Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

Math 11/L

ENGINEERING CALCULUS 1
MODULE 5: Application of Derivatives

Source: https://medium
...
EDWIN L
...


Overview

The topics in this module are
included in this section
...
It
helps you explore new ideas and
capabilities
...

It is a job aimed at evaluating
your mastery in acquiring learning
skills
...


Answer Key

This contains answers to all
activities in the module
...


MODULE 5 APPLICATION OF DERIVATIVES

1

ENGINEERING CALCULUS 1

OVERVIEW
COURSE DESCRIPTION:

[DATE]








CONTENTS:
The Differentials
The Differentials (Applications)
Parametric Equations
Indeterminate Forms
Maxima and Minima Problems
Related Rates Problems

An introductory course covering the core concepts of
limit, continuity and differentiability of functions involving one
or more variables
...


COURSE OUTCOME:
At the end of this course, you must be able to solve the problem that involve the application of
derivatives using the appropriate formula
...

INTRODUCTION
Good day future engineers! Welcome to Engineering Course
...
As you continue to do so in this
module, you can learn about the various ways in which the derivatives can be apply also in real life
situation
...
Just
like addition, subtraction, multiplication and division derivative is another concept in the field of
mathematics
...


LESSON PROPER
The differential of any function is equal to its derivative multiplied by the differential of the
independent variable
...

It follows from the above definition that all the fundamental formulas for derivatives become
differential formulas if we merely multiply through by 𝑑𝑥
...

EXAMPLE 5
...
1
Find the differential of function 𝑦 = 𝑥 3 − 2𝑥
SOLUTION:
𝑑𝑦 = 3𝑥 3−1 𝑑𝑥 − 2𝑥 1−1 𝑑𝑥
𝑑𝑦 = 3𝑥 2 𝑑𝑥 − 2𝑑𝑥
EXAMPLE 5
...
2
𝑧 2 −1

Find the differential of function 𝑦 = 𝑧 2 +1
SOLUTION:
𝑑𝑦 =

(𝑧 2 +1)2𝑧𝑑𝑧−(𝑧 2 −1)2𝑧𝑑𝑧
(𝑧 2 +1)2
4𝑧𝑑𝑧

=(𝑧 2 +1)2
EXAMPLE 5
...
3
Find the differential of function 𝑦 3 + 2𝑥𝑦 = 3
SOLUTION:
3𝑦 2 𝑑𝑦 + 2𝑥𝑑𝑦 + 2𝑦𝑑𝑥 = 0
−2𝑦𝑑𝑥

𝑑𝑦 = 3𝑦2+2𝑥

MODULE 5 APPLICATION OF DERIVATIVES

3

ENGINEERING CALCULUS 1

[DATE]

4
...
𝑦 = 3𝑥 4 − 15𝑥 3 + 𝑥 − 4
2
...
𝑢 = √2𝑦 − 𝑦 2
1

1

4
...
Please answer it all first before checking, this is important for
self-assessment
...

If you got 2 – 3 items correct, good, you understand most of the concepts, you only need to practice
for a few times
...
Compare your answer to the answer given, then try to answer
the questions again before you proceed to the next lesson
...


LESSON PROPER
Approximate Formulas
Very often we wish to compute, or to estimate within safe limits, the change in the value of a
function caused by a small change in the value of the independent variable
...

In any approximate computation, the amount by which the computed value of the function differs
from the true is called error of the computation
...

Example 5
...
1
Find an approximate formula for the area of a narrow circular ring
...
2
...
73
SOLUTION
Put 𝑦 = √𝑥, from which
𝑑𝑥

𝑑𝑦 = 2

√𝑥

For 𝑥 we choose which is close to 8
...
Choose
𝑥=9

𝑎𝑛𝑑

∆𝑥 = 𝑑𝑥 = −0
...
73
...
27
2√9

=

−0
...
045

√8
...
045 = 𝟐
...
To five decimal places the correct value is 2
...
2
...
, using
1
boards 2 𝑖𝑛
...

SOLUTION
The volume of cube of a length 𝑥 is,
𝑉 = 𝑥 3 , from which
𝑑𝑉 = 3𝑥 3 𝑑𝑥 = 3𝑥 3 ∆𝑥
Assume that 𝑑𝑥 = ∆𝑥 =

0
...


