By:
Elizabeth Carlson
Gina Liu
The Hundred Acre Inc. ©
1
Table of Contents
What Are Parametric Equations? ……………................................................. 3
Analytical Example …..…………………………………..…………………….. 3
Combining Parametric with Vectors (Analytical Example Part 2)………… 4
AP Level Free Response (Graphing Calculator) ……………….…………..... 5
Real World Application ……………………..……………………………….... 8
AP Level Multiple Choice …..…………………………………………………. 8
Conceptual Examples ………………………………………………………….. 9
Famous Person Who Worked with Parametric Equations …….………… 10
Vector/ Force Problem (just for fun) ………………………………….…….. 10
Works Cited…………………………………………………………………….13
2
What Are Parametric Equations?
Parametric equations describe curves in a different way than normal x-y functions; they relate
three variables to describe a curve, as opposed to two. Usually, they consist of two components,
where 𝑥(𝑡) describes the horizontal path of an object and 𝑦(𝑡) describes the vertical path the
object, both in terms of time 𝑡. Similarly, 𝑥 ′ (𝑡) describes the horizontal velocity and 𝑦 ′ (𝑡)
describes the vertical velocity.
Tip:
Parametric equations are not very different
from normal functions; you can think of 𝑥(𝑡)
as 𝑦1 , and (𝑡) 𝑎𝑠 𝑦2 . One equation does
not affect the other, but in this case, they
relate to describe a single curve with respect
to t.
Part One: Analytical Example
𝑑𝑦
1. Find 𝑑𝑥 for the following set of parametric equations:
𝑥(𝑡) = 𝑡 4 + 5 sin 𝑡
𝑦(𝑡) = √𝑡 +cos t
First, take the derivative of the two parametric equations separately:
𝑑𝑥
= 4𝑡 3 + 5 cos 𝑡
𝑑𝑡
𝑑𝑦
1
=
− sin 𝑡
𝑑𝑡
2 √𝑡
Great! So now we have the change in x and the change in y with respect to t. Wait a
second…that sounds familiar. Change in x and change and y relate to find slope! Isn’t that what
𝑑𝑦
𝑑𝑥
is? In general,
∆𝑦 = 𝑑𝑦
∆𝑥 = 𝑑𝑥
3
Since
∆𝑦
∆𝑥
=
𝑑𝑦 𝑑𝑦
,
𝑑𝑥 𝑑𝑥
=
𝑑𝑦/𝑑𝑡
𝑑𝑥/𝑑𝑡
When applied to this problem,
1
− sin 𝑡
𝑑𝑦
2√𝑡
= 3
𝑑𝑥
4𝑡 + 5 cos 𝑡
Tip: When solving this kind of problem, it may be easier to not
simplify your final answer for
𝑑𝑦
𝑑𝑥
further than the problem is
simplified in the example. If you were to evaluate the derivative of
the previous example specific t-value, it would be less time
consuming to just plug the specific t value into a less simplified
equation (rather than spending so much time simplifying).
Part Two:
Parametric
Combining
with Vectors
2. So how does taking the derivative of parametric equations compare to taking the
derivative of a vector-valued function? The two processes are surprisingly similar. To
illustrate, if we were to use the same equations from part one to describe the x and y
components of a vector, position would be described by the following vector:
< 𝑡4 + 5 sin 𝑡 , √𝑡 + cos 𝑡 >
Describes vertical position
Describes horizontal position
𝑑𝑥
When finding a velocity vector, take the derivative of each component separately (see part1); 𝑑𝑡
will be the velocity in the x direction,
𝑑𝑦
𝑑𝑡
will be the velocity in the y direction. Therefore, the
velocity vector would be the following:
1
< 4𝑡 3 + 5 cos 𝑡, 2
√𝑡
Describes horizontal motion
− sin 𝑡 >
Describes vertical motion
*A graphing calculator may be necessary to solve the next example problem*
4
3. Pooh is struggling to find his pot of honey and searches along a path; his path can be
described by the following set of equations when 0≤ 𝑡 ≤ 2.1:
𝑥(𝑡) = −𝑡 3 + 4𝑡
𝑦(𝑡) = 2𝑡 4 + 1
a) Piglet is helping Pooh in his honey finding endeavors, and is searching on a linear path.
