FACULTY OF CIVIL AND ENVIRONMENTAL ENGINEERING
ENGINEERING MATHEMATICS III BFC 24103
SEKSYEN 2
SEM 2 SESSION 2016/2017
APPLICATION OF VECTOR VALUED FUNCTION
MUHAMMAD RIDHWAN BIN MOHD RASID DF150077 MUHAMAD HANIF BIN ROS KAMAR DF150087 MUHAMMAD IKRAM BIN RAHAMAN CF 150022
LECTURER’S NAME: MISS NORFATIN BINTI SALAM
DATE OF SUBMISSON: 22 MAY 2017 CONTENT
Table of Content
1
Introduction
2
Fila Table
3
Problem Statement
4
Discussion
5
Conclusion
14
References
16
Minutes of Meeting
17
1
INTRODUCTION In the Western world prior to the Sixteenth Century, it was generally assumed that the acceleration of a falling body would be proportional to its mass - that is, a 10 kg object was expected to accelerate ten times faster than a 1 kg object. It was an immensely popular work among academicians and over the centuries it had acquired a certain devotion verging on the religious. A plane curve is defined as the set of ordered pairs (f(t), g(t)) together with their defining parametric equations x = f(t) and y = g(t) Where f and g are continuous functions of t on an interval I. A space curve C is the set of all ordered triples (f(t), g(t), h(t)) together with their defining parametric equations x = f(t), y = g(t), and z = h(t) Where f, g, and h are continuous functions of t on an interval I. A new type of function, called a vector-valued function, is introduced. This type of function maps real numbers to vector. Technically, a curve in the plane or in space consists of a collection of points and the defining parametric equations. Two different curves can have the same graph. For instance, each of the curves given by r(t) = sin t i + cos t j and r(t) = sin t2 i + cos t2 j has the unit circle as its graph, but these equations do not represent the same curve because the circle is traced out in different ways on the graphs. Be sure you see the distinction between the vector-valued function r and the real-valued functions f, g, and h. All are functions of the real variable t, but r(t) is a vector, whereas f(t), g(t), and h(t) are real numbers (for each specific value of t). Vector-valued functions serve dual roles in the representation of curves. By letting the parameter t represent time, you can use a vector-valued
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function to represent motion along a curve. Or, in the more general case, you can use a vectorvalued function to trace the graph of a curve. Facts
Idea
Learning Issue
Action Plan
The position function
This information can
How to find the
Acceleration can be
of Keira’s 1977
be used to find the
acceleration in mph
calculated by using
Camaro moving in a
acceleration in mph
of her Camaro?
vector valued
straight line is r(t) =
of her Camaro.
function formula.
2t³ - 10t² + 5.
a(t) = dv(t)/dt = d2r(t)/dt2
FILA TABLE
3
PROBLEM STATEMENT We know that given the position function of an object that the velocity of the object is the first derivative of the position function and the acceleration of the object is the second derivative of the position function. So, given this it shouldn’t be too surprising that if the position function of an object is given by the vector function r(t) then the velocity and acceleration of the object is given by, v(t) = r’(t)
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a(t) = v’(t)
DISCUSSION Example Question 1: If the acceleration of an object is given by a = I + 2j + 6k. find the object velocity and position function given that the initial velocity is v (0) = j – k and the initial position is r (0) = I – 2j +3k. Solution: v(t) = ∫ a(t) dt = ∫ I + 2j + 6tk dt = ti + 2tj + 3t2k + c
J – k = v(0) = c v(t) = ti + 2tj +3t2k + j – k = ti + (2t + 1)j + (3t2 – 1 )k
r(t) = ∫ v(t) dt = ∫ ti + (2t + 1)j + (3t2 – 1)k dt =
1 2 2 3 2 t i + (t + t)j + (t – t)k + c
I – 2j + 3k = r(0) = c So, the position function is, 1 r(t) = ( 2 t2 + 1)I + (t2 + t – 2)j + (t3 – t +3)k 5
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Application Question 2: Daniel Bego dives into the pool with a velocity equation v(t) = t² - t – 4. Find the location of Daniel Bego after five seconds in meter.
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Solution: 5
Positon of Daniel Bego after five seconds, r(t) =
∫ t 2−t−4 dt 0
5 1 3 1 2 [ t − t −4 t] = 3 2 0
125 25 − −20 ) – ( 0 ) 3 2
=(
= 9.166 meter Question 3: A roller coaster accelerates with equation a(t) = 6t – 4 and has a velocity of 80mph at 5 seconds. Is the roller coaster speeding up or slowing down at 10 seconds?
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Solution: Acceleration of the roller coaster, a(t) = 6t – 4 a(10) = 6(10) – 4 = 56 mph Velocity of the roller coaster at 10 seconds, v(t) = =
∫ a ( t ) dt ∫ 6 t−4 dt 2
= 3 t −4 t +c v(5) = 80 mph 2 80 = 3 ( 5 ) −4 ( 5 ) +c
c = 25 2 v(t) = 3 t −4 t +25
so, both acceleration and velocity have positive sign which is meaning that the roller coaster is speeding up.
