Mr. Groden
Parametric
AP Calculus BC
Zach Yeger & Syed Masood
http://en.wikipedia.org/wiki/Parametric
_equation#mediaviewer/File:Butterfly_
trans01.svg
14
Contents
Rational ........................................................................................................................................... 3
Real World Applicability ................................................................................................................ 4
Math History ................................................................................................................................... 5
Basic analytical Example ................................................................................................................ 6
AP Level Multiple Choice Example ............................................................................................... 8
AP Level Conceptual Example ..................................................................................................... 10
AP Level Conceptual Example Solution ...................................................................................... 11
Example utilizing Graphs as part of Solution ............................................................................... 12
AP Level Free Response Question (FRQ) Example ..................................................................... 14
AP Level Free Response Question (FRQ) Example Solution ...................................................... 15
AP Level Free Response Question (FRQ) Scoring Guideline ...................................................... 17
Analatical Exercises ...................................................................................................................... 18
AP Level Multiple Choice Exercises ............................................................................................ 21
Analatical Exercises Solutions...................................................................................................... 25
AP Level Multiple Choice Exercises Solutions ............................................................................ 33
Work Cited .................................................................................................................................... 38
Rational
a) Zach and I chose parametric for our section of study because of our level of study in the
fields of physics. Classes such as AP Physics B and AP Physics C have shaped our
thoughts and attitudes towards a greater understanding of vectors in more than 1
dimensional space. Physics is the study of motions in 2 and/or 3 dimensions space and 1
dimension of time. Parametric then becomes the math for physics in how it relates to the
study of motion in relation of time in the form of system of equations.
b) Because of the time and effort taken to create this project the amount of effort will be
massive, thus considering that we are in an AP class, the effort can not, should not, and
will not be in vein. That effort put into this project will be towards ensuring a better
generation of tomorrow, considering that it the true purpose of every generation.
According to Darwin every generation acts in order to succeed and further the next thus
the effort put in will not be in vein. This will be done in the form of example and step by
step indication on specific question will yet giving a general summarization of the
concept as whole. By maintain a high level of expectance, yet creating a foundation from
the base, the progression will be natural yet intense, but due to the fact of the natural
progression, it will not seem tense.
c) The textbook, AP central, other websites, former BC students, and most importantly our
great intellect as mathematicians will be our tools of choice.
d) We enjoy how parametric can relate to physics. Physics involves real world application
which we find very interesting. The way kinematics is designed to function in order to
explain the motion observed in the real world known as reality, parametric becomes the
origins for its design. Parametric being the method of deriving the equations taken by
granted by physics students is actually a high level of calculus. That realization is why
parametric is so important, it was essentially defines movement in reality, in the form of
velocity, acceleration, and displacement, and applied rotation.
e) We will convey our enjoyment by teaching with enthusiasm and sincerity. By maintain a
high level of expectance, yet creating a foundation from the base, the progression will be
natural yet intense, but due to the fact of the natural progression, it will not seem tense.
f) We will come up with our own style textbook. Rather than creating a textbook for a
general audience, our textbook will be specifically designed for Pennsburg AP Calculus
students. We will also include more personal related concepts for Pennsburg, and will be
more complete in each individual section as we are only responsible for parametric.
Real World Applicability
The world is vectors. The applications of vectors define the existence of the world. The
understanding of the functionality of the world thus revolves around the understanding of the
functionality of vectors. Therefore parametric being the means of vector analysis is critical in
developing a comprehension of the physical reality we live in.
Physics being the means to which understanding of the physical world is reached based around
the usage of vectors thus inevitability based around the understanding of parametric. The
quintessential examples are the Kinematics formula being formed via calculus and their
derivation around parametric. But parametric just limited to classical physics, Quantum
Mechanics and Quantum Chromodynamics are based around particle position, velocity,
acceleration and the understanding of gluons and force carrier particles all based upon the
foundations of parametric and vector analysis.
Aside from the world of Quantum Theory, Parametric is a vital component in understanding the
behaviors of galaxies and stars. The rotation of plants, orbits and optics (I know it’s a fragment).
The movement of light and its characteristics as a wave or particle are all dependent on vector
analysis, for which parametric is a form of.
Furthermore, parametric is used in computer science to simulate vital physics, and run complex
programs involving multivariable analysis in more than 1 or 2 dimensional analysis.
As I have mentioned before the applications of vectors are limitless, fields such as chemistry,
biology, political science seem distant from parametric, but all involve the understanding of
vectors but instead of direction of space, the values indicate direction in accordance to their
respective fields, allowing parametric to hold value.
Parametric is the understanding of vectors, vectors is how the physical reality of the world
functions, therefore Parametric is how the world is understood. Vectors by extension are the
world; therefore Parametric is the definition of the physical world.
Math History
Parametric Equations are topic in the sub-branch of Calculus known as Vector Calculus. The
establishment of Calculus is credited to the legendary mathematician Isaac Newton.
Having been alive during the 17th Century and establishing Calculus, Isaac Newton created the
mean by which parametric would be established in how to understand the vectors that exist in the
physical world. Newton also created the foundation of Newton Mechanics, the understanding of
the physical world via vectors. Thus Newton only founded the means by which parametric would
exist but also its major purpose as a tool for vector analysis.
Newton foresaw that in order to understand the physical that vectors analysis would be critical as
numbers are numbers if exist or not hold no meaning if the numbers themselves hold no value.
However values with meaning, direction, imply the importance as they can be used to describe
the physical world. The birth of Parametric thus became the means in understanding the physical
world taking the form of Physics in the form on Newtonian Mechanics.
Having published works in the field of Math, Physics, Astronomy, and Economics, the
understanding of vectors was a vital component of how he analysis his findings. The birth of
parametric occurred in this form, simple vector analysis with roots in the foundation of Calculus.
In his book Philosophiae Naturalis Principia Mathematica, this translates to Mathematical
Principles of Natural Philosophy, Newton takes a closer look into the physical reality of this
world to realize that math is a part of nature that explained by vectors giving rise to the need to
understand vectors. With the study of change, Calculus, alongside with the new need to
comprehend vectors, parametric was born.
Isaac Newton contributed a lot to Humanity and its progression into modern society, however
Parametric Equations is one of his most valuable contributions.
Basic analytical Example
I)
Find the acceleration vector of a scuba diver given that 𝑣𝑥 = 5𝑡 3 + 7𝑡 and
𝑎(𝑡)𝑦 = 8𝑡 2 at 𝑡 = 1
𝑥 ′ (𝑡) = 5𝑡 3 + 7𝑡
𝑦 ′ (𝑡) = 16𝑡
𝑥 ′′ (𝑡) = 15𝑡 + 7
𝑥 ′′ (1) = 21
𝑦 ′′ (𝑡) = 16
𝑦 ′′ (1) = 16
= ⟨21,16⟩
The acceleration vector is the derivative of the velocity vector, and because the velocity vector is
broken down into its components, the acceleration is also broken down into its components as
each component indicates magnitude and direction. Thus first take each individual derivative of
each component and solve the derivate at the required time to find the acceleration vector for that
instance.
