Here are the steps to solve this problem:
(a) Let t = time and y = height. Then the differential equation is:
dy/dt = -32 ft/sec^2
Integrate both sides:
∫dy = ∫-32 dt
y = -32t + C
Initial conditions: at t = 0, y = 0
0 = -32(0) + C
C = 0
Therefore, the equation is: y = -32t
When y = 0 (the maximum height), t = 0.625 sec
(b) Put t = 0.625 sec into the equation:
y = -32(0.625) = -20 ft
Continuity says that the limit of a function at a point equals the value of the function at that point, or, that small changes in the input give only small changes in output. This has important implications, such as the Intermediate Value Theorem.
Integration by substitution is the chain rule in reverse.
NOTE: the final location is section specific. Section 1 (morning) is in SILV 703, Section 11 (afternoon) is in CANT 200
Basic Calculus 11 - Derivatives and Differentiation RulesJuan Miguel Palero
It is a powerpoint presentation that discusses about the lesson or topic of Derivatives and Differentiation Rules. It also encompasses some formulas, definitions and examples regarding the said topic.
Continuity says that the limit of a function at a point equals the value of the function at that point, or, that small changes in the input give only small changes in output. This has important implications, such as the Intermediate Value Theorem.
Integration by substitution is the chain rule in reverse.
NOTE: the final location is section specific. Section 1 (morning) is in SILV 703, Section 11 (afternoon) is in CANT 200
Basic Calculus 11 - Derivatives and Differentiation RulesJuan Miguel Palero
It is a powerpoint presentation that discusses about the lesson or topic of Derivatives and Differentiation Rules. It also encompasses some formulas, definitions and examples regarding the said topic.
Lesson 8: Derivatives of Polynomials and Exponential functionsMatthew Leingang
Some of the most famous rules of the calculus of derivatives: the power rule, the sum rule, the constant multiple rule, and the number e defined so that e^x is its own derivative!
01. Differentiation-Theory & solved example Module-3.pdfRajuSingh806014
Total No. of questions in Differentiation are-
In Chapter Examples 31
Solved Examples 32
The rate of change of one quantity with respect to some another quantity has a great importance. For example the rate of change of displacement of a particle with respect to time is called its velocity and the rate of change of velocity is
called its acceleration.
The following results can easily be established using the above definition of the derivative–
d
(i) dx (constant) = 0
The rate of change of a quantity 'y' with respect to another quantity 'x' is called the derivative or differential coefficient of y with respect to x.
Let y = f(x) be a continuous function of a variable quantity x, where x is independent and y is
(ii)
(iii)
(iv)
(v)
d
dx (ax) = a
d (xn) = nxn–1
dx
d ex =ex
dx
d (ax) = ax log a
dependent variable quantity. Let x be an arbitrary small change in the value of x and y be the
dx
d
(vi) dx
e
(logex) = 1/x
corresponding change in y then lim
y
if it exists, d 1
x0 x
is called the derivative or differential coefficient of y with respect to x and it is denoted by
(vii) dx
(logax) =
x log a
dy . y', y
dx 1
or Dy.
d
(viii) dx (sin x) = cos x
So, dy dx
dy
dx
lim
x0
lim
x0
y
x
f (x x) f (x)
x
(ix) (ix)
(x) (x)
d
dx (cos x) = – sin x
d (tan x) = sec2x
dx
The process of finding derivative of a function is called differentiation.
If we again differentiate (dy/dx) with respect to x
(xi)
d (cot x) = – cosec2x
dx
d
then the new derivative so obtained is called second derivative of y with respect to x and it is
Fd2 y
(xii) dx
d
(xiii) dx
(secx)= secx tan x
(cosec x) = – cosec x cot x
denoted by
HGdx2 Jor y" or y2 or D2y. Similarly,
d 1
we can find successive derivatives of y which
(xiv) dx
(sin–1 x) = , –1< x < 1
1 x2
may be denoted by
d –1 1
d3 y d4 y
dn y
(xv) dx (cos x) = –
,–1 < x < 1
dx3 ,
dx4 , ........, dxn , ......
d
(xvi) dx
(tan–1 x) = 1
1 x2
Note : (i)
y is a ratio of two quantities y and
x
(xvii) (xvii)
d (cot–1 x) = – 1
where as dy
dx
dy
is not a ratio, it is a single
dx
d
(xviii) (xviii)
(sec–1 x) =
1 x2
1
|x| > 1
quantity i.e.