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Question 4: The position function of Keira’s 1977 Camaro moving in a straight line is r(t) = 2t³ - 10t² + 5. Find the acceleration in mph of her Camaro at 8.6 seconds.
Solution: 3
2
Position vector of Camaro, r(t) = 2t −10 t +5 2 Velocity of Camaro, r’(t) = v(t) = 6 t −20 t
Acceleration of Camaro, r” (t) = a(t) = 12t−20 Acceleration of Camaro at 8.6 seconds, a (8.6) = 12(8.6) – 20 = 83.2 mph
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Methodology Question 1: Vector calculus plays an important role in differential geometry and in the study of partial differential equations. It is used extensively in physics and engineering, especially in the description of electromagnetic fields, gravitational fields and fluid flow. First get the velocity. To do this we need to integrate the acceleration. v(t) = ∫ a(t) dt = ∫ I + 2j + 6tk dt = ti + 2tj + 3t2k + c To get velocity completely we will need to determine the constant of integration. We can use the initial velocity to get this. J – k = v(0) = c The velocity of the object is then, v(t) = ti + 2tj +3t2k + j – k = ti + (2t + 1)j + (3t2 – 1 )k We will find the position function by integrating the velocity function. r(t) = ∫ v(t) dt = ∫ ti + (2t + 1)j + (3t2 – 1)k dt =
1 2 2 3 2 t i + (t + t)j + (t – t)k + c
Using initial position, I – 2j + 3k = r(0) = c So, the position function is,
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1 r(t) = ( 2 t2 + 1)I + (t2 + t – 2)j + (t3 – t +3)k Question 2: For the particle moving along the vector-valued function or curve r(t), the magnitude of the first derivative r(t) which is r’(t) measures the size of the displacement vector per unit time, i.e. the velocity v(t) = r’(t) So we need to integrate the velocity, v(t) to get the position vector, r(t). r ( t ) =∫ v (t) 5
Positon of Daniel Bego after five seconds, r(t) =
∫ t 2−t−4 dt 0
Since this is definite integral which has a limit, so we don’t have to put the c after we integrate the velocity of Daniel Bego. 5 1 3 1 2 [ t − t −4 t] = 3 2 0
Substitute the t value into the positon vector equation which is 5 then we get the answer. =(
125 25 − −20 ) – ( 0 ) 3 2
= 9.166 meter
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Question 3: Substitute the value of t which 10 seconds into the acceleration equation of the roller coaster to know whether the outcome acceleration is positive valued or negative valued. Acceleration of the roller coaster, a(t) = 6t – 4 a(10) = 6(10) – 4 = 56 mph Note that the velocity is also the tangent vector and hence it is in the direction of tangent line. The acceleration of the particle is defined as the derivative of the velocity. a(t) = v’(t) = r’’(t) Hence, we need to integrate the acceleration of the roller coaster to find its velocity. Since this integration is not a definite integral so we must add c. Velocity of the roller coaster at 10 seconds, v(t) = =
∫ a ( t ) dt ∫ 6 t−4 dt
2 = 3 t −4 t +c
To find the value of c we need to compare that product from integration with the given velocity at 5 second which is 80 mph. v(5) = 80 mph 2
80 = 3 ( 5 ) −4 ( 5 ) +c c = 25 Substitute the value of c into the equation. Then we get the velocity equation of the roller coaster. 2 v(t) = 3 t −4 t +25
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so, both acceleration and velocity have positive sign which is meaning that the roller coaster is speeding up.
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Question 4: For this problem, we need to differentiate the position vector of the Camaro and differentiate again to find the acceleration of Camaro.
a(t) = dv(t)/dt = d2r(t)/dt2
3 2 Position vector of Camaro, r(t) = 2t −10 t +5
2 Velocity of Camaro, r’(t) = v(t) = 6 t −20 t
Acceleration of Camaro, r” (t) = a(t) = 12t−20
Then, substitute the value of t which is 8.6 seconds to find its acceleration of Camaro.