II)
If acceleration of the scuba diver is given by 𝑎(𝑡) = ⟨5𝑡 3 + 7𝑡, 8𝑡 2 ⟩ what is the
minimum acceleration of the diver vertically?
Maximum accerlation 𝑦-direction
𝑑
𝑑𝑡
𝑑
𝑑𝑡
2
𝑎(𝑡)𝑦 = 8𝑡 = 0
𝑎(𝑡)𝑦 = 8𝑡 2 = 16𝑡 = 0
𝑡=0
𝑎′′(𝑡)𝑦 = 16
𝑎′′(0)𝑦 = 16
𝑎′′(𝑡)𝑦 is postive 𝑡 = 0 is a relative minimum
To find the minimum you first use
the 1st derivative test, thus find the
critical point in the 𝑦- direction.
Thus by using the 2nd derivative
test you can verify that the critical
point is a minimum. However since
this is an absolute minimum you
must test endpoints as well
III)
8𝑡
Given that acceleration of the diver is 𝑎(𝑡) = 5𝑡 2 +7 what is 𝑣(𝑡)𝑦 if the velocity function
5
in the 𝑥-direction of the diver is given by 𝑣(𝑡)𝑥 = 4 𝑡 4 − 4 from 𝑡 = 0 𝑡𝑜 𝑡 = 1
8𝑡
𝑑𝑦 𝑣 ′ (𝑡)𝑦
𝑎(𝑡) = 2
=
=
5𝑡 + 7 𝑑𝑥 𝑣 ′ (𝑡)𝑥
𝑎(𝑡) =
8𝑡
5𝑡 2 +7
=
𝑣 ′ (𝑡)𝑦
𝟎
5𝑡 2 + 7
𝟎
𝒗(𝑡)𝑦 = −𝟏. 𝟓𝟔𝟖
.
𝑥
′
5
8𝑡 (4 𝑡 4 − 4)
∫ 𝑣′ (𝑡)𝑦 𝒅𝒕 = ∫
𝑣′ (𝑡)𝑥
𝑣 ′ (𝑡)𝑦
4
𝟏 8𝑡
𝑣′ (𝑡)𝑦
set equation equal: 𝑣′ (𝑡) = 𝑎(𝑡).
5 4
𝑡 −4
4
5
𝟏
𝑑𝑦
velocity, thus 𝑑𝑥 =
Given 𝑣(𝑡)𝑥 , you can find 𝑣′(𝑡)𝑥 ,
and since you already know 𝑎(𝑡),
[5𝑡 2 + 7]𝑣 ′ (𝑡)𝑦 = 8𝑡 ( 𝑡 4 − 4)
𝑣 ′ (𝑡)𝑦 =
The acceleration is the derivative of
5
(4 𝑡4 − 4)
5𝑡2 + 7
𝒅𝒕 = −𝟏. 𝟓𝟔𝟖
Then just solve for 𝑣 (𝑡)𝑦 . This
gives you the acceleration in the 𝑦 direction. Now by integrating (the
reverse of differentiation)𝑡 =
0 𝑡𝑜 𝑡 = 1 you find the velocity of
the diver over that given interval.
AP Level Multiple Choice Example
Find the slope of the parametric equations 𝑥(𝑡) = 3𝑡 2 + 𝑡 and 𝑦(𝑡) = 10𝑡 at 𝑡 = 6
I)
a)
37
10
b) 6
10
c) 37
d)
10
7
e) 1
I)
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑡
𝑑𝑦
𝑑𝑡
The answer is not a) because that
𝑑𝑥
𝑑𝑥
= 𝑥 ′ (𝑡) = 6𝑡 + 1
𝑑𝑡
𝑑𝑦
𝑑𝑦 𝑑𝑡
10
=
=
𝑑𝑥 𝑑𝑥 6𝑡 + 1
𝑑𝑡
𝑑𝑦
= 𝑦 ′ (𝑡) = 10
𝑑𝑡
represents 𝑑𝑦 ≠ 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒
The answer is not b) because that is
just the time which you are
required to find the slope
The answer is not d) because it was
probably a simple algebraic
mistake in which you did not
multiply the time initially by the
𝑑𝑥
𝑡=6
co-efficient of 6 in determining 𝑑𝑡
The answer is not e) because the
slope when calculated via the left,
𝑑𝑦
10
=
𝑑𝑥 6(6) + 1
𝑑𝑦 10
=
𝑑𝑥 37
The anwser is C
10
is not 1 but in fact c) 37
II)
The velocity of a deep sea diver is given by 𝑣(𝑡) = ⟨cos 𝑡 , 𝑒 𝑡 ⟩, which graph best
represents the position of the diver
a)
d)
b)
e)
c)
https://www.desmos.com/calculator
II)
𝑣(𝑡) = ⟨cos 𝑡 , 𝑒 𝑡 ⟩
𝑣(𝑡)𝑥 = cos 𝑡
𝑣(𝑡)𝑦 = 𝑒 𝑡
𝑥(𝑡)𝑥 = sin 𝑡 + 𝐶
𝑥(𝑡)𝑦 =
1
𝑥(𝑡) = ⟨sin 𝑡, 𝑒 𝑡 ⟩
𝑡
1
𝑥(𝑡)𝑦 = sin−1 𝑥(𝑡) 𝑒 sin
𝑥
1 𝑡
𝑒 +𝐶
𝑡
sin−1 𝑥(𝑡)𝑥 = 𝑡
−1 𝑥(𝑡)
𝑥
graph on calculator
The answer is D
AP Level Conceptual Example
𝑑
Given that the graphs below represent 𝑑𝑡 𝑎(𝑡), when 𝑡 = 0 𝑣(𝑡) = ⟨4, −9⟩ is the velocity of the
scuba diver increasing, decreasing, or constant, at an increasing, decreasing, or constant rate in
what direction horizontally and vertically when lim 𝑓(𝑡)? How is 𝑥(𝑡) being affect given that
𝑡→4
𝑡 = 0, 𝑥(𝑡) = ⟨0,0⟩ ?
𝑑
𝑑𝑡
𝑑
𝑑𝑡
𝑎(𝑡)𝑥
𝑎(𝑡)𝑦
https://www.desmos.com/calculator
AP Level Conceptual Example Solution
𝑡 = 4,
𝑎′ (𝑡)𝑥 = 3.152
𝑡 = 4,
𝑎′ (𝑡)𝑦 = −0.662
𝑑
Thus the acceleration in the 𝑦-direction is increasing at a decreasing rate, since 𝑑𝑡 𝑎(𝑡)𝑦 is
concave down and positive.