dx dy÷ dx
dx x x2 1
(ii)
dy is
dx
d (y) in which d/dx is simply a symbol
dx
(xix)
d (cosec–1 x) = – 1
dx
of operation and not 'd' divided by dx.
d
(xx) dx
(sinh x) = cosh x
d
(xxi) dx
d
(cosh x) = sinh x
Theorem V Derivative of the function of the function. If 'y' is a function of 't' and t' is a function of 'x' then
(xxii) dx
d
(tanh x) = sech2 x
dy =
dx
dy . dt
dt dx
(xxiii) dx
d
(xxiv) dx
d
(coth x) = – cosec h2 x (sech x) = – sech x tanh x
Theorem VI Derivative of parametric equations If x = (t) , y = (t) then
dy dy / dt
=
(xxv) dx
(cosech x) = – cosec hx coth x
dx dx / dt
(xxvi) (xxvi)
(xxvii) (xxvii)
d (sin h–1 x) =
The Roman Empire A Historical Colossus.pdfkaushalkr1407
The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
The Indian economy is classified into different sectors to simplify the analysis and understanding of economic activities. For Class 10, it's essential to grasp the sectors of the Indian economy, understand their characteristics, and recognize their importance. This guide will provide detailed notes on the Sectors of the Indian Economy Class 10, using specific long-tail keywords to enhance comprehension.
For more information, visit-www.vavaclasses.com
How to Create Map Views in the Odoo 17 ERPCeline George
The map views are useful for providing a geographical representation of data. They allow users to visualize and analyze the data in a more intuitive manner.
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
Students, digital devices and success - Andreas Schleicher - 27 May 2024..pptxEduSkills OECD
Andreas Schleicher presents at the OECD webinar ‘Digital devices in schools: detrimental distraction or secret to success?’ on 27 May 2024. The presentation was based on findings from PISA 2022 results and the webinar helped launch the PISA in Focus ‘Managing screen time: How to protect and equip students against distraction’ https://www.oecd-ilibrary.org/education/managing-screen-time_7c225af4-en and the OECD Education Policy Perspective ‘Students, digital devices and success’ can be found here - https://oe.cd/il/5yV
How to Make a Field invisible in Odoo 17Celine George
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
We all have good and bad thoughts from time to time and situation to situation. We are bombarded daily with spiraling thoughts(both negative and positive) creating all-consuming feel , making us difficult to manage with associated suffering. Good thoughts are like our Mob Signal (Positive thought) amidst noise(negative thought) in the atmosphere. Negative thoughts like noise outweigh positive thoughts. These thoughts often create unwanted confusion, trouble, stress and frustration in our mind as well as chaos in our physical world. Negative thoughts are also known as “distorted thinking”.
2. Find all possible functions
F(x) whose derivative is
f(x) = 2x+1
F(x) = x2 + x + 5
F(x) = x2 + x - 1000
F(x) = x2 + x + 1/8
F(x) = x2 + x - π
F(x) =
x2 + x
3. Definition
A function F is called an antiderivative (also an
indefinite integral) of a function f in the
interval I if
F '( x)
f ( x)
for every value x in the interval I.
The process of finding the antiderivative of a
given function is called antidifferentiation or
integration.
4. Find all antiderivatives
F(x) of f(x) = 2x+1
F(x) = x2 + x + 5
F(x) = x2 + x - 1000
F(x) = x2 + x + 1/8
F(x) = x2 + x - π
In fact, any function of the form F(x) =
x2 + x + c where c is a constant is an
antiderivative of 2x + 1
F(x) =
x2 + x
5. Theorem
If F is a particular antiderivative of f on an
interval I, then every antiderivative of f on I
is given by
F ( x) c
where c is an arbitrary constant, and all the
antiderivatives of f on I can be obtained by
assigning particular values for c. .
6. Notation
4 The symbol
denotes the operation of
antidifferentiation, and we write
f ( x)dx
F ( x) c
where F’(x)=f(x), and c is an arbitrary constant.
This is read “The indefinite integral of f(x)
with respect to x is F(x) + c".
7. f ( x)dx
F ( x) c
In this notation,
is the integral sign;
f(x) is the integrand;
dx
is the differential of x which denotes
the variable of integration; and
c
is called the constant of integration.
4 If the antiderivative of the function on interval
I exists, we say that the function is integrable
over the interval I.
