

(a)

The position at $t = 3$ seconds is:

s (3) = 16 (3)^{2} = 144 ft .

The average velocity during the first 3 seconds is:

\begin{aligned} v_{\text{avg}} &= \frac{s(3) - s(0)}{3 - 0} \\ &= \frac{144 - 0}{3} \\ &= 48\text{ ft/s}. \end{aligned}

(b)

The average velocity from $t = 2$ to $t = 3$ is:

\begin{aligned} v_{\text{avg}} &= \frac{s(3) - s(2)}{3 - 2} \\ &= \frac{16(3)^2 - 16(2)^2}{1} \\ &= 16(9 - 4) \\ &= 80\text{ ft/s}. \end{aligned}

Instantaneous Velocity

To define velocity at a single instant $t_{0}$ , we observe that the average velocity over a very short time span $Δ t$ closely approximates instantaneous velocity. Taking the limit as $Δ t \to 0$ :

v (t_{0}) = lim_{Δ t \to 0} v_{avg} = lim_{Δ t \to 0} \frac{Δ s}{Δ t} = lim_{Δ t \to 0} \frac{f (t_{0} + Δ t) - f (t_{0})}{Δ t} .

This limit is called the derivative of the position function with respect to time. The absolute value of instantaneous velocity is instantaneous speed:

speed = | v (t_{0}) | .

For the dropped stone with position $s = f (t) = 16 t^{2}$ , find the instantaneous velocity at $t = 3 s$ .

Solution

Evaluating the limit algebraically:

\begin{aligned} \lim_{\Delta t\to 0} \frac{\Delta s}{\Delta t} &= \lim_{\Delta t\to 0} \frac{f(3 + \Delta t) - f(3)}{\Delta t} \\ &= \lim_{\Delta t\to 0} \frac{16(3 + \Delta t)^2 - 16(3)^2}{\Delta t} \\ &= \lim_{\Delta t\to 0} \frac{16(9 + 6\Delta t + (\Delta t)^2) - 144}{\Delta t} \\ &= \lim_{\Delta t\to 0} \frac{96\Delta t + 16(\Delta t)^2}{\Delta t} \\ &= \lim_{\Delta t\to 0} (96 + 16\Delta t) \\ &= 96\text{ ft/s}. \end{aligned}

Thus, the instantaneous velocity at $t = 3 s$ is $96 ft/s$ .

Rate of Change

More generally, if two quantities $x$ and $y$ are related by $y = f (x)$ , the (instantaneous) rate of change of $y$ with respect to $x$ at $x_{0}$ is defined as:

lim_{Δ x \to 0} \frac{f (x_{0} + Δ x) - f (x_{0})}{Δ x},

provided that this limit exists. Often we omit "instantaneous" and simply refer to the "rate of change of $y$ with respect to $x$ ."

Exercises

A particle moves along a straight line with position function $s (t) = 3 t^{2} - 2 t + 1$ , where $s$ is in meters and $t$ is in seconds.

(a) Find the position of the particle at $t = 1$ and $t = 4$ .

(b) Calculate the average velocity $v_{avg}$ over the time interval $[1, 4]$ .

Answer

(a) $s (1) = 2 m$ , $s (4) = 41 m$

(b) $v_{avg} = 13 m/s$

Solution

(a)

Evaluating the position function:

\begin{aligned} s(1) &= 3(1)^2 - 2(1) + 1 \\ &= 3 - 2 + 1 \\ &= 2\text{ m}. \end{aligned}\begin{aligned} s(4) &= 3(4)^2 - 2(4) + 1 \\ &= 3(16) - 8 + 1 \\ &= 48 - 8 + 1 \\ &= 41\text{ m}. \end{aligned}
(b)
The average velocity over

[1, 4]

is the displacement divided by the elapsed time: \begin{aligned} v_{\text{avg}} &= \frac{\Delta s}{\Delta t} \\ &= \frac{s(4) - s(1)}{4 - 1} \\ &= \frac{41 - 2}{3} \\ &= \frac{39}{3} \\ &= 13\text{ m/s}. \end{aligned}

A rocket is launched straight up into the air, and its height $h (t)$ (in feet) after $t$ seconds is given by $h (t) = 120 t - 16 t^{2}$ .

