3.4: The Derivative as a Rate of Change - Part 1: Motion ... · Part 1 Motion Along a Line Classic Example (Yet Again) Let’s look once again at the stoplight problem. The graph

Part 1

3.4: The Derivative as a Rate of Change

Part 1: Motion Along a Line

MATH 165: Calculus I

Department of Mathematics

Iowa State University

Paul J. Barloon

MATH 165 Section 3.4

Part 1 Motion Along a Line

Classic Example (Yet Again)

Let’s look once again atthe stoplight problem.

The graph actually showsthe car’s positionaccording to the positionfunction s(t) = 2t2.

5 10 15 20 25 30

300

600

900

1,200

1,500

1,800 (30,1800)

(10,200)

(0,0)

time(sec)

distance(feet)



Classic Example (Yet Again)

Let’s look once again atthe stoplight problem.

The graph actually showsthe car’s positionaccording to the positionfunction s(t) = 2t2.

5 10 15 20 25 30

300

600

900

1,200

1,500

1,800 (30,1800)

(10,200)

(0,0)

time(sec)

distance(feet)



Displacement

We found the averagevelocity over the first 30seconds by first finding thedisplacement of the carover that period:

1800 = s(0 + 30)� s(0)

In general, displacementover a time period �t isgiven by:

�s = s(t +�t)� s(t)

5 10 15 20 25 30

300

600

900

1,200

1,500

1,800 (30,1800)

(10,200)

(0,0)

time(sec)

distance(feet)



Displacement


1800 = s(0 + 30)� s(0)


�s = s(t +�t)� s(t)

5 10 15 20 25 30

300

600

900

1,200

1,500

1,800 (30,1800)

(10,200)

(0,0)

time(sec)

distance(feet)



Displacement


1800 = s(0 + 30)� s(0)


�s = s(t +�t)� s(t)

5 10 15 20 25 30

300

600

900

1,200

1,500

1,800 (30,1800)

(10,200)

(0,0)

time(sec)

distance(feet)



Average Velocity

Then, the average velocitywas:

1800

30=

s(0 + 30)� s(0)

30� 0

In general,average velocity over aperiod �t is given by:

�s

�t=

s(t +�t)� s(t)

�t

5 10 15 20 25 30

300

600

900

1,200

1,500

1,800 (30,1800)

(10,200)

(0,0)

time(sec)

distance(feet)



Average Velocity


1800

30=

s(0 + 30)� s(0)

30� 0


�s

�t=

s(t +�t)� s(t)

�t

5 10 15 20 25 30

300

600

900

1,200

1,500

1,800 (30,1800)

(10,200)

(0,0)

time(sec)

distance(feet)



Average Velocity


1800

30=

s(0 + 30)� s(0)

30� 0


�s

�t=

s(t +�t)� s(t)

�t

5 10 15 20 25 30

300

600

900

1,200

1,500

1,800 (30,1800)

(10,200)

(0,0)

time(sec)

distance(feet)



Velocity

We then let �t “get small” to find the velocity function for thecar.

Definition

If a body’s position at time t is given by s = f (t), then the body’svelocity (or instantaneous velocity) at time t is:

v(t) =ds

dt= lim

�t!0

f (t +�t)� f (t)

�t



EXAMPLE 1a: Suppose that a body’s position along a straightline is given by the function s(t) = t2 � 3t + 2 for 0 t 2.

What is the body’s displacement over the given interval?



EXAMPLE 1b: Suppose that a body’s position along a straightline is given by the function s(t) = t2 � 3t + 2 for 0 t 2.

What is the body’s average velocity over the given interval?



EXAMPLE 1c: Suppose that a body’s position along a straightline is given by the function s(t) = t2 � 3t + 2 for 0 t 2.

What is the body’s instantaneous velocity at t = 1 second?



EXAMPLE 1d: Suppose that a body’s position along a straightline is given by the function s(t) = t2 � 3t + 2 for 0 t 2.

At what time does the body change direction?