Piglet’s path is tangent to Pooh’s at t= 2; find the equation of the line that describes
Piglet’s path.
First, find the x-y coordinate of Pooh’s path at t=2:
𝑥(2) = −(2)3 + 4(2) = 0
𝑦(2) = 2(2)4 + 1 = 33
𝑑𝑦
𝑑𝑥
Next, find the slope of the tangent line,𝑑𝑥 , starting with finding 𝑑𝑡 and
𝑑𝑦
𝑑𝑡
𝑑𝑥
= −3𝑡 2 + 4
𝑑𝑡
𝑑𝑦
= 8𝑡 3
𝑑𝑡
Next, divide
𝑑𝑦
𝑑𝑡
𝑑𝑥
𝑑𝑦
by 𝑑𝑡 to obtain𝑑𝑥 :
𝑑𝑦⁄
𝑑𝑦
𝑑𝑡
=
𝑑𝑥 𝑑𝑥⁄
𝑑𝑡
𝑑𝑦
8𝑡 3
=
𝑑𝑥 −3𝑡 2 + 4
Now, to find the slope of the tangent line at 2, substitute 2 in for t into the equation for
𝑑𝑦
𝑑𝑥
:
𝑑𝑦
8(2)3
|
=
𝑑𝑥 𝑡=2 −3(2)2 + 4
𝑑𝑦
64
|
=
= −8
𝑑𝑥 𝑡=2 −8
Lastly, write the equation for the tangent line using all of the information that has just been
calculated:
𝑦 − 33 = −8(𝑥 − 0)
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b) Find the time at which Pooh is the furthest to the right.
First, decide what the question is actually asking for. This question must relate only to x (t) since
left/right only describes horizontal position, as opposed to vertical position. If Pooh is the
furthest to the right, he is at his most positive horizontal position. Therefore, the point the
question is looking for is the absolute maximum of x (t) over the given time interval.
Since we are trying to find the absolute maximum of x (t), find the derivative of x (t):
𝑑𝑥
= −3𝑡 2 + 4
𝑑𝑡
Plug this equation into a graphing calculator to compute a graph. The window should be set so
only the time interval being questioned is showing. The graph should look like this (the x-values
in this equation are like the y-values of an x-y equation, t-values like the x-values of an x-y
equation):
6
4
2
0
-2
0
0.5
1
1.5
2
2.5
-4
-6
-8
-10
The points to be considered as possible values of an absolute maximum include points where
changes from positive to negative, and the endpoints 0 and 2.1.
𝑑𝑥
𝑑𝑡
𝑑𝑥
𝑑𝑡
changes from positive to
negative at t=1.1547 (which can be found by using your graphing calculator’s zero feature), so
that is a possibility. However, since
𝑑𝑥
𝑑𝑡
is positive at all points before 1.1547 and negative all
points after 1.1547, x (t) is increasing at all points before and decreasing at all points after
1.1547. Therefore, 1.1547 is the only viable option for and is the absolute maximum over the
interval.
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Tip: In order to obtain precise answers on free-response problems, take full advantage of
your calculator’s store feature. This feature allows you to work with un-rounded, more
precise answers from previous calculations. This guarantees that you will not get a
𝑡=
1.1547
wrong answer just because you rounded throughout the problem.