(a) Find the average velocity of the rocket over the time interval $[2, 3]$ .

(b) Use the limit definition of instantaneous velocity to find the rocket's velocity at $t = 2$ seconds.

(c) At what time $t$ does the rocket reach its maximum height (where its instantaneous velocity is zero)?

Answer

(a) $v_{avg} = 40 ft/s$

(b) $v (2) = 56 ft/s$

(c) $t = 3.75 s$

Solution

(a)

Evaluating height at $t = 2$ and $t = 3$ :

\begin{aligned} h(2) &= 120(2) - 16(2)^2 \\ &= 240 - 64 \\ &= 176\text{ ft}. \end{aligned}\begin{aligned} h(3) &= 120(3) - 16(3)^2 \\ &= 360 - 144 \\ &= 216\text{ ft}. \end{aligned}

Average velocity over $[2, 3]$ :

\begin{aligned} v_{\text{avg}} &= \frac{h(3) - h(2)}{3 - 2} \\ &= \frac{216 - 176}{1} \\ &= 40\text{ ft/s}. \end{aligned}
(b)
Instantaneous velocity at

t = 2

using limit definition:

v (2) = lim_{Δ t \to 0} \frac{h (2 + Δ t) - h (2)}{Δ t} .

Calculating

h (2 + Δ t)

: \begin{aligned} h(2 + \Delta t) &= 120(2 + \Delta t) - 16(2 + \Delta t)^2 \\ &= 240 + 120\Delta t - 16[4 + 4\Delta t + (\Delta t)^2] \\ &= 240 + 120\Delta t - 64 - 64\Delta t - 16(\Delta t)^2 \\ &= 176 + 56\Delta t - 16(\Delta t)^2. \end{aligned} Subtracting

h (2) = 176

h (2 + Δ t) - h (2) = 56 Δ t - 16 (Δ t)^{2} .

Taking the limit: \begin{aligned} v(2) &= \lim_{\Delta t\to 0} \frac{56\Delta t - 16(\Delta t)^2}{\Delta t} \\ &= \lim_{\Delta t\to 0} (56 - 16\Delta t) \\ &= 56\text{ ft/s}. \end{aligned}
(c)
General instantaneous velocity formula at time

t

: \begin{aligned} v(t) &= \lim_{\Delta t\to 0} \frac{120(t + \Delta t) - 16(t + \Delta t)^2 - (120t - 16t^2)}{\Delta t} \\ &= \lim_{\Delta t\to 0} \frac{120\Delta t - 32t\Delta t - 16(\Delta t)^2}{\Delta t} \\ &= \lim_{\Delta t\to 0} (120 - 32t - 16\Delta t) \\ &= 120 - 32t. \end{aligned} Setting

v (t) = 0

: \begin{aligned} 120 - 32t &= 0 \\ 32t &= 120 \\ t &= \frac{120}{32} \\ &= 3.75\text{ s}. \end{aligned}

The area $A$ of a circle depends on its radius $r$ according to $A (r) = π r^{2}$ .

(a) Calculate the average rate of change of the area with respect to the radius as $r$ increases from $5 cm$ to $5.1 cm$ .

(b) Use the limit definition to find the instantaneous rate of change of area $A$ with respect to $r$ at $r = 5 cm$ .

(c) What geometric feature of the circle does this instantaneous rate of change represent?

Answer

(a) $10.1 π \approx 31.73 {cm}^{2} / cm$

(b) $10 π \approx 31.42 {cm}^{2} / cm$

(c) The circumference of the circle ( $C = 2 π r$ ).

Solution

(a)

Evaluating area at $r = 5$ and $r = 5.1$ :

A (5) = π (5)^{2} = 25 π .

A (5.1) = π (5.1)^{2} = 26.01 π .

Average rate of change:

\begin{aligned} \text{Average rate} &= \frac{A(5.1) - A(5)}{5.1 - 5} \\ &= \frac{26.01\pi - 25\pi}{0.1} \\ &= \frac{1.01\pi}{0.1} \\ &= 10.1\pi \approx 31.73\text{ cm}^2/\text{cm}. \end{aligned}
(b)
Instantaneous rate of change at

r = 5

Instantaneous rate = lim_{Δ r \to 0} \frac{A (5 + Δ r) - A (5)}{Δ r} .

