4.1 derivatives as rates linear motions

Derivatives as Rates and Linear Motions

Let’s investigate derivatives as “rates” of changes.
Mathematics formulas appearing in the real world
have specified input unit and output unit.

The input variable time, t, is often used in applications
with units such as seconds, hours, years, etc..

The output unit could be in meters (m) for distance
measurement as in D = D(t) = distance of a moving
object from a fixed location.
D(t)

The output unit could be in centigrade (co) for
temperature measurement as in
T = T(t) = temperature of an object at time t.

The output unit could be in centigrade (co) for
temperature measurement as in
T = T(t) = temperature of an object at time t.
The output might be W kilograms (kg) for weight
measurement as in
W = W(t) = weight of an object at time t.

Linear Motions

Linear Motions
We begin with Linear Motions, which are problems
related to the movements of a point on a straight line.

Linear Motions
We start with an example from physics.
If a stone is thrown straight up on Earth
then D = –16t2 + vt where
D = height in feet,
t = time in seconds
v = initial speed (ft /sec)

Linear Motions
D = –16t2 + vt
D = height in feet,
t = time in seconds

Linear Motions
D = –16t2 + vt
D = height in feet,
t = time in seconds
A stone is thrown straight up at a
speed of 96 ft/sec, after t seconds
it reaches the height D where D(t) =

Linear Motions
D = –16t2 + vt
D = height in feet,
t = time in seconds
it reaches the height D where D(t) = –16t2 + 96t.

Linear Motions
D = –16t2 + vt
D = height in feet,
t = time in seconds
it reaches the height D where D(t) = –16t2 + 96t.
We make a few observations with some simple algebra.

The amount of time the stone will
stay in the air is the time between it
leaveing the ground till it falls back to D = -16t2 + 96t
the ground, both with D = 0.

Solve D = –16t2 + 96t = 0

Solve D = –16t2 + 96t = 0
–16t(t – 6) = 0
or t = 0, 6.

Solve D = –16t2 + 96t = 0
–16t(t – 6) = 0
or t = 0, 6.
Hence it will stay in air for 6 seconds.
It reaches the maximum height in half of that time or
3 seconds.

Solve D = –16t2 + 96t = 0
–16t(t – 6) = 0
or t = 0, 6.
3 seconds. The corresponding maximum height it
reaches is D(3) = –16(3)2 + 96(3) = 144 ft.

Solve D = –16t2 + 96t = 0
–16t(t – 6) = 0
or t = 0, 6.
3 seconds. The corresponding maximum height it
reaches is D(3) = –16(3)2 + 96(3) = 144 ft.
Note that the speed of the stone slows from
96 ft/sec to 0 ft/sec as it reaches the height of 144 ft.

The speed or rate of the moving stone is
different at any given time t.
D = -16t2 + 64t

different at any given time t. We call this
“the instantaneous rate (of change in D)” D = -16t2 + 64t
or simply “the rate” at time t.

“the instantaneous rate (of change in D)” D = -16t2 + 64t
For example, the instantaneous rate at
t = 3 is 0 ft/sec when the stone reaches
the maximum height of 144 ft.

“the instantaneous rate (of change in D)” D = -16t + 64t
2

This “instantaneous rate” is the derivative D'(t).

2

To see this, consider the “average
rate of change of a function f(t)
from the time t = a to t = b”.

2

To see this, consider the “average
rate of change of a function y = f(t) (b, f(b))

from the time t = a to t = b” which is (a, f(a)) f(b)–f(a) = Δy
b-a=Δx
f(b) – f(a) = Δy = the slope of the
b–a Δt
cord as shown. a b

The average rate of D(t) = –16t 2 + 96t
from t = 1 to t = 3 is
D(3) – D(1)
3–1

from t = 1 to t = 3 is
D(3) – D(1) = 144 – 80
3–1 2
= 32ft/sec

t=3
from t = 1 to t = 3 is 64 ft in 2 sec
= 32 ft/sec
D(3) – D(1) = 144 – 80
3–1 2 t=1
= 32ft/sec

t=3
= 32 ft/sec
D(3) – D(1) = 144 – 80
3–1 2 t=1
= 32ft/sec
On the graph of D = D(t) this is
the slope of the cord from
(1, 80) to ( 3, 144).

t=3
= 32 ft/sec
D(3) – D(1) = 144 – 80
3–1 2 t=1
= 32ft/sec
the slope of the cord from
(1, 80) to ( 3, 144).
D
D(t) = -16t2 + 96t

t
1 3 6

t=3
= 32 ft/sec
D(3) – D(1) = 144 – 80
3–1 2 t=1
= 32ft/sec
the slope of the cord from slope =32 (3,
144)
= avg. rate
(1, 80) to ( 3, 144).
D
D(t) = -16t2 + 96t