0
...
) ( 12 𝑓𝑡
...
2
...
with a maximum error of 0
...
Find
the approximate maximum error in the computed volume
...
1 𝑖𝑛
...
2 (0
...
𝟑
Propagated and Relative Error
If a measured value x is used to compute another value 𝑓(𝑥), the difference of 𝑓(𝑥 + ∆𝑥) and
𝑓(𝑥) or ∆𝑦 is the propagated error
...

If the propagated error is given in relative terms if we compare 𝑑𝑉 with 𝑉
...


𝑑𝑉
𝑉

is called the

Example 5
...
5
The radius of the ball bearing is measured to be 0
...
If the measurement is correct within
0
...

SOLUTION
a)

4

𝑉 = 3 𝜋𝑟 3 → 𝑑𝑉 = 4𝜋𝑟 2 𝑑𝑟

where 𝑟 = 0
...
02

𝑑𝑉 = 4𝜋(0
...
02)
= ±𝟎
...
02)
0
...
𝟓%

MODULE 5 APPLICATION OF DERIVATIVES

6

ENGINEERING CALCULUS 1

[DATE]

Example 5
...
6
A right circular cylinder has a height of 24 ft
...
If the error measuring is 3 inches,
a) what is the possible error in volume b) what is the relative error?
SOLUTION
a)
...
25𝑓𝑡) + 𝜋𝑟 2 (±0
...
25𝑓𝑡) + 𝜋(82 𝑓𝑡 2 )(±0
...
𝟖 𝒇𝒕𝟑
b)
...
5𝑓𝑡 3
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑒𝑟𝑟𝑜𝑟 =
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑒𝑟𝑟𝑜𝑟 =

𝑑𝑉
𝑉
±351
...
5 𝑓𝑡 3

= ±𝟎
...
2 Practice Problem
Answer the following problems:
1
...

2
...

3
...

4
...
01 𝑖𝑛
...
Please answer it all first before checking, this is important for
self-assessment
...

If you got 2 – 3 items correct, good, you understand most of the concepts, you only need to practice
for a few times
...
Compare your answer to the answer given, then try to answer
the questions again before you proceed to the next lesson
...


LESSON PROPER
3
...
These equation is called parametric equation of a
curve
...
When dealing parametric
equations, it is convenient to use differentials in finding derivatives, particularly the derivatives of higher
order
...
3
...
3
...
3a Practice Problem
Find the first and second derivatives of 𝑦 with respect to 𝑥 from the parametric equations given
...
𝑥 = 1 + 𝑡 2 , 𝑦 = 4𝑡 − 3
2
...
𝑥 = 𝑡 2 + 1, 𝑦 = 𝑡 2 + 1
4
...
Please answer it all first before checking, this is important for
self-assessment
...

If you got 2 – 3 items correct, good, you understand most of the concepts, you only need to practice
for a few times
...
your
Take
a breather
beforegiven,
you proceed
the
If you got 0 –You
1, it is
okay,finished
do not worry
...


lesson 4
...
2 Parametric Equation (Application)
3
...
1 Rectilinear Motion
Another application of finding derivatives is to compute velocities and accelerations of moving
objects
...
That is, the
acceleration is the derivative of the velocity with respect to time 𝑡
...
So we
have
𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 𝑎 =

𝑑𝑣
𝑑𝑡

=

𝑑2 𝑥
𝑑𝑡 2

𝑑𝑣

= 𝑣 𝑑𝑥

Example 5
...
3
The position of an object at time t is given by 𝑥 = 3𝑡 3 − 6𝑡 2 + 4𝑡 − 2, where 𝑥 is in feet and 𝑡 is
in seconds
...
2
...
An important instance
of uniformly accelerated motion arise when a body moves near the earth’s surface in a vertical
straight line
...

𝑎=

𝑑𝑣
𝑑𝑡

=𝑔;

1

𝑥 = 2 𝑔𝑡 2 + 𝑣𝑜 𝑡;

𝑣 2 = 𝑣𝑜 2 + 2𝑔𝑥

Example 5
...
4
A ball is dropped from the balloon at a height of 640𝑓𝑡
...