8. Integration Rules
1. Constant Rule. If k is any real number, then
the indefinite integral of k with respect to x is
kdx
kx C
2. Coefficient Rule. Given any real number
coefficient a and integrable function f,
af ( x)dx
a f ( x)dx
9. Integration Rules
3. Sum and Difference Rule. For integrable
functions f and g,
[ f1 ( x)
f 2 ( x)]dx
f1 ( x)dx
f 2 ( x)dx
4. Power Rule. For any real number n,
where n ≠ -1, the indefinite integral xn of is,
n 1
n
x dx
x
C
n 1
10. Example 1.
(5 x 7)dx
5 xdx
7dx
5 xdx
7dx
1
2
5( x
5
2
x
2
2
C1 ) 7 x C2
7x C
13. Integration Formulas for
Trigonometric Functions
sin x dx
cos x dx
2
sec x dx
cos x C
sin x C
tan x C
csc2 x dx
cot x C
sec x tan x dx
csc x cot x dx
sec x C
csc x C
14. Example 4.
2
(3 csc x cot x 7 sec x)dx
2
3 csc x cot xdx 7 sec xdx
3 csc x C1
7 tan x C2
3 csc x 7 tan x C
15. Example 5.
2
3 tan
4 cos
d
cos
1
3
tan d
4 cos d
cos
3 sec tan d
4 cos d
3 sec
C
4 sin
16. Exercises:
3
1. y ( 2 y
2.
3.
2
3)dy
x ( x 1)dx
y4
2 y2 1
dy
y
sin x
4.
dx
2
cos x
5. (2 cot2
3 tan 2 )d
17. Integration by Chain Rule/Substitution
For integrable functions f and g
f ( g ( x))[ g '( x)dx] F ( g ( x)) C
where is an F antiderivative of f and C is an
arbitrary constant.
18. Example 6.
36 x
24
2
3
6x
4
6x
2 (6 x
2
8
5
(6 x
(6 x
5 dx
3
2
5 (18 x dx)
5) 4 (18 x 2 dx)
3
3
g’(x)=18x2
1
3
5)
5
Let g(x) = 6x3+5
5
4
C
4
5)
5
4
C
19. Example 6. Take 2!
36 x 2 4 6 x 3 5 dx 2
6 x 3 5 (18 x 2 dx)
4
4
2
u du
1
Let u = 6x3 + 5
2 (u ) du
2
8
5
8
5
u
5
u
5
4
du = 18x2 dx
4
C
4
5
C
4
(6 x
3
5)
5
4
C
20. Let g(t) = t4 + 2t
g’(t) = 4t3 + 2
Example 7.
2t 3 1
t4
2t
7
dx
= 2(2t3 + 1)
1 2(2t 3 1) dx
7
4
2
t 2t
1
2
1 t
2
t
4
4
12 t 4
7
2t
2t
6
C
6
1
2t
2(2t 3 1)dx
6
C
21. Example 8.
5
2
x (x
Let u = x2 -1
12
1) 2dx
du = 2x dx
x
2 2
(x
2
2
x2 = u+1
12
1) 2 xdx
12
(u 1) u du
(u
14
2u
13
1
15
u15
1
15
( x 2 1)15
2
14
12
u )du
u14
1
13
1
7
u13 C
( x 2 1)14
1
13
( x 2 1)13 C
22. Example 9.
sin 2 x 2 cos 2 x dx
1
2
2 cos 2 x (2 sin 2 xdx)
1
2
u (du )
1
2
Let u = 2 – cos2x
1/ 2
u (du )
1 2
2 3
1
3
du = 0 – (-sin2x)(2dx)
u 3/ 2
=2sin2xdx
C
(2 cos 2 x)
3/ 2
C
23. Example 10.
(tan 2 x cot 2 x) 2 dx
sin 2 x
cos 2 x
2
cos 2 x
sin 2 x
2
sin 2 x cos 2 x
cos 2 x sin 2 x
1
dx
2
2
cos 2 x sin 2 x
sec2 2 x csc2 2 xdx
2
dx
2
dx
24. Example 10.
sec2 2 x csc2 2 xdx
sec2 2 x(cot2 2 x 1)dx
2
2
sec 2 x cot 2 xdx
sec 2 xdx
(tan 2 x) 2 sec2 2 xdx
(tan 2 x) 2 2 sec2 2 xdx
1
2
1
2
(tan 2 x)
1
2
cot 2 x
1
1
2
1
2
2
sec2 2 xdx
1
2
tan 2 x C
tan 2 x C
2 sec2 2 xdx
25. Exercises:
1. 7 x(2 x
2
6
1) dx
3
2 2
2. 5 x (9 4 x )
2r
3.
dr
7
(1 r )
2
2
4. y csc3 y cot 3 y dy
cos3x
5.
dx
1 2 sin 3x
2
x(3x 1)dx
6.