(1, 80)

t
1 3 6

t=3
= 32 ft/sec
D(3) – D(1) = 144 – 80
3–1 2 t=1
= 32ft/sec
144)
= avg. rate
(1, 80) to ( 3, 144).
D
The average rate from t = 1 to t = 2 D(t) = -16t2 + 96t

is D(2) – D(1)= 128 – 80 = 48 ft/sec (1, 80)

2–1 1
t
1 3 6

t=3
= 32 ft/sec
D(3) – D(1) = 144 – 80
3–1 2 t=1
= 32ft/sec
144)
= avg. rate
(1, 80) to ( 3, 144).
D
The average rate from t = 1 to t = 2 D(t) = -16t2 + 96t

is D(2) – D(1)= 128 – 80 = 48 ft/sec (1, 80)

2–1 1
which is as expected because it’s t
traveling faster in the beginning. 1 3 6

Given y = f(x), we define the instantaneous rate at
time t = a as the derivative f '(a) = lim f(a + h) – f(a)
0 h
if it exists.

0 h
if it exists.
With the above example D(t) = –16t2 + 96t, we have
D'(t) = –16*2t + 96 = –32t + 96

0 h
if it exists.
D'(t) = –16*2t + 96 = –32t + 96
Hence D'(1) = 64 ft/sec or that the upward speed of
the stone is 64 ft/sec at t = 1.

0 h
if it exists.
D'(t) = –16*2t + 96 = –32t + 96
The rate of 64 ft/sec means that if the gravity
disappeared at the instant t = 1, then the stone would
go upward at a constant speed of 64 ft/sec.

0 h
if it exists.
D'(t) = –16*2t + 96 = –32t + 96
The rate of 64 ft/sec means that if the gravity
disappeared at the instant t = 1, then the stone would
go upward at a constant speed of 64 ft/sec.
It also means that at the instant t = 1,
per unit change in the input would give 64 units of
change in output, i.e. 1 sec. in time t to 64 ft. in height.

Example A. A point moves up and down and its
position at time t is y(t) = –t2–t+12.
a. Find the position of the
point at t = 0. For what
value(s) of t is y = 0?

point at t = 0. For what
When t = 0 the position is
y(0) = 12.

point at t = 0. For what y
y= 12 at
When t = 0 the position is t=0
y(0) = 12.
0

y = –t2–t+12

y= 12 at
y(0) = 12.
0
If the position y = 0, then
–t2 – t + 12 = 0 or

y = –t2–t+12

y= 12 at
y(0) = 12.
0
If the position y = 0, then
–t2 – t + 12 = 0 or
t2 + t –12 = 0
y = –t2–t+12
(t – 3) (t + 4) = 0
or when t = 3 or –4 that y = 0.

point at t = 0. For what y y
y= 12 at
y(0) = 12.
0 y= 0 at 0
If the position y = 0, then t= –4

–t2 – t + 12 = 0 or
t2 + t –12 = 0
y = –t2–t+12
(t – 3) (t + 4) = 0

point at t = 0. For what y y y
y= 12 at
y(0) = 12.
0 y= 0 at 0 0 y= 0 at
If the position y = 0, then t= –4 t= 3

–t2 – t + 12 = 0 or
t2 + t –12 = 0
y = –t2–t+12
(t – 3) (t + 4) = 0

b. Find the rate of change in y at t = –4, 0, 3.
Is the point going up or down at those points?

The rate at t is y'(t) =–2t – 1 the derivative at t.

The rates at t = –4, 0, 3 are
y'(–4) = 7
y'(0) = –1
y'(3) = –7

y'(–4) = 7 y
y'(0) = –1
y'(3) = –7
Respectively, the point is
0 y= 0 at
going up when t = –4 as it t= –4
passed y = 0.

y = –t2–t+12

y'(–4) = 7 y y
y'(0) = –1
y= 12 at
y'(3) = –7 t=0
0 y= 0 at 0
going up when t = –4 as it t= –4
passed y = 0. It’s on the
way down when t = 0 as it
passes y =12. y = –t2–t+12

y'(–4) = 7 y y y
y'(0) = –1
y= 12 at
y'(3) = –7 t=0
0 y= 0 at 0 0 y= 0 at
going up when t = –4 as it t= –4 t= 3
passed y = 0. It’s on the
way down when t = 0 as it
passes y =12. When t = 3, y = –t2–t+12
it passes y = 0 going down.

c. When does y reach the maximum value and
what is the maximum y value?

When y reaches the maximum value, its rate y'(t)
must be 0 because the rate can’t be + or –
(it’s not going up nor down at that instance).