SOLUTION
With 𝑔 = 32
1

𝑥 = 2 𝑔𝑡 2 + 𝑣𝑜 𝑡
1

𝑥 = 2 (−32)𝑡 2 + 96𝑡, from which
𝑑𝑥 = (−32𝑡 + 96)𝑑𝑡 =

𝑑𝑥
𝑑𝑡

= 𝑣 = −32𝑡 + 96

At the highest point, 𝑣 = 0; hence 𝑡 = 3,
1
𝑥 = (−32)(3)2 + 96(3) = 144𝑓𝑡
2

Thus the height above the starting point is 144𝑓𝑡
...
:
1

𝑥 = 2 𝑔𝑡 2 + 𝑣𝑜 𝑡
640 = 16𝑡 2 − 96𝑡
16(𝑡 2 − 6𝑡 − 40) = 0; 𝑡 = −4
Thus the ball is in the air for 𝟏𝟎 𝒔𝒆𝒄
...
2
...

With the starting point as origin and the y-axis is positive upward, the initial condition are;
𝑥 = 0,
𝑦 = 0,
𝑣𝑥 = 𝑣𝑜 cos 𝛼,
𝑣𝑦 = 𝑣𝑜 sin 𝛼 when 𝑡 = 0
The force of gravity acts vertically downward; there is no horizontal force
...
3
...
How far does an arrow flies?
SOLUTION:
When the arrow hit the ground at maximum distance
𝑦 = 0
...
8 𝑠𝑒𝑐
𝑥 = 𝑣𝑜 𝑡 cos 𝛼
𝑥 = 200(8
...
𝟓𝒇𝒕
3
...
4 Velocity and Acceleration in Curvilinear Motion
If a point moves in a plane curve, its coordinates are functions of time:
𝑥 = 𝑓(𝑡),
𝑦 = 𝑔(𝑡)
Thus the total velocity is the vector sum of the velocities parallel to the axis
𝑑𝑥

𝑑𝑦

𝑑𝑡

𝑑𝑡

𝑣 = √𝑣𝑥 2 + 𝑣𝑦 2 = √( )2 + ( )2
inclined to the x-axis at an angle 𝛼 such that
𝑣𝑦

tan 𝛼 = 𝑣

𝑥

The acceleration is the vector 𝛼 whose components, parallel to the axis are
𝑑𝑦 2

𝑑2 𝑥

𝑎 = √𝑎𝑥 2 + 𝑎𝑦 2 = √ 𝑑𝑡 2 + 𝑑𝑡 2

inclined to the x-axis at an angle 𝛽 such that
𝑎𝑦

tan 𝛽 = 𝑎

𝑥

Example 5
...
6
Given the parametric equation below determine the total velocity and acceleration at 𝑡 = 2
𝑥 (𝑡) = 𝑡 3 − 3𝑡 2 + 𝑡

𝑦(𝑡) = 𝑡 2 − 2𝑡 + 3

and

SOULUTION:
For velocity
𝑥 (𝑡) = 𝑡 3 − 3𝑡 2 + 𝑡 from which

𝑑𝑥

𝑦(𝑡) = 𝑡 2 − 2𝑡 + 3 from which

𝑑𝑦

𝑣 = √(

𝑑𝑡

𝑑𝑡

= 3𝑡 2 − 6𝑡 + 1
= 2𝑡 − 2

𝑑𝑥 2
𝑑𝑦
) + ( )2 = √(3(2)2 − 6(2) + 1)2 + (2(2) − 2)2 = √5
𝑑𝑡
𝑑𝑡

MODULE 5 APPLICATION OF DERIVATIVES

11

ENGINEERING CALCULUS 1

[DATE]

For Acceleration
𝑑𝑥
𝑑𝑡
𝑑𝑦
𝑑𝑡

= 3𝑡 2 − 6𝑡 + 1 from which
= 2𝑡 − 2 from which

𝑑2 𝑥
𝑑𝑡 2

𝑑2 𝑥
𝑑𝑡 2

= 6𝑡 − 6

=2

2
𝑑 2 𝑥 𝑑𝑦
𝑎 = √ 2 + 2 = √(6(2) − 6)2 + 22 = 𝟒√𝟐
𝑑𝑡
𝑑𝑡

5
...
The position of an object at time t is given by 𝑥 = 34𝑡 2 − 8𝑡 + 2, where 𝑥 is in feet and 𝑡 is in
seconds
...
A ball is thrown upward and rises 9 𝑓𝑡
...
Find the total time taken for the ball to
return to the starting point
...
A point describe parabola 𝑦 2 = 4𝑥 + 1, with a constant velocity, 𝑣𝑦 = 4
...