𝑑
Thus acceleration in the 𝑥- direction is increasing at an increasing rate, since 𝑑𝑡 𝑎(𝑡)𝑥 is concave
up and positive.
The 𝑣(𝑡)𝑥 is positive and the 𝑎(𝑡)𝑥 is positive, thus the 𝑣(𝑡)𝑥 is increasing
The 𝑣(𝑡)𝑦 is negative and 𝑎(𝑡)𝑦 is positive, thus magnitude of 𝑣(𝑡)𝑦 is decreasing as the 𝑎(𝑡)𝑦
opposed 𝑣(𝑡)𝑦
Thus the diver is moving horizontally to the right at an increasing positive velocity while also
moving vertically down with a decreasing velocity.
Since 𝑡 = 0, 𝑥(𝑡) = ⟨0,0⟩ and 𝑡 = 0 𝑣(𝑡) = ⟨4, −9⟩, the diver is increasing his displacement
towards the positive direction, as 𝑥(𝑡)𝑥 = 0 and 𝑣(𝑡)𝑥 is positive. Also the diver is increasing
his displacemnt towards the negative direction, as 𝑥(𝑡)𝑦 = 0 and 𝑣(𝑡)𝑦 is negative.
Thus the diver is swimming towads the right and diving further down.
Example utilizing Graphs as part of Solution
I)
The position of a deep sea diver when resurfacing is given by the following equations:
𝑥(𝑡) = sin 𝑡
𝑦(𝑡) =
4𝑡
3𝑡
2𝑡
− 3! + 2!
4!
-3𝜋 ≤ 𝑡 ≤ 3𝜋
a)
Using a graphing calculator, sketch the parametric curve
b)
Using your graphing calculator, determine at points of with a vertical tangent line
c)
Determine the Surface area of the curve when rotated around the 𝑦- axis from
0 ≤ 𝑡 ≤ 3𝜋
I)
Solution
a)
𝑥(𝑡) = sin 𝑡
𝑦(𝑡) =
4𝑡
3𝑡
2𝑡
− 3! + 2!
4!
-3𝜋 ≤ 𝑡 ≤ 3𝜋
https://www.desmos.com/
calculator
b)
𝑥(𝑡) = sin 𝑡
Vertical tangent lines occur when
𝑑
𝑑𝑡
𝑥(𝑡) = cos 𝑡 = 0
𝑑
𝑑𝑡
𝑥(𝑡) = 0
3𝜋 ≤ 𝑡 ≤ 3𝜋
−5𝜋 −3𝜋 −𝜋 𝜋 3𝜋 5𝜋
,
,
, ,
,
2
2
2 2 2 2
𝑡=
The values are 𝑡 can be verified at vertical tangents by comparing to the graph in part a
c)
Rotation around 𝑦- axis
𝑏
𝑆𝐴 = 2𝜋 ∫ 𝑥(𝑡)√(
𝑎
𝑑𝑥
= cos 𝑡
𝑑𝑡
𝑑𝑥 2
𝑑𝑦 2
) + ( ) 𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑦 1 1 1
= − +
𝑑𝑡 3! 2! 1
3𝜋
𝑆𝐴 = 2𝜋 ∫
0
1 1 1 2
2
√
sin 𝑡 (cos 𝑡) + ( − + ) 𝑑𝑡 = 10.888 𝑢𝑛𝑖𝑡𝑠 2
3! 2! 1
AP Level Free Response Question (FRQ) Example
CALCULUS BC
SECTION IIX
Time—15 minutes
Number of problems—1
1.
A scuba diver is the Pacific Ocean visiting the Great Barrier Reef of the coast of
Australia. The diver has an oxygen tank that can hold 15 Liters of a mixture of oxygen
𝑑𝑎
𝑡2
𝑡3
and helium. The rate in hours at which the mixture is consumed is
=𝑡− −
𝑑𝑡
2!
3!
, “𝑎” is the amount of air.
a)
Given that the diver’s vertical position can accounted for via equation 𝑦(𝑡) = 2𝑡 2 − 3,
what is the maximum depth the diver can reach before having to resurface?
b)
The diver’s velocity vector at 𝑡 = 2 is ⟨3,4⟩, if 𝑥 ′ (𝑡) the velocity of the diver in the 𝑥direction is 2𝑡 + 1 and 𝑥(2) = 8, then what is the 𝑥-coordinate of the diver at 𝑡 = 4?
c)
What is the total distance the diver has traveled in the time before entering and
resurfacing, in the time he has before running out of air?
AP Level Free Response Question (FRQ) Example Solution
a)
∫
𝑑𝑎
𝑑𝑡 = 𝑎
𝑑𝑡
∫𝑡 −
2
𝑡2 𝑡3
−
𝑑𝑡 = 𝑎
2! 3!
3
4
𝑡
𝑡
𝑡
−
−
+𝑐 =𝑎
2 18 24
𝑎 = 15,
𝑡=0
02 03 04
−
−
+ 𝑐 = 15
2 18 24
𝐶 = 15
𝑡2 𝑡3 𝑡4
−
−
+ 15 = 𝑎
2 18 24
𝑎=0
𝑡2 𝑡3 𝑡4
−
−
+ 15 = 0
2 18 24
𝑡 = −5.570 ℎ𝑜𝑢𝑟𝑠,
4.697 ℎ𝑜𝑢𝑟𝑠
𝑡 = 4.697 ℎ𝑜𝑢𝑟𝑠
𝑡
= 𝑡𝑖𝑚𝑒 𝑓𝑜𝑟 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑑𝑒𝑝𝑡ℎ
2
𝑡 = 2.348 ℎ𝑜𝑢𝑟𝑠
𝑦(𝑡) = 2𝑡 2 − 3
𝑦(2.348) = 2(2.348)2 − 3
𝑦(2.348) = 8.029 𝑚𝑒𝑡𝑒𝑟𝑠 𝑑𝑒𝑒𝑝
The second fundamental theorem of
Calculus says the integral of the
derivate is the function itself,
𝑑𝑎
therefore ∫ 𝑑𝑡 𝑑𝑡 = 𝑎. By using the
initial condition given the equation
for “𝑎” can be determined. Now by
finding the critical points you can
determine relative maximums and
minimums, but since you cannot
have negative time, −5.570 ℎ𝑜𝑢𝑟𝑠 is
not a feasible answer; therefore
4.697 ℎ𝑜𝑢𝑟𝑠 is the maximum time
the diver has. Now because it’s only
on way down, the time is halved.