Speed

As we just saw, the sign of a velocity value is meaningful.Sometimes, though, we are interested in the unsigned rate atwhich position is changing.

Definition

Speed is the absolute value of velocity.

Speed = |v(t)| =��ds

dt

��



EXAMPLE 1e: Suppose that a body’s position along a straightline is given by the function s(t) = t2 � 3t + 2 for 0 t 2.

What is the body’s speed at t = 1 second?



Acceleration

The (instantaneous) rate at which velocity changes is a body’sacceleration.

Definition

If a body’s position at time t is given by s = f (t), then the body’sacceleration at time t is:

a(t) =dv

dt=

d2s

dt2

ora(t) = v 0(t) = s 00(t)



EXAMPLE 1f: Suppose that a body’s position along a straightline is given by the function s(t) = t2 � 3t + 2 for 0 t 2.

What is the body’s acceleration at t = 1 second?



Quiz Yourself

The graph at the right show abody’s velocity at time t.

What is the body’s accelerationat time t = 2 seconds?

A) 4

B) 2

C) 0

D) �2

E) �4



The End


Part 2

3.4: The Derivative as a Rate of Change

Part 2: Related Rates

MATH 165: Calculus I

Department of Mathematics

Iowa State University

Paul J. Barloon


Part 2 Related Rates

Introduction

In our car (and generic “body”) examples, the independentvariable has always been time (t):

Position = s(t)

Velocity = v(t) = s 0(t)

Acceleration = a(t) = v 0(t) = s 00(t)

We now consider situations in which the independent variable issomething else . . .



Introduction


Position = s(t)






Introduction


Position = s(t)






Introduction


Position = s(t)






Introduction


Position = s(t)






A Cube

The volume of a cube may beconsidered to be a function ofthe length of the sides l :

V (l) = l3

So, it makes sense to ask aboutthe instantaneous rate of changeof the volume with respect to theside length:

dV

dl=

d

dl[l3] = 3l2



A Cube

The volume of a cube may beconsidered to be a function ofthe length of the sides l :

V (l) = l3

So, it makes sense to ask aboutthe instantaneous rate of changeof the volume with respect to theside length:

dV

dl=

d

dl[l3] = 3l2



EXAMPLE 1a: At what ratedoes the volume of a cubechange with respect to sidelength when l = 5 ft?



EXAMPLE 1b: Byapproximately how much doesthe volume increase when theside length changes from 5 to 5.1feet?



Note that this isjust the slope of thetangent at l = 5when volume isgraphed on thevertical axis andside length on thehorizontal:



Quiz Yourself

Suppose that a spill is forming a circular oil slick around a leakingtanker. How fast does the area of the slick change with respect toits radius when that radius is 90 feet?

A) 45⇡ ft2/ft

B) 90⇡ ft2/ft

C) 180⇡ ft2/ft

D) 810⇡ ft2/ft

E) 8100⇡ ft2/ft



The End


MATH 165 14–19,32–37 Warm-Up Question – Sep. 14, 2018

Suppose that a body’s position along a straight line

is given by the function s(t) = t2 � 3t + 2 for

0 t 2 (t in seconds).

At what time(s) does the body change direction?

A) t = 0 secs

B) t = 1 sec

C) t = 32 secs

D) t = 2 secs

E) t = 1 and t = 2 secs

F) The body does not change direction

MATH 165 14–19,32–37 Warm-Up Question – Sep. 14, 2018

Suppose that a body’s position along a straight line

is given by the function s(t) = t2 � 3t + 2 for

0 t 2 (t in seconds).

At what time(s) does the body change direction?

A) t = 0 secs

B) t = 1 sec

*C) t = 32 secs

D) t = 2 secs

E) t = 1 and t = 2 secs

F) The body does not change direction

Documents

3.4: The Derivative as a Rate of Change - Part 1: Motion ... · Part 1 Motion Along a Line Classic Example (Yet Again) Let’s look once again at the stoplight problem. The graph