Acceleration of Camaro at 8.6 seconds, a (8.6) = 12(8.6) – 20 = 83.2 mph
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CONCLUSIONS A vector-valued function is a function which produces a vector for the output. Since a vector can be broken down into its component parts this function can be described by how it is behaving in each component, i.e. F(t) = hf(t), g(t), h(t)i = f(t)i + g(t)j + h(t)k. We can ask calculus questions about these functions, i.e., limits, derivatives, antiderivatives and definite integrals. The key is that we will define these items in the same way and since we know that vector addition and scaling works component-by-component and calculus in the end boils down to fancy additions and scaling we conclude that it suffices to work component by component (to be fair this should be carefully checked; but we will leave those annoying details to the math majors). On a side note, when finding the anti-derivative of a vector valued function we will have a separate constant for each component. Of course we can combine these constants into a single vector and conclude that two antiderivatives of a vector-valued function differ by a constant vector. The various derivative rules that we have learned for real-valued functions also generalize as we would expect. As an application, suppose we have a parametric curve (x (t), y (t), z (t)). Then we can think of this as a vector valued function by putting a vector from the origin to the location of the particle at time t, i.e. r (t) =[x(t), y(t), z(t)]. Then we have that v (t) = r’(t) represents the velocity of the particle at time t and a(t) = v’(t) = r’’(t) equals the acceleration of the particle at time t. This can be used to describe how the behavior of the particle (i.e., location, direction of motion, acceleration, and so on). By taking antiderivatives we can also find position given some information about velocity and/or acceleration. Note that the magnitude of velocity is speed, and if we want to find distance traveled by the particle from t = a to t = b we can do this by finding the integral of speed. So we have
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v ( t ) dt =¿ Length =
b
∫¿
b
∫ r ' ( t ) dt a
a
By expanding this we see we get the same expression that we had before for length.
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REFERENCES -
Chudinov S. Peter, (2011). Approximate Analytical Investigation of Projectile Motion in a Medium with Quadratic Drag Force. International Journal of Sports Science and
-
Engineering Vol. 05, No. 01, pp. 027-042 Morales A. Daniel (2011). A generalization on projectile motion with linear resistance.
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Canadian Journal of Physics 89(12) Warburton R. D. H. and Wang J. (2004). Analysis of asymptotic projectile motion with air
-
resistance using the Lambert function. American Journal of Physics. Volume 72, Issue 11 Warburton R. D. H. and Wang J. (2010). Analytic Approximations of Projectile Motion
-
with Quadratic Air Resistance. Journal of service science and Management. Pp 98-105 Robinson G. and Robinson I, (2013).The motion of an arbitrarily rotating spherical projectile and its application to ball games. Physica Scripta.
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MINUTES OF MEETING
Project Name: Problem – Application of Vector Valued Function Date
: 9/5/2017 (Tuesday)
Venue
: Pusat Niaga Siswa (PNS), UTHM
Time
: 5-6 p.m.
Attendance
: 1. Muhammad Ikram Rahaman CF150022 2. Muhammad Ridhwan bin Mohd Rasid DF15007 3. Muhamad Hanif bin Ros Kamar DF150087
Agenda Understanding the question
Topic discussed The ways of solving the
Pre-work/preparation All the members should
given.
questions. - Research in detail about
prepare the formulas needed
the Application of Vector Valued Function topic in the calculus -
module. Study the example of look-alike questions.
What the question is all about? Making of introduction.
All the members gave their suggestions and all the ideas combined. Introduction was done.
Prepared by,
Confirmed by,
19
solve this questions and at l 1 look-alike question.
…………………………….
……………………………
IKRAM RAHAMAN
RIDHWAN RASID
20
SECOND MINUTES OF MEETING
Project Name: Problem – Application of Vector Valued Function Date
: 13/5/2017 (SATURDAY)
Venue
: Tungku Tun Aminah Library, UTHM
Time
: 5-6 p.m.
Attendance
: 1. Muhammad Ikram Rahaman CF150022 2. Muhammad Ridhwan bin Mohd Rasid DF15007 3. Muhamad Hanif bin Ros Kamar DF150087
Agenda Solving the question
Topic discussed The arrangement of methods
Pre-work/preparation All the members should bring
given.
and formulas used in order to
the complete set of solution.
solve the questions. - All the formulas compiled and the members chose the -
needed formulas The methods arranged based on the research in the books.
All the members start solving the problem by own. Prepared by,
Confirmed by,
…………………………….
……………………………
IKRAM RAHAMAN
RIDHWAN RASID
21
MINUTES OF MEETING
Project Name: Problem – Application of Vector Valued Function Date
: 19/5/2017 (FRIDAY)
Venue
: Tungku Tun Aminah Library, UTHM
Time
: 5-7 p.m.
Attendance
:
1. Muhammad Ikram Rahaman CF150022 2. Muhammad Ridhwan bin Mohd Rasid DF15007 3. Muhamad Hanif bin Ros Kamar DF150087
Agenda Finalize the solution and
Topic discussed The accurate solution and ways
Pre-work/preparation Ridhwan was assigned to p
presentation slide and
of making report and
the report and bring it on th
report making.
presentation slide - All the solution gathered
day of presentation.
and best solution and -
accurate answer selected. Fila table and other contents needed for the
-
report prepared. Presentation slide was prepared together.
Prepared by,
Confirmed by,
…………………………….
……………………………
IKRAM RAHAMAN
RIDHWAN RASID
22