Then by using the equation for
𝑦(𝑡)given you are able to determine
the height.
b)
c)
∫ 𝑥 ′ (𝑡)𝑑𝑡 = 𝑥(𝑡)
𝑏
𝑑𝑥 2
𝑑𝑦 2
𝐴𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ = ∫ √( ) + ( ) 𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑎
∫ 2𝑡 + 1𝑑𝑡 = 𝑡 2 + 𝑡 + 𝑐 = 𝑥(𝑡)
𝑡 2 + 𝑡 + 𝑐 = 𝑥(𝑡)
𝑡 = 2,
𝑥(𝑡) = 8
22 + 2 + 𝑐 = 8
𝑐=2
2
4 + 4 + 2 = 𝑥(4) = 22 𝑚𝑒𝑡𝑒𝑟𝑠
𝑦(𝑡) = 2𝑡 2 − 3
𝑑𝑦
= 4𝑡,
𝑑𝑡
𝑑𝑥
= 2𝑡 + 1
𝑑𝑡
𝑡 = 0 ℎ𝑜𝑢𝑟𝑠 𝑡𝑜 𝑡 = 4.697 ℎ𝑜𝑢𝑟𝑠
𝐴𝑟𝑐 𝐿𝑒𝑛𝑔𝑡ℎ
4.697
=∫
√(2𝑡 + 1)2 + (4𝑡)2 𝑑𝑡
0
𝐴𝑟𝑐 𝐿𝑒𝑛𝑔𝑡ℎ = 𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝐴𝑟𝑐 𝐿𝑒𝑛𝑔𝑡ℎ = 739.649 𝑚𝑒𝑡𝑒𝑟𝑠
Velocity is the derivative of the
position thus by using the Second
Fundamental Theorem of Calculus
you are able to find the position of
the diver in the 𝑥 − direction
The total Distance would be the
Arc Length, because the path of the
Diver is curved. Thus by using the
Arc Length formula, are will be
able find the total distance. The
Limits will be the total amount of
time available using the time found
in part a), 𝑡 = 0 𝑡𝑜 𝑡 =
4.697 ℎ𝑜𝑢𝑟𝑠. Now because
velocity is the derivative of
position in respect to time
𝑑𝑦
𝑑𝑡
𝑎𝑛𝑑
𝑑𝑥
𝑑𝑡
are the velocity of the
diver in each respective direction.
Then by using the equation the total
distance can be determined.
AP Level Free Response Question (FRQ) Scoring Guideline
Part A – 3 Points
𝑑𝑎
1 Point for ∫ 𝑑𝑡 𝑑𝑡 = 𝑎 – Knowing and Using the 2nd Fundamental Theorem of Calculus
1 Point for 𝑡 = 2.348 ℎ𝑜𝑢𝑟𝑠 – Calculating and using the correct time
1 Point for 𝑦(2.348) = 8.029 𝑚𝑒𝑡𝑒𝑟𝑠 𝑑𝑒𝑒𝑝 – Calculating the answer correctly with Units
Part B – 3 Points
1 Point for ∫ 𝑥 ′ (𝑡)𝑑𝑡 = 𝑥(𝑡) – Knowing and Using the 2nd Fundamental Theorem of Calculus
1 Point for 22 + 2 + 𝑐 = 8 𝑐 = 2 – Calculating and Using Initial Conditions
1 Point for 42 + 4 + 2 = 𝑥(4) = 22 𝑚𝑒𝑡𝑒𝑟𝑠 – Determining the Answer with Units
Part C – 3 Points
1 Point for
𝑑𝑦
𝑑𝑡
= 4𝑡,
𝑑𝑥
𝑑𝑡
= 2𝑡 + 1 – Finding the both velocity components of the Diver
4.697
1 Point for 𝐴𝑟𝑐 𝐿𝑒𝑛𝑔𝑡ℎ = ∫0
√(2𝑡 + 1)2 + (4𝑡)2 𝑑𝑡 -- Setting and Evaluating the Correct
Integral
1 Point for 𝐴𝑟𝑐 𝐿𝑒𝑛𝑔𝑡ℎ = 739.649 𝑚𝑒𝑡𝑒𝑟𝑠 – Calculating the Correct Answer with Units
Analatical Exercises
All exercises are non-graphing calculator unless states otherwise. Also all are
in order of difficulty. All directions for each individual exercise will be given in
each individual exercise.
𝑑𝑦
I)
Find 𝑑𝑥 given the following set of equations: 𝑥 = 𝑡 2 , 𝑦 = 𝑡 3 + 1
II)
Find
III)
Find the Velocity Vector when 𝑡 = 4 given the set of equations:
𝑑𝑦
𝑑𝑥
given the following set of equations: 𝑥 = cos 𝜃, 𝑦 = (sin 3𝜃)2
𝑥 = 3𝑡 2 , 𝑦 = 2𝑡 + 3
IV)
𝑑𝑦
Using your graphing calculator, find 𝑑𝑥 when 𝑡 = 0 given the set of equations:
𝑥(𝑡) = 2 sin 𝑡 , 𝑦(𝑡) = 3 sin 3𝑡
V)
Using your graphing calculator, determine the interval for which the velocity of a scuba
diver is negative if his position is defined by:
𝑥(𝑡) = sin 𝑡 , 𝑦(𝑡) = cos 𝑡
VI)
Using your graphing calculator, determine the length of the curve from 𝑡 = 0 to 𝑡 = 3 of
the following set of equations:
𝑥(𝑡) = 3𝑡 − 𝑡 2 , 𝑦(𝑡) = 𝑡 + 4
VII)
Find the point of vertical tangency for the equation 𝑥(𝑡) = 2𝑡 2 − 𝑡 and 𝑦(𝑡) = 5𝑡 3
IIX)
Find the time where the diver is not moving if his position is defined by the following set
1
of equations: 𝑥(𝑡) = 2 𝑡 2 , 𝑦(𝑡) = 3𝑡 2 − 5𝑡
IX)
Set up the integral that represents the area of the surface generated by revolving the curve
about the 𝑥 – axis given the following set of equations:
1
𝑥(𝑡) = 𝑡 2 , 𝑦(𝑡) = 2𝑡 + 3
2
X)
0≤𝑡≤1
Graph the following set of equations:
𝑥(𝑡) = 2𝑡 + 1 ,
XI)
given
𝑦(𝑡) = 𝑡 2
The path of a diver can modeled by the equations 𝑥(𝑡) = 25𝑡 and 𝑦(𝑡) = 15𝑡 − 16𝑡 2 ,
how far away is the diver at the highest point of his path?
You may use your graphing calculator
XII)
A diver is 75 meters underwater initially. The equations 𝑥 = 𝑡 − 1 and 𝑦 =
𝑡2
8
− 75
represent his position. The diver will get the bends if he reaches the surface in less than
15 minutes. Will the diver be okay?