Expanding

A (5 + Δ r)

: \begin{aligned} A(5 + \Delta r) &= \pi(5 + \Delta r)^2 \\ &= \pi[25 + 10\Delta r + (\Delta r)^2] \\ &= 25\pi + 10\pi\Delta r + \pi(\Delta r)^2. \end{aligned} Subtracting

A (5) = 25 π

A (5 + Δ r) - A (5) = 10 π Δ r + π (Δ r)^{2} .

Taking the limit: \begin{aligned} \text{Instantaneous rate} &= \lim_{\Delta r\to 0} \frac{10\pi\Delta r + \pi(\Delta r)^2}{\Delta r} \\ &= \lim_{\Delta r\to 0} (10\pi + \pi\Delta r) \\ &= 10\pi \approx 31.42\text{ cm}^2/\text{cm}. \end{aligned}
(c)
For general radius

r

, the instantaneous rate of change of area is:

lim_{Δ r \to 0} \frac{π (r + Δ r)^{2} - π r^{2}}{Δ r} = 2 π r .

This formula

2 π r

is exactly the **circumference** of the circle.

A particle moves along the $x$ -axis according to the position function $x (t) = t^{3} - 6 t^{2} + 9 t$ , where $x$ is in meters and $t \geq 0$ is in seconds.

(a) Find the instantaneous velocity function $v (t) = lim_{Δ t \to 0} \frac{x (t + Δ t) - x (t)}{Δ t}$ .

(b) Determine the time intervals during which the particle is moving to the right ( $v (t) > 0$ ) and moving to the left ( $v (t) < 0$ ).

(c) Calculate the particle's speed at $t = 2$ seconds.

Answer

(a) $v (t) = 3 t^{2} - 12 t + 9$

(b) Moving right on $[0, 1) \cup (3, \infty)$ ; moving left on $(1, 3)$

(c) $Speed = 3 m/s$

Solution

(a)

Using the limit definition of instantaneous velocity:

\begin{aligned} x(t + \Delta t) &= (t + \Delta t)^3 - 6(t + \Delta t)^2 + 9(t + \Delta t) \\ &= t^3 + 3t^2\Delta t + 3t(\Delta t)^2 + (\Delta t)^3 - 6[t^2 + 2t\Delta t + (\Delta t)^2] + 9t + 9\Delta t. \end{aligned}

Subtracting $x (t) = t^{3} - 6 t^{2} + 9 t$ :

x (t + Δ t) - x (t) = 3 t^{2} Δ t - 12 t Δ t + 9 Δ t + 3 t (Δ t)^{2} - 6 (Δ t)^{2} + (Δ t)^{3} .

Dividing by $Δ t$ and taking the limit:

\begin{aligned} v(t) &= \lim_{\Delta t\to 0} (3t^2 - 12t + 9 + 3t\Delta t - 6\Delta t + (\Delta t)^2) \\ &= 3t^2 - 12t + 9. \end{aligned}
(b)
Factoring the velocity expression: \begin{aligned} v(t) &= 3(t^2 - 4t + 3) \\ &= 3(t - 1)(t - 3). \end{aligned} Setting

v (t) = 0

gives critical times

t = 1

and

t = 3

.
- On

[0, 1)

(t - 1) < 0

and

(t - 3) < 0 ⟹ v (t) > 0

(moving right).
- On

(1, 3)

(t - 1) > 0

and

(t - 3) < 0 ⟹ v (t) < 0

(moving left).
- On

(3, \infty)

(t - 1) > 0

and

(t - 3) > 0 ⟹ v (t) > 0

(moving right).
(c)
Evaluating velocity at

t = 2

: \begin{aligned} v(2) &= 3(2)^2 - 12(2) + 9 \\ &= 12 - 24 + 9 \\ &= -3\text{ m/s}. \end{aligned} The particle's speed is the absolute value of velocity:

speed = | v (2) | = | - 3 | = 3 m/s .

Quick Reference

Average Velocity

Instantaneous Velocity

Rate of Change

Exercises

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