4
2
4
(3x 2 x 1)
26. Applications of
Indefinite Integrals
1. Graphing
Given the sketch of the graph of the function,
together with some function values, we
can sketch the graph of its antiderivative
as long as the antiderivative is
continuous.
27. Example 11. Given the sketch of the function f
=F’(x) below, sketch the possible graph of F if it is
continuous, F(-1) = 0 and F(-3) = 4.
F(x)
X<-3
F’(x)
F’’(x) Conclusion
+
-
Increasing,
Concave down
0
-
Relative maximum
-3<x<-2
-
-
Decreasing,
Concave down
X=-2
-
0
Decreasing,
Point of inflection
-2<x<-1
-
+
Decreasing
Concave up
0
+
Relative minimum
+
+
Increasing,
Concave up
5
X=-3
4
4
3
2
1
-5
-4
-3
-2
-1
0
1
2
3
4
5
-1
-2
-3
-4
-5
X=-1
X>-1
0
29. Applications of
Indefinite Integrals
1. Boundary/Initial Valued Problems
There are many applications of indefinite integrals
in different fields such as physics, business,
economics, biology, etc.
These applications usually desire to find particular
antiderivatives that satisfies certain conditions
called initial or boundary conditions, depending
on whether they occur on one or more than one
point.
30. Example 11.
Suppose we wish to find a particular
antiderivative satisfying the equation
dy
dx
6x 1
and the initial condition y=7 when x =2.
31. Sol’n of Example 11
dy
(6 x 1)dx
dy
(6 x 1)dx
2
y 3x
but x
x C
2 when
7 3(2)
2
3 C
y
C
7, then
7
Thus the particular antiderivative desired,
y
3x
2
x 7
32. Example 12.
The volume of water in a tank is V
cubic meters when the depth of water is h
meters. The rate of change of V with
respect to h is π(4h2 +12h + 9), find the
volume of water in the tank when the
depth is 3m.
33. Sol’n of Example 12
dV
dh
4h 2 12h
4h 2 12h
dV
V
4h3
3
0
4(03 )
6(02 )
3
C
6h 2
9 dh
9h
Volume V=0 if depth
h =0
C
9(0)
C
0
Thus V
V
9
4h3
3
4(33 )
6(32 )
3
6h 2
9(3)
9h
207
m3
34. The Differential Equations
Equation containing a function and its derivative or just its
derivative is called differential equations.
Applications occur in many diverse fields such as physics,
chemistry, biology, psychology, sociology, business,
economics etc.
The order of a differential equation is the order of the
derivative of highest order that appears in the equation.
The function f defined by y= f(x) is a solution of a
differential equation if y and its derivatives satisfy the
equation.
35. dy
6x 1
dx
dy (6 x 1)dx
dy
(6 x 1)dx
y 3x 2 x C
but x 2 when
7 3(2) 2 3 C
y
C
7, then
7
Thus find the particular solution
y
3x
2
x 7
36. If each side of the differential equations
involves only one variable or can be
reduced in this form, then, we say that these
are separable differential equations.
Complete solution (or general solution)
y = F(x) + C
Particular solution – an initial condition is
given
37. Example 13. Find the complete
solution of the differential equation
d2y
2
dx
d2y
dx 2
let
2
d y dy
4x 3
2
dx
dx
dy (4 x 3)dx
dy
y
(4 x 3)dx
2x
2
dy
dx
y
y
dy
dx
dy
dx
2 x 2 3 x C1
dy
3 x C1
d
dx
4x 3
(2 x
2
3
x3
3
2
2
3 x C1 )dx
x 2 C1 x C2
38. Example 14. Find the particular solution of
the differential equation in Ex. 13 for which
y=2 and y’=-3 when x=1.
y
2x
2
2
3 x C1
2
3
x3
2
3 x C1
3 2(1)
C1
y
2
3
(1)3
C2
8
y
2
3
x
3
3
2
x
2
8x
3
2
x 2 C1 x C2
3
2
(1) 2 8(1) C2
47
6
47
6
39. Example 16.
A stone is thrown vertically upward from the
ground with an initial velocity of 20ft/sec.
(a) How long will the ball be going up?
Ans. 0.625 sec
(b) How high will the ball go?
Ans. 6.25 ft
(c) With what velocity will the ball strike the
ground?
Ans. 20 ft/sec