1
XIII) A diver’s position at 𝑡 = 0 is ⟨0,0, ⟩. If the diver swims east at 2 𝑚𝑒𝑡𝑒𝑟/𝑠𝑒𝑐𝑜𝑛𝑑𝑠 and
1
increases his speed 2 𝑚𝑒𝑡𝑒𝑟/𝑠𝑒𝑐𝑜𝑛𝑑𝑠 every 𝑠𝑒𝑐𝑜𝑛𝑑, what is his position at 𝑡 = 4 ?
XIV) Consider a flying fish that launched out of the water at 45° and landed 10 𝑚𝑒𝑡𝑒𝑟𝑠 away
from where it took off. How fast was the fish swimming?
AP Level Multiple Choice Exercises
I)
The acceleration of a scuba diver is given by the following graph. Which of the following
is the best set of equations defining the movement of the diver
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calculator
1
a)
𝑥 ′ (𝑡) = 2 𝑡 2
𝑦 ′ (𝑡) = cos 𝑡
b)
𝑥 ′′ (𝑡) = 𝑡 2
𝑦(𝑡) = sin 𝑡
c)
𝑥(𝑡) = 𝑡 3
𝑦(𝑡) = tan 𝑡
d)
𝑥(𝑡) = 𝑡 2
𝑦(𝑡) = −𝑡
e)
𝑥 ′′ (𝑡) = 2 𝑡 2
1
𝑦 ′′ (𝑡) = 𝑡
2
II)
The position of the Diver is given ⟨4𝑡 3 , 𝑒 𝑡 ⟩ (𝑖𝑛 𝑚𝑒𝑡𝑒𝑟𝑠) when 𝑡 = 0 𝑠𝑒𝑐𝑜𝑛𝑑𝑠, what is
the speed of the diver when 𝑡 = 4 𝑠𝑒𝑐𝑜𝑛𝑑𝑠?
a)
‖𝑣‖ = 0
b)
‖𝑣‖ = ⟨12𝑡, 2𝑡𝑒 𝑡 ⟩
c)
‖𝑣‖ = √48 + 8𝑒 16
d)
‖𝑣‖ = √(48)2 + (8𝑒 16 )2
e)
‖𝑣‖ = −24
III)
What is the Total Distance a diver travels from 𝑡 = 0 𝑡𝑜 𝑡 = 4 given that his position can
be defined by the following set of equations:
𝑚𝑒𝑡𝑒𝑟𝑠
𝑠𝑒𝑐𝑜𝑛𝑑
𝑚𝑒𝑡𝑒𝑟𝑠
2
𝑥(𝑡) = 4𝑡 3
𝑠𝑒𝑐𝑜𝑛𝑑
𝑚𝑒𝑡𝑒𝑟𝑠
𝑠𝑒𝑐𝑜𝑛𝑑
𝑚𝑒𝑡𝑒𝑟𝑠
𝑠𝑒𝑐𝑜𝑛𝑑
𝑚𝑒𝑡𝑒𝑟𝑠
𝑠𝑒𝑐𝑜𝑛𝑑
𝑦(𝑡0 = 𝑒 𝑡
2
2
a)
4
2
∫0 √(4𝑡 3 )2 + (𝑒 𝑡 ) 𝑑𝑡 = 𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
b)
4
2
∫0 (4𝑡 3 ) (√(4𝑡 3 )2 + (𝑒 𝑡 ) ) 𝑑𝑡 = 𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
c)
∫0 (12𝑡 2 ) (√(12𝑡 2 )2 + (2𝑡𝑒 𝑡 ) ) 𝑑𝑡 = 𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
d)
𝑒
∫0 √(12𝑡 2 ) + (2𝑡𝑒 𝑡 )𝑑𝑡 = 𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
e)
4
2
∫0 √(12𝑡 2 )2 + (2𝑡𝑒 𝑡 ) 𝑑𝑡 = 𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
2
4
2
2
4
2
IV)
a)
Given the velocity of the diver 𝑣(𝑡) = ⟨− sin 𝑡 , 𝑡 2 ⟩, which graph best represents the
position of the diver?
d)
b)
e)
c)
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V)
What is the position of a diver given the acceleration at 𝑡 = 0, 𝑎(𝑡) = ⟨9𝑡, − cos 𝑡⟩, when
𝑡 = 4?
When
𝑡 = 0,
𝑣(𝑡) = ⟨0,0⟩,
𝑥(𝑡) = 0,
𝑦(𝑡) = 0
a)
𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 = ⟨0, cos 4 − 1⟩
b)
𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 = ⟨96, cos 4 − 1⟩
c)
𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 = ⟨−96, cos 4 + 1⟩
d)
𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 = ⟨2 𝑡 3 , cos 𝑡⟩
e)
𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 = ⟨3𝑡, − sin 𝑡⟩
3
Analatical Exercises Solutions
I)
𝑥 = 𝑡2
𝑦 = 𝑡3 + 1
𝑥 ′ (𝑡) = 2𝑡
𝑦 ′ (𝑡) = 3𝑡 2
𝑑𝑦 𝑦 ′ (𝑡) 3𝑡 2
=
=
𝑑𝑥 𝑥 ′ (𝑡)
2𝑡
𝑑𝑦 3
= 𝑡
𝑑𝑥 2
II)
𝑥 = cos 𝜃
𝑦 = sin 3𝜃
𝑥 ′ = − sin 𝜃
𝑦 ′ = 3 cos 3𝜃
𝑑𝑦
𝑑𝑥
III)
=
3 cos 3𝜃
− sin 𝜃
𝑥 = 3𝑡 2
𝑦 = 2𝑡 + 3
𝑡=4
𝑥 ′ = 6𝑡
𝑦′ = 2
𝑡=4
𝑥 ′ (4) = 6(4) = 24
𝑦 ′ (4) = 2
𝑡=4
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = ⟨ 24 , 2 ⟩
IV)
𝐸𝑛𝑡𝑒𝑟 𝑃𝑎𝑟𝑎𝑚𝑒𝑡𝑟𝑖𝑐 𝑀𝑜𝑑𝑒 𝑜𝑛 𝐺𝑟𝑎𝑝ℎ𝑖𝑛𝑔 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑜𝑟
𝐺𝑟𝑎𝑝ℎ 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠:
𝑥(𝑡) = 2 sin 𝑡
𝑦(𝑡) = 3 sin 3𝑡
𝑆𝑒𝑙𝑒𝑐𝑡 𝐶𝑎𝑙𝑐 𝑎𝑛𝑑 𝑠𝑒𝑙𝑒𝑐𝑡
𝑑𝑦
= 4.500
𝑑𝑥
𝑑𝑦
𝑑𝑥
when 𝑡 = 0
V)
𝑥(𝑡) = sin 𝑡
𝑦(𝑡) = cos 𝑡
𝑥 ′ (𝑡) = cos 𝑡
𝑦 ′ (𝑡) = − sin 𝑡
𝑑𝑦
− sin 𝑡
=
>0
𝑑𝑥
cos 𝑡
(0,
VI)
𝜋
3𝜋
) 𝑢 (𝜋,
)
2
2
𝑥(𝑡) = 3𝑡 − 𝑡 2
𝑦(𝑡) = 𝑡 + 4
𝑥 ′ (𝑡) = 3 − 2𝑡
𝑦 ′ (𝑡) = 1
2
2
𝐴𝑟𝑐 𝐿𝑒𝑛𝑔𝑡ℎ = ∫ √(𝑥 ′ (𝑡)) + (𝑦 ′ (𝑡)) 𝑑𝑡
𝐴𝑟𝑐 𝐿𝑒𝑛𝑔𝑡ℎ = 𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
3
𝐴𝑟𝑐 𝐿𝑒𝑛𝑔𝑡ℎ = ∫ √(3 − 2𝑡)2 + (1)2 𝑑𝑡
0
𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 5.653 𝑢𝑛𝑖𝑡𝑠
VII)
𝑥(𝑡) = 2𝑡 2 − 𝑡
𝑦(𝑡) = 5𝑡 3
𝑥 ′ (𝑡) = 4𝑡 − 1
𝑦 ′ (𝑡) = 15𝑡 2
𝑥 ′ (𝑡) = 0
4𝑡 − 1 = 0
𝑡=
1
4
IIX)
𝑥(𝑡) = 15
𝑦(𝑡) = 3𝑡 2 − 5𝑡
𝑥 ′ (𝑡) = 0
𝑦 ′ (𝑡) = 6𝑡 − 5
𝑥 ′ (𝑡) = 0
𝑦 ′ (𝑡) = 0
6𝑡 − 5 = 0
𝑡=
5
6
, 𝑥 ′ (𝑡) = 0 , 𝑦 ′ (𝑡) = 0
1
IX)
𝑥(𝑡) = 𝑡 2
𝑦(𝑡) = 2𝑡 + 3
𝑥 ′ (𝑡) = 𝑡
𝑦 ′ (𝑡) = 2
2
0≤𝑡≤1
𝑑𝑥 2
𝑑𝑦 2
𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 𝑟𝑒𝑣𝑜𝑙𝑣𝑒𝑑 𝑎𝑟𝑜𝑢𝑛𝑑 𝑥 − 𝑎𝑥𝑖𝑠 = 2𝜋 ∫ 𝑦√( ) + ( ) 𝑑𝑡
𝑑𝑡
𝑑𝑡
1
𝑆𝐴𝑥−𝑎𝑥𝑖𝑠 = 2𝜋 ∫ (2𝑡 + 3)√(𝑡)2 + (2)2 𝑑𝑡
0
𝑦(𝑡) = 𝑡 2
𝑥(𝑡) = 2𝑡 + 1
X)
𝒕
𝒙(𝒕)
𝒚(𝒕)
0
1
0
1
3
1
2
5
4
3
7
9
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XI)
𝑥(𝑡) = 25𝑡
𝑦(𝑡) = 15𝑡 − 16𝑡 2
𝑥 ′ (𝑡) = 25
𝑦 ′ (𝑡) = 15 − 32𝑡
𝑑𝑦 𝑦 ′ (𝑡)
=
𝑑𝑥 𝑥 ′ (𝑡)
𝑑𝑦 15 − 32𝑡
=
=0
𝑑𝑥
25
𝑡=
15
32
15
15
𝑥 ( ) = 25 ∗
32
32
15
𝑥 ( ) = 11.719 𝑚𝑒𝑡𝑒𝑟𝑠
32
XII)
𝑥(𝑡) = 𝑡 − 1
𝑦(𝑡) =
𝑡2
8
− 75
𝑡 < 15 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 (𝑡ℎ𝑒 𝑑𝑖𝑣𝑒𝑟 𝑔𝑒𝑡𝑠 𝑡ℎ𝑒 𝑏𝑒𝑛𝑑𝑠)
Initally the diver is 75 meters underwater
𝑦(0) = −75 𝑚𝑒𝑡𝑒𝑟𝑠
𝑡2
𝑦(𝑡) = − 75
8
𝑡2
0 = − 75
8
(𝑤𝑜𝑢𝑙𝑑 𝑏𝑒 𝑤ℎ𝑒𝑛 𝑡ℎ𝑒 𝑑𝑖𝑣𝑒𝑟 ℎ𝑖𝑡 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒)
𝑡2
= 75
8
𝑡 2 = 600
𝑡 = 24.495 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
The diver will not get the bends.
XIII) position at 𝑡 = 0 is ⟨0,0, ⟩
𝑣𝑥 (𝑡) = −
1 𝑚𝑒𝑡𝑒𝑟𝑠
2 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
𝑎𝑥 (𝑡) = −
1 𝑚𝑒𝑡𝑒𝑟𝑠
2 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 2
𝑡 = 4 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
∫ 𝑎𝑥 (𝑡)𝑑𝑡 = 𝑣𝑥 (𝑡)
𝑣𝑥 (𝑡) = (𝑎𝑥 (𝑡))(𝑡) + 𝑣0 (𝑡)
∫ 𝑣𝑥 (𝑡) = 𝑥(𝑡)
1
𝑥(𝑡) = (𝑎𝑥 (𝑡))𝑡 2 + 𝑣0 (𝑡)(𝑡) + 𝑥0 (𝑡)
2
𝑥(4) =
1
1
1
(− ) 42 − (4) + 0
2
2
2
𝑥(4) = 6 𝑚𝑒𝑡𝑒𝑟𝑠
= 𝑓𝑖𝑠ℎ
XIV)
45° = 𝜃
𝑥 = 10 𝑚𝑒𝑡𝑒𝑟𝑠
𝑦 = ? ? 𝑚𝑒𝑡𝑒𝑟𝑠
𝑎𝑦 (𝑡) = 9.8
𝑚𝑒𝑡𝑒𝑟𝑠
𝑠𝑒𝑐𝑜𝑛𝑑 2
∫ 𝑎𝑦 (𝑡) 𝑑𝑡 = 𝑣𝑦 (𝑡)
45° = 𝜃
𝑥 = 10 𝑚𝑒𝑡𝑒𝑟𝑠
𝑣𝑦 (𝑡) = 𝑎𝑦 (𝑡)(𝑡) + 𝑣𝑦 0 (𝑡)
∫ 𝑣𝑦 (𝑡) 𝑑𝑡 = 𝑦(𝑡)
𝑦(𝑡) =
1
𝑎𝑦 (𝑡)(𝑡)2 + 𝑣𝑦 0 (𝑡)(𝑡) + 𝑦0 (𝑡)
2
𝑦0 (𝑡) = 0 𝑚𝑒𝑡𝑒𝑟𝑠
tan 𝜃 =
𝑦
𝑥
𝑥 = 10 𝑚𝑒𝑡𝑒𝑟𝑠
𝑦 = 𝑥 tan 𝜃
𝑎𝑦 (𝑡) = 0
𝑚𝑒𝑡𝑒𝑟𝑠
𝑠𝑒𝑐𝑜𝑛𝑑 2
∫ 𝑎𝑥 (𝑡) 𝑑𝑡 = 𝑣𝑥 (𝑡)
𝑣𝑥 (𝑡) = 𝑎𝑥 (𝑡)(𝑡) + 𝑣𝑥 0 (𝑡)
𝑎𝑥 (𝑡) = 0
𝑚𝑒𝑡𝑒𝑟𝑠
𝑠𝑒𝑐𝑜𝑛𝑑 2
𝑣𝑥 (𝑡) = 𝑣𝑥 0 (𝑡)(𝑡)
∫ 𝑣𝑥 (𝑡) 𝑑𝑡 = 𝑥(𝑡)
𝑦 = 10 tan 45°
𝑦 = 10 𝑚𝑒𝑡𝑒𝑟𝑠
1
10 = (9.8)(𝑡)2 + 𝑣𝑦 0 (𝑡)(𝑡)
2
2
10
10
10 = 4.9 (
) + 𝑣𝑦 0 (𝑡)(
)
𝑣𝑥 0 (𝑡)
𝑣𝑥 0 (𝑡)
tan 𝜃 =
10
2
𝑦′
𝑥′
𝑥(𝑡) = 𝑣𝑥 0 (𝑡)(𝑡) + 𝑥0 (𝑡)
𝑥0 (𝑡) = 0 𝑚𝑒𝑡𝑒𝑟𝑠
𝑥(𝑡) = 𝑣𝑥 0 (𝑡)(𝑡)
𝑥(𝑡) = 10 𝑚𝑒𝑡𝑒𝑟𝑠
10
=𝑡
𝑣𝑥 0 (𝑡)
2
10
10 = 4.9 (
) + (10tan 45)
𝑣𝑥 0 (𝑡)
𝑣𝑥 0 (𝑡) =
AP Level Multiple Choice Exercises Solutions