Check your answer in the answer key
...

If you got 3 items correct, very good, you already understand the concept
...

If you got 0 – 1, it is okay, do not worry
...


MODULE 5 APPLICATION OF DERIVATIVES

12

ENGINEERING CALCULUS 1

[DATE]

Lesson

4

Indeterminate Forms

In this lesson you will learn how to evaluate the limits of indeterminate forms equations using
derivatives
...
If 𝑓 (𝑎) = 𝑓(𝑏), then there
must be at least one number 𝑐 in (𝑎, 𝑏) such that 𝑓 ′(𝑐 ) = 0
...
1 Example diagram for Rolle’s Theorem
In the first two diagrams, there is only one possible value of 𝑐 such that 𝑓 ′(𝑐 ) = 0
...
The fourth diagram shows a constant function, so its derivative is always 0
...
Now, let’s look at some pictures where Rolle’s
Theorem does not apply:

Figure 5
...
2, in all three cases the derivative is never been zero
...
1
Find the point where the derivative is equal to zero of the equation 𝑓 (𝑥 ) = 𝑥 2 − 2𝑥 + 1 and
between points (−1,3)
...
Then there’s at least one
number 𝑐 in (𝑎, 𝑏) such that
𝑓 ′ (𝑐 ) =

𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎

Figure 5
...
In the above picture, there are actually
two tangents that work—the x-coordinates are at 𝑐𝑜 and 𝑐1
...

Example 4
...
Use Mean Value Theorem
...

Example 4
...
The product is then said to take the
indeterminate form 0⋅ ∞
...

0

Example 4
...

Example 4
...
The transformation required are simple:
1
sin3 𝑥
1 − sin3 𝑥
lim+ (sec3 𝑥 − tan3 𝑥) = lim+ ( 3 −
)
=
lim
𝜋
𝜋
𝜋+ cos 3 𝑥
cos 𝑥 cos 3 𝑥
𝑥→
𝑥→
𝑥→
2

2

2

By using theorem 23
lim+

𝜋
𝑥→
2

−3 sin2 𝑥 cos 𝑥
= lim tan 𝑥 = −∞
−3 cos 2 𝑥 sin 𝑥 𝑥→𝜋+
2

The Indeterminate Form 00 , ∞0 , 1∞
Consider the function 𝑦 = 𝑓(𝑥)𝐹(𝑥) , the function is said, in the respective cases, assume the
indeterminate form 00 , ∞0 , 1∞
...

Example 4
...
Then
ln 𝑦 =

ln(1−cos 𝑥)
ln 𝑥

Therefore,
lim+ ln 𝑦 = lim+

𝑥→0

𝑥→0

ln(1−cos 𝑥)
ln 𝑥
sin 𝑥

𝑥
= lim+ 1−cos
1

𝑥→0

𝑥

𝑥 sin 𝑥

= lim+ 1−cos 𝑥
𝑥→0

= lim+
𝑥→0

sin 𝑥+𝑥 cos 𝑥
sin 𝑥

= 1+1 = 2
From 𝑙𝑛𝑦 → 2 it follows that 𝑦 = 𝑒 2
...
4 Practice Problem
Evaluate the limit of the following:
1
...
lim sin−1 𝛽
𝛽→0

1

1

3
...
lim (cot 𝑥 − 𝑒𝑥 −1)
𝑥→0

Check your answer in the answer key
...

If you got 4 - items correct, very well, you already understand the concept
...

If you got 0 – 1, it is okay, do not worry
...


MODULE 5 APPLICATION OF DERIVATIVES

17

ENGINEERING CALCULUS 1

[DATE]

Lesson

Maxima and Minima Problems

5

In this lesson you will learn how to evaluate the derivatives of a Hyperbolic functions
...
1 Maxima and Minima
Max-Min Theorem: If 𝑓 is continuous on (𝑎, 𝑏), then 𝑓 has at least one maximum and one minimum
on (𝑎, 𝑏)
...