(it’s not going up nor down at that instance). So it
reaches the max. when y'(t) = -2t –1 = 0 or t = –½,

(it’s not going up nor down at that instance). So it
reaches the max. when y'(t) = -2t –1 = 0 or t = –½,
at y(–½ ) = –(–½)2 – (–½) + 12 = 12¼.
y y y (–1/2,12.25)
y
y'(–1/2)=0
y= 12 at (0,12)
t=0 y'(0)=–3

0 y= 0 at 0 0 y= 0 at t
(–4, 0) (3,0)
t= –4 t= 3 y'(–4)=7 y'(3)= –7

y = –t2–t+12 y = –t2–t+12

We may interpret the movement as the shadow cast
onto the y–axis as the point moves along the
parabola.

parabola.
y (–1/2,12.25)
y
Max at y'(–1/2)=0
y=12.25 (0,12)
t =–1/2 y'(0)=–3

cast
0 y= 0 at t
t= –4
shadow (–4, 0) (3,0)
y'(–4)=7 y'(3)= –7

y = –t2–t+12
y = –t2–t+12

parabola.
y y (–1/2,12.25)
y
Max at y'(–1/2)=0
y=12.25 y= 12 at (0,12)
t =–1/2 t=0 y'(0)=–3

cast
0 y= 0 at 0 t
t= –4
shadow (–4, 0) (3,0)
y'(–4)=7 y'(3)= –7

y = –t2–t+12
y = –t2–t+12

parabola.
y y y (–1/2,12.25)
y
Max at y'(–1/2)=0
y=12.25 y= 12 at (0,12)
t =–1/2 t=0 y'(0)=–3

cast
0 y= 0 at 0 0 y=0 t
t= –4
shadow (–4, 0) (3,0)
at t=3 y'(–4)=7 y'(3)= –7

y = –t2–t+12

parabola.
y y y (–1/2,12.25)
y
Max at y'(–1/2)=0
y=12.25 y= 12 at (0,12)
t =–1/2 t=0 y'(0)=–3

cast
0 y= 0 at 0 0 y=0 t
t= –4
shadow (–4, 0) (3,0)
at t=3 y'(–4)=7 y'(3)= –7

y = –t2–t+12
We define the velocity of the linear motion as the
1st derivative y' of the position function y.

parabola.
y y y (–1/2,12.25)
y
Max at y'(–1/2)=0
y=12.25 y= 12 at (0,12)
t =–1/2 t=0 y'(0)=–3

cast
0 y= 0 at 0 0 y=0 t
t= –4
shadow (–4, 0) (3,0)
at t=3 y'(–4)=7 y'(3)= –7

y = –t2–t+12
We define the velocity of the linear motion as the
1st derivative y' of the position function y.
The velocity y' gives the rates of changes in the
positions y as in the phrase “the velocity is 60 mph”.

The 2nd derivative y'' of y is the acceleration
(or deceleration depending on the signs of y'').

The acceleration gives the rates of changes in the
velocities.

velocities. Hence if the deceleration is 11 ft /sec2,
it would take 8 seconds for a car traveling at 60 mph
to come to a complete stop. (1 m = 5280 ft)

Example B. A point moves along the x–axis and its
position is given as x(t) = t4/4 – t3/3.

.

a. Find its velocity x'. When is it heading right and
.
when is it heading left?

.
when is it heading left? When does it change
direction?

.
direction? What are the right most and left most
positions of the point on the x–axis?

.
b. Find the acceleration x'' and when does x'' = 0?
What’s the meaning of x'' = 0?

The velocity is x' (t) = t3 – t2.

The velocity is x' (t) = t3 – t2. It’s heading right when
the velocity is positive and heading left when the
velocity is negative.

velocity is negative. Solve x' (t) = 0 and draw the
sign–chart:

sign–chart: t3 – t2 = 0  t2(t – 1) = 0  t = 0, 1

sign–chart: t3 – t2 = 0  t2(t – 1) = 0  t = 0, 1
Sign–chart – – + + + x'(t) = t3–t2
0 1

sign–chart: t3 – t2 = 0  t2(t – 1) = 0  t = 0, 1
Sign–chart – – + + + x'(t) = t3–t2
0 1

Hence the point is heading left when t < 1, i.e. from
“t = –∞” up to t = 1,

sign–chart: t3 – t2 = 0  t2(t – 1) = 0  t = 0, 1
Sign–chart – – + + + x'(t) = t3–t2
0 1

“t = –∞” up to t = 1, except at t = 0, it paused (why?).

sign–chart: t3 – t2 = 0  t2(t – 1) = 0  t = 0, 1
Sign–chart – – + + + x'(t) = t3–t2
0 1

“t = –∞” up to t = 1, except at t = 0, it paused (why?).
After t = 1, it changes direction and heads to the
right. Here is the corresponding graphic projection.