I)
The graph on the interval from ( 0 , ∞ ) the 𝑥- value is increasing. The 𝑦 – value of the
graph are symmetrically distributed above and below the 𝑥- axis, thus eliminating choice
“d” and “e” as “d” would be increasing only, and “e” would be decreasing only at time
passed.
Because the graph represents acceleration, the 2nd derivative of 𝑦(𝑡) = tan 𝑡 is needed;
which is 𝑦 ′′ (𝑡) = 2 sec 2 𝑡 tan 𝑡 . is 𝑦 ′′ (𝑡) = 2 sec 2 𝑡 tan 𝑡 does not give systematic
𝑦 ′′ (𝑡) values above and below the 𝑦 -axis. Therefore choice “c” is not the best choice,
leaving “a” and “b.”
Also, due to the fact “a” is defined in the linear in the 𝑥-direction in relation to 𝑡 will “b”
is defined parabolic when both are converted to acceleration by taking different levels of
derivatives; the answer cannot be “a” because both have the same defining function of
acceleration in the 𝑦 - direction but because when standardized, the ±√𝑥 = 𝑡 of “b”
allows for the symmetry above and below the 𝑥-axis, something a linear function cannot
do.
Also graphing all possible solutions would have also yielded the answer:
Choice “a”
Choice “b”
Choice “c”
Choice “d” and Choice “e” are not
included as the graphs are straight lines
due to both components being either
constant, linear, or zero, thus not able to
show graphically due to limitations with
graphing program
https://www.desmos.com/calculator
II)
2
Position of the Diver ⟨4𝑡 3 , 𝑒 𝑡 ⟩ (𝑖𝑛 𝑚𝑒𝑡𝑒𝑟𝑠) when 𝑡 = 0 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
2
𝑥(𝑡) = 4𝑡 3
𝑦(𝑡) = 𝑒 𝑡
𝑥 ′ (𝑡) = 12𝑡
𝑦 ′ (𝑡) = 2𝑡𝑒 𝑡
𝑥 ′ (4) = 48
𝑚𝑒𝑡𝑒𝑟𝑠
𝑠𝑒𝑐𝑜𝑛𝑑
2
𝑦 ′ (4) = 8𝑒 16
𝑚𝑒𝑡𝑒𝑟𝑠
𝑠𝑒𝑐𝑜𝑛𝑑
‖𝑣‖ = √(𝑥 ′ (𝑡))2 + (𝑦 ′ (𝑡))2
‖𝑣‖ = √(48)2 + (8𝑒 16 )2
𝑚𝑒𝑡𝑒𝑟𝑠
𝑠𝑒𝑐𝑜𝑛𝑑
𝑇ℎ𝑒 𝑎𝑛𝑠𝑤𝑒𝑟 𝑖𝑠 𝑐ℎ𝑜𝑖𝑐𝑒 "𝑑"
Choice “e” cannot be correct because speed is only magnitude, the negative indicates direction
thus the option given is a velocity. Therefore, by the definition of what is the question is requiring
choice “e” is not a viable option.
Furthermore Choice “a” is incorrect because by the position is not defined by linear functions,
therefore the velocity is not zero, as it varies with time and has magnitude. Therefore choice “a”
indicating that the speed is zero indicates velocity is zero, which it is not; thus Choice “a” is
incorrect.
Choice “b” is incorrect as it does not give speed by the general terms for velocity is each
respective direction. To gain the value for speed at 𝑡 = 4, each individual velocity function must
be evaluated for 𝑡 = 4 then the use of the Pythagorean Theorem.
Choice “c” is incorrect as the Pythagorean Theorem was used incorrectly, as both components are
squared and the square root is taken of the addition of squares. Choice “c” did not square each
component thus option results an incorrect outcome. Therefore, Choice “c” is incorrect.
Therefore only leaving Choice “d” to be the correct answer.