In the first graph, the function attains its maximum at 𝑥 = 𝑐 and its minimum at 𝑥 = 𝑑
...
The third
graph has a maximum at 𝑥 = 𝑏 but the minimum is at both 𝑥 = 𝑐 and 𝑥 = 𝑑
...
Finally, the fourth graph shows a
constant function, which is continuous; in fact, every point in the interval (𝑎, 𝑏) is both a maximum and
a minimum, since the function never goes above or below the constant value 𝑐
...
5
...
6
...
from an observer, rises vertically at the rate of 10𝑓𝑡/𝑠𝑒𝑐
...
?
SOLUTION
In time 𝑡, the balloon rises a distance 10𝑡, so that
𝑠 = √3600 + 100𝑡 2
𝑑𝑠
𝑑𝑡

100𝑡

= √3600+100𝑡2

When 𝑡 = 8,
𝑑𝑠
𝑑𝑠

=

800
√3600+6400

= 𝟖𝒇𝒕/𝒔𝒆𝒄

Example 5
...
2
As a man walks across a bridge at the rate of 5𝑓𝑡/𝑠𝑒𝑐, a boat passes directly beneath him at
10𝑓𝑡/𝑠𝑒𝑐
...
above the water, how fast are the man and boat separating 3 𝑠𝑒𝑐
...
, the man covers a distance 5𝑡, the boat a distance 10𝑡
...


MODULE 5 APPLICATION OF DERIVATIVES

21

ENGINEERING CALCULUS 1

[DATE]

𝑑𝑠
375
𝟐𝟓
=
=
𝒇𝒕/𝒔𝒆𝒄
𝑑𝑡 √2025
𝟑
Example 5
...
3
A man on a wharf 20𝑓𝑡 above the water pulls in a rope, to which a boat is attached, at the rate
of 4𝑓𝑡/𝑠𝑒𝑐
...
Then, given
𝑑𝑥

𝑑𝑟
𝑑𝑡

, we have to

find 𝑑𝑡
...
6
...
Find how fast the surface is rising when the water is 8𝑓𝑡
...

SOLUTION
The volume of water is
1

𝑉 = 𝜋𝑟 2 ℎ
3

By similar triangles,
𝑟

5

1

= 20,
ℎ

𝑟 = 4ℎ

Hence
𝑉=

𝜋ℎ2

,
48

𝑑𝑉
𝑑𝑡

=

𝜋ℎ2 𝑑ℎ
16 𝑑𝑡

The rate of change of the volume of water with respect to
time is
𝑑𝑉
𝑑𝑡

= 15

So that
𝜋ℎ 2 𝑑ℎ
16 𝑑𝑡

= 15,

𝑑ℎ
𝑑𝑡

240

= 𝜋ℎ2

When ℎ = 8,
𝑑ℎ
𝑑𝑡

15

= 4𝜋 = 𝟏
...
2 Exponential Growth and Decay
Exponential growth equation:

𝑃(𝑡) = 𝑃𝑜 𝑒 𝑘𝑡

Exponential decay equation:

𝑃(𝑡) = 𝑃𝑜 𝑒 −𝑘𝑡

for radioactive decay with half-life 𝑡 = 1/2:

𝑃 (𝑡) = 𝑃𝑜 𝑒 −𝑘𝑡 with 𝑘 =

𝑙𝑛2
𝑡1
2

Example 5
...
5
The population of rabbits started 3 𝑦𝑒𝑎𝑟𝑠 ago at 1000, but now has grown to 64,000, then what
will the population be 𝑜𝑛𝑒 𝑦𝑒𝑎𝑟 from now? Also, what is the total time it will take for the population to
grow from 1000 to 400,000?
SOLUTION:
Well, we have 𝑃𝑜 = 1000, since that’s the initial population
...
The problem is, we don’t know what 𝑘 is
...
𝟑𝟐 𝒚𝒆𝒂𝒓𝒔
So although it takes 4 𝑦𝑒𝑎𝑟𝑠 to get up to a population of 256,000, it only takes approximately
1
two-sevenths of a year more - about 3 2 - to get up to 400,000
...


MODULE 5 APPLICATION OF DERIVATIVES

23

ENGINEERING CALCULUS 1

[DATE]

5
...
Water is flowing into a vertical cylindrical tank at the rate of 24𝑓𝑡 3 /𝑚𝑖𝑛
...
, how fast is the surface rising?
2
...
Find the radius of the
tank
...
A man 6𝑓𝑡 tall walks away from a lamp post 16𝑓𝑡 high at the rate of 5 𝑚𝑖/ℎ𝑟
...
A rectangular trough is 10𝑓𝑡 long and 3𝑓𝑡 wide
...
Please answer it all first before checking, this is important for selfassessment
...