The graph of x(t) = t4/4 – t3/3 x
x(t) = t4/4 – t3/3
is shown to the right.

t

The graph of x(t) = t /4 – t /3
4 3
x
x(t) = t4/4 – t3/3
Below is the projection of this
graph (a compressed version) t
onto the x–axis.

x

4 3
x
x(t) = t4/4 – t3/3
onto the x–axis. The point is coming from the right,
pauses (why?) at x = 0 when t = 0,

x=0
x
at t =0

4 3
x
x(t) = t4/4 – t3/3
pauses (why?) at x = 0 when t = 0, continues on to the
left. At t = 1 it reaches the left most position x = –1/12,

x = –1/12 x=0
x
at t =1 at t =0

4 3
x
x(t) = t4/4 – t3/3
pauses (why?) at x = 0 when t = 0, continues on to the
left. At t = 1 it reaches the left most position x = –1/12,
then it changes direction and heads right after t = 1.

x = –1/12 x=0
x
at t =1 at t =0

The acceleration is x''(t) = 3t2 – 2t.

Set 3t2 – 2t = 0  t = 0, 2/3.

Set 3t2 – 2t = 0  t = 0, 2/3.
Sign–chart + – + x'' (t) = 3t –2t
2

0 2/3

Set 3t2 – 2t = 0  t = 0, 2/3.
Sign–chart + – + x'' (t) = 3t –2t
2

0 2/3
Hence both of them are inflection points of x(t).

Set 3t2 – 2t = 0  t = 0, 2/3.
Sign–chart + – + x'' (t) = 3t –2t
2

0 2/3
The sign–chart tells us that the point “accelerates”
from t = –∞ to t = 0,

x=0
x
at t =0

Set 3t2 – 2t = 0  t = 0, 2/3.
Sign–chart + – + x'' (t) = 3t –2t 2

0 2/3
from t = –∞ to t = 0, i.e. as it goes left it slows down.
The point is “accelerating” because
the 2nd derivative x''(t) > 0,
however the speed of the point is decreasing,
or “decelerating” in daily usage of the word.
It’s heading left and slowing down is a more
apt answer.
x=0
x
at t =0

Set 3t2 – 2t = 0  t = 0, 2/3.
Sign–chart + – + x'' (t) = 3t –2t
2

0 2/3
from t = –∞ to t = 0, i.e. as it goes left it slows down.
At t = 0, it stops at x = 0, and the acceleration is 0.

x=0
x
at t =0

Then it “decelerates” from t = 0 to t = 2/3,

+ – + x'' (t) = 3t2–2t
0 2/3

t=2/3 x=0
x
at t =0

Then it “decelerates” from t = 0 to t = 2/3,

Again, the point is “decelerating” because
the 2nd derivative x''(t) < 0,
however the speed of the point is increasing,
or “accelerating” in daily usage of the word.
It’s heading left and speeding up is a more
apt answer.

+ – + x'' (t) = 3t2–2t
0 2/3

t=2/3 x=0
x
at t =0

Then it “decelerates” from t = 0 to t = 2/3, that is, it
speeds up the leftward motion from t = 0 until t = 2/3.

+ – + x'' (t) = 3t2–2t
0 2/3

t=2/3 x=0
x
at t =0

After t = 2/3 it “accelerates”, i.e. slowing down the
leftward motion.

+ – + x'' (t) = 3t2–2t
0 2/3

t=2/3 x=0
x
at t =0

leftward motion. At t = 1 the “acceleration” slows the
leftward motion or the velocity to 0 with x-min. at -1/12.

+ – + x'' (t) = 3t2–2t
0 2/3

x = –1/12 t=2/3 x=0
x
at t =1 at t =0

Then the “acceleration” causes the point to change
direction and head right after t = 0.
+ – + x'' (t) = 3t2–2t
0 2/3

x = –1/12 t=2/3 x=0
x
at t =1 at t =0

Then the “acceleration” causes the point to changes
direction and head right, speeding up after t = 1.
At t = 0, 2/3, + – + x'' (t) = 3t2–2t
the acceleration 0 2/3
becomes deceleration
and vice versa.

x = –1/12 t=2/3 x=0
x
at t =1 at t =0

Here is a summary about the acceleration x''(t).
1. Given that x''(t) > 0 and
a. its velocity is negative, i.e. x' (t) < 0,
it means its leftward motion is slowing down.
b. its velocity is positive, i.e. x' (t) > 0,
it means its rightward motion is speeding up.
2. Given that x''(t) < 0 and
a. its velocity is negative, i.e. x' (t) < 0,
it means its leftward motion is speeding up.
b. its velocity is positive, i.e. x' (t) > 0,
it means its rightward motion is slowing down.
3. Given that x''(t) = 0. If it’s an inflection point, then
the acceleration ceased and it begins to decelerate,
or vice versa. If it’s not an inflection point, it’s a pause.

4.1 derivatives as rates linear motions

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