III)
2
𝑥(𝑡) = 4𝑡 3
𝑦(𝑡) = 𝑒 𝑡
𝑥 ′ (𝑡) = 12𝑡
𝑦 ′ (𝑡) = 2𝑡𝑒 𝑡
𝑎𝑡 𝑡 = 0
𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑤ℎ𝑒𝑛 𝑡 = 4
2
𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝐴𝑟𝑐 𝐿𝑒𝑛𝑔𝑡ℎ
(the path is curved thus the addition length due to the and extra variable must taken in account)
𝑏
𝐴𝑟𝑐 𝐿𝑒𝑛𝑔𝑡ℎ = ∫ √(
𝑎
𝑑𝑥 2
𝑑𝑦 2
) + ( ) 𝑑𝑡
𝑑𝑡
𝑑𝑡
4
2
2
𝐴𝑟𝑐 𝐿𝑒𝑛𝑔𝑡ℎ = ∫ √(12𝑡)2 + (2𝑡𝑒 𝑡 ) 𝑑𝑡
0
𝑇ℎ𝑒 𝑎𝑛𝑠𝑤𝑒𝑟 𝑖𝑓 𝑐ℎ𝑜𝑖𝑐𝑒 "𝑒"
Choice “a” is incorrect because it uses position function instead of the velocity functions, thus
obtain an incorrect value, due the velocity being the derivative of position, the 2nd Fundamental
Theorem of Calculus is used to integrate back to distance.
Choice “b” is incorrect because it adds an additional 𝑥(𝑡) into the integral, thus changing the Arc
Length formula to the integral of Surface Area around the 𝑦-axis however without being
multiplied by 2𝜋.
Choice “c” is incorrect as it adds and additional 𝑥 ′ (𝑡) into the integral, thus changing the Arc
Length formula to an incorrect integral.
Choice “d” is incorrect because the velocities with the square root are not squared. Choice “d” is
incorrect as the Pythagorean Theorem was used incorrectly, as both components are squared and
the square root is taken of the addition of squares. Choice “d” did not square each component thus
option results an incorrect outcome. Therefore, Choice “d” is incorrect.
Therefore only leaving Choice “e” to be the correct answer.
IV)
𝑣(𝑡) = ⟨− sin 𝑡 , 𝑡 2 ⟩
∫ 𝑣𝑥 (𝑡)𝑑𝑡 = 𝑥(𝑡)
∫ − sin 𝑡 𝑑𝑡 = 𝑥(𝑡)
𝑥(𝑡) = cos 𝑡 + 𝐶
∫ 𝑣𝑦 (𝑡)𝑑𝑡 = 𝑦(𝑡)
∫ 𝑡 2 𝑑𝑡 = 𝑦(𝑡)
𝑦(𝑡) =
1 3
𝑡 +𝐶
3
Because Initial Conditions are not known, and “C” represents a constant to distinct the particular
function from a family of functions, in this case “C” can be dropped, as the parent function
would be suitable for determining the best graph.
𝑥(𝑡) = cos 𝑡
1
𝑦(𝑡) = 𝑡 3
3
cos−1 𝑥(𝑡) = 𝑡
1
𝑦(𝑡) = 𝑡 3
3
𝑦=
1
(cos −1 𝑥)3
3
By graphing this function or by making a table the resultant graph is Choice “a”
Choice “e” is an incorrect answer because it is oscillating in the 𝑦-direction, the vertical
component of the velocity is parabolic, thus the position via integration should be cubic.
Therefore the oscillation is invalid, thus resulting in the choice being incorrect.
Choice “c” is incorrect because the graph shows an even function (it’s even because it rises on
both ends of the graph); 𝑦(𝑡) the position function is cubic, which is odd. Therefore it is
incorrect because the 𝑦 component is incorrect.
Choice “b” is incorrect because it intersects the 𝑥-axis twice, indicating it has two zeros,
therefore requiring another term in its 𝑦 component. Because the parent function only has the
one and the constant would shift the graph vertically but not add an extra intersection, choice “b”
is incorrect because its 𝑦 component is incorrect.
Choice “d” is incorrect because while it fits all conditions met so far, the ends of the curve are
vertical indicating vertical asymptotes. Because vertical asymptotes are not in the domain either
component of the velocity, there cannot be vertical asymptotes in the domain of the position;
therefore excluding choice “d” from being a correct answer.
V)
𝑡=0
𝑎(𝑡) = ⟨ 9𝑡 , − cos 𝑡 ⟩ 𝑣(𝑡) = ⟨ 0, 0⟩
𝑥(0) = 0
𝑦(0) = 0
𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑤ℎ𝑒𝑛 𝑡 = 4
𝑏
𝑏
∫ 𝑎𝑥 (𝑡) 𝑑𝑡 = 𝑣𝑥 (𝑡)
∫ 𝑎𝑦 (𝑡) 𝑑𝑡 = 𝑣𝑦 (𝑡)
𝑎
𝑎
∫ 9𝑡 𝑑𝑡 = 𝑣𝑥 (𝑡)
∫ − cos 𝑡 𝑑𝑡 = 𝑣𝑦 (𝑡)
9 2
𝑡 + 𝐶 = 𝑣𝑥 (𝑡)
2
− sin 𝑡 + 𝐶 = 𝑣𝑦 (𝑡)
9 2
𝑡 = 𝑣𝑥 (𝑡)
2
− sin 𝑡 = 𝑣𝑦 (𝑡)
𝑏
𝑏
∫ 𝑣𝑥 (𝑡) 𝑑𝑡 = 𝑥(𝑡)
∫ 𝑣𝑦 (𝑡) 𝑑𝑡 = 𝑦(𝑡)
𝑎
4
∫
0
𝑎
9 2
𝑡 𝑑𝑡 = 𝑥(𝑡)
2
96 = 𝑥(𝑡)
4
∫ − sin 𝑡 𝑑𝑡 = 𝑦(𝑡)
0
cos 4 − 1 = 𝑦(𝑡)
𝑃𝑜𝑠𝑡𝑖𝑜𝑛 = ⟨ 96 , cos 4 − 1 ⟩
𝑇ℎ𝑒 𝑎𝑛𝑠𝑤𝑒𝑟 𝑖𝑠 𝐶ℎ𝑜𝑖𝑐𝑒 "𝑏"
Choice “a” is incorrect because taking the while the initial position in the 𝑥 direction is so; the
velocity and acceleration are not. Therefore, the position is constantly changing, and when 𝑡 = 4
the position is not zero, therefore choice “a” is incorrect.
Choice “c” is incorrect because the position in the 𝑥 direction is in magnitude 96 units, direction
is not correct. The acceleration of the diver was positive and the initial velocity was zero,
therefore the velocity at 𝑡 = 4 would be positive resulting in the position being positive as it was
to zero initially. Therefore, choice “c” is incorrect.
Choice “d” and “e” are incorrect because they are attempts for the general solutions for position.
The problem required a specific solution when 𝑡 = 4, therefore because they attempt to provide
the general solutions and not what the question asked for they are incorrect.
Work Cited
"Desmos Graphing Calculator." Desmos Graphing Calculator. Web. 12 June 2014.
Larson, Ron, Robert P. Hostetler, and Bruce H. Edwards. Calculus of a Single Variable. Boston:
Houghton Mifflin, 2006. Print.
Wikipedia. Wikimedia Foundation. Web. 12 June 2014.