If you got 2-3 items correct, good, you understand most of the concepts, you only need to practice for a few
times
...
Compare your answer to the answer given, then try to answer the
questions again before you proceed to the next lesson
...
Use pen
to write your answer
...
Take note that all pages must have 0
...
Copy the questions then answer
...
Please avoid erasure
...
5:
Find the differentials of the following:
1
...
𝑟 = √

1+𝑠

Find the first and second derivative of 𝑦 with respect to 𝑥 from the parametric equations given
...
𝑥 = (𝑡+1)2; 𝑦 = 𝑡 2 + 3
4
...
𝑥 = √𝑡 + 2; 𝑦 = 𝑡 2 − 3
Evaluate the limit of the following
6
...
lim

tan−1 𝛽

𝛽→0

𝛽
1

8
...
The base of a right triangle is fixed at 3𝑓𝑡
...
long and subject to change
...

10
...
1 𝑖𝑛
...

11
...

12
...
What are the object’s velocity, acceleration and total distance travel at time 𝑡 = 1?
13
...
Show that they strike the ground at the same time
...
Given the parametric equation below determine the total velocity and acceleration at 𝑡 = 2
1

𝑥 (𝑡) = √𝑡 3 − 𝑡 2 + 𝑡

and

𝑦(𝑡) = 𝑡 2 − 2√𝑡 + 3

15
...


MODULE 5 APPLICATION OF DERIVATIVES

25

ENGINEERING CALCULUS 1

[DATE]

16
...
How much of the material can be saved?
17
...

18
...

What should be the relative dimensions of the field to make the amount of fencing?
19
...
Find the length of the shortest ladder that will reach the
house, when one end rest on the ground outside the wall
...
A triangular trough is 10𝑓𝑡 long, 6 𝑓𝑡 across the top, and 3 𝑓𝑡 deep
...
3 /𝑚𝑖𝑛, find how fast the surface is rising when the water is 6 𝑖𝑛
...

21
...
If the top slides downward at the rate of 2𝑓𝑡/𝑠𝑒𝑐, how
fast the lower ends is moving when it is 16𝑓𝑡 from the wall
...
A man 6𝑓𝑡 tall walks away from a lamp post 16𝑓𝑡 high at the rate of 5 𝑚𝑖/ℎ𝑟
...
A train, starting at noon, travels north at 40 𝑚𝑖/ℎ𝑟
...

24
...
A man walk along the path at 6𝑓𝑡/𝑠𝑒𝑐
...
1 Practice Problem:
1
...
𝑑𝑧 = 3(1 − 2𝑣 3)(1 − 2𝑣 + 𝑣 4)−2 𝑑𝑣
1

3
...
𝑑𝑥 = (𝑡 2 − 2

2

− 𝑡 2 ) 𝑑𝑡
𝑡

√

Answer for 5
...
𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑥 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠
2
...
25
...
4
...
3a Practice Problem:
1
...

3
...


2 𝑑2 𝑦

𝑑𝑦

1

= − 𝑡 ,𝑑𝑥2 = − 𝑡 3
𝑑𝑥

𝑑𝑦

𝑑2 𝑦

𝑑𝑥

= 6,𝑑𝑥2 = 0

𝑑𝑦

𝑑2 𝑦

= 1,𝑑𝑥2 = 0
𝑑𝑥

𝑑𝑦
𝑑𝑥

=−

2𝑡 4 −𝑡 3 𝑑2 𝑦
2

,

𝑑𝑥 2

= −(

8𝑡 3 −3𝑡 2
2

𝑡3

)(− )
2

Answer for 4
...
𝑣 = 60 𝑓𝑡/𝑠𝑒𝑐, 𝑎 =

68𝑓𝑡
𝑠𝑒𝑐 2

,𝑥 = 28𝑓𝑡

2
...
5 𝑠𝑒𝑐
3
...
4 Practice Problem:
1
...
1
3
...
1/2
Answer for 5
...
1/2
2
...
4 𝑘 3
4
...
3
MODULE 5 APPLICATION OF DERIVATIVES

27

ENGINEERING CALCULUS 1

[DATE]

Answer for 5
...
0
...
2
...
8 𝑚𝑖/ℎ𝑟
4
...
4𝑓𝑡/𝑚𝑖𝑛

References
1
...
Clyde E
...
Rainville, Differential and integral calculus 6th edition
3

---