c) Find Pooh’s speed at t=1.25
𝑑𝑦
When finding speed, be careful!! |𝑑𝑥 | does not describe speed;
𝑑𝑦
𝑑𝑥
is the slope of the curve at an
instant. Well then, what is the speed? Speed is found differently with parametric equations (see
figure for more details). Speed can be described by the following equation:
Speed = √[𝑥′ (𝑡)]2 + [𝑦′ (𝑡)]2
Once you recognize that this is the right equation to use, you could take a few approaches to this
problem. You could do this problem by finding the derivative of each equation and plugging in
1.25, but using you graphing calculator to full potential would be more efficient in this case. Plug
x (t) and y (t) into 𝑌1 and 𝑌2 of your graphing calculator. Then calculate the derivative of both
equations at t= 1.25
If 𝑌1 = x (t), when calculating x’ (t), this is what your calculator screen should look like:
nDeriv (𝑌1 , x, 1.25)
-.687501
If 𝑌2 = y (t), when calculating y’ (t), this is what your calculator screen should look like:
nDeriv (𝑌2 , x, 1.25)
15.62501
Now that we have values for both x’ (t) and y’ (t), we can plug them into the equation and solve
for speed:
Speed = √(−.687501)2 + (15.62501)2 = 15.6401
Real World Application
7
Just think, would it be easier to describe a curve with two variables or three variables?
Parametric equations have way more applications to real world than normal functions because they
consist of three variables. In the real world, not all things move in a straight line; normal functions
sometimes cannot accurately describe the motions of the object we want to observe. With parametric
equations, one can easily describe the path, velocity, and other behaviors of any object. Parametric
equation not only tells you where the object is, it also tells you where the object was at any given time.
Parametric equations are often closely connected with vectors in physics. For example, scientists use
parametric equations and vectors to calculate the exact force needed for space shuttles to land; scientists
also use parametric equations to calculate the behaviors of different particles. It is also very useful in
finding the arc length of a curve. By find the arc length, scientists can trace out the length of the path of
the object.
Furthermore, parametric equations are around us every day, we just don’t realize it. Why do you
think Michael Jordan is so good at basketball? I will share the secret with you: he used calculus when he
was making those famous shots!! With parametric equations, Jordan can calculate the exact path of the
ball with relation to time, thus accurately throwing the ball into the basket almost every time.
Also, why do you think Tigger can accurately jump on Pooh every time? Don’t make the mistake
of thinking he is stupid; he knows calculus!
4. Tigger is on his way to Pooh’s house. The position of Tigger is defined by a parametric
curve where 𝑥(𝑡) =
1 3
𝑡
3
− 2𝑡 2 + 3𝑡 and 𝑦(𝑡) = 𝑡 3 − 3𝑡. What are all times when
Tigger is at rest?
A) 𝑡 = ±1
B) 𝑡 = ±1 , 3
C) 𝑡 = 0
D) 𝑡 = 1
E) 𝑡 = 3
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𝑥 ′ (𝑡) describes Tigger’s horizontal motion and 𝑦 ′ (𝑡) describes his vertical motion, so Tigger is at
rest when both 𝑥 ′ (𝑡) and 𝑦 ′ (𝑡) = 0.
Given that 𝑥(𝑡) =
1 3
𝑡
3
− 2𝑡 2 + 3𝑡 , so by taking the derivative, 𝑥 ′ (𝑡) = 𝑡 2 − 4𝑡 + 3. Set
𝑥 ′ (𝑡) = 0 and you can find that 𝑡 = 1, 3. Apply the same method, 𝑦 ′ (𝑡) = 3𝑡 2 + 3. When
𝑦 ′ (𝑡) = 0, 𝑡 = ±1. Since both 𝑥 ′ (𝑡) and 𝑦 ′ (𝑡) = 0 at 𝑡 = 1, the object is at rest at that point.
The correct answer is D 
5. Given that
𝑑𝑥
𝑑𝑡
= −2 when 𝑡 = −13 and that
𝑑𝑦
𝑑𝑡
𝑑𝑥
= 𝑡( 𝑑𝑡 + 3) for all values where 𝑡 ≤ 5.
What is the concavity of the curve at 𝑡 = −13
𝑑2 𝑦
𝑑𝑦
𝑑𝑥
The second derivative describes the concavity of a curve. 𝑑𝑥 2 = 𝑑𝑥 / 𝑑𝑡
𝑑2 𝑦
Thus, 𝑑𝑥 2 |
𝑡=−13
=
(−13)(−2+3)
−2
1
(−2)
=
−13
4
𝑑2 𝑦
Since 𝑑𝑥 2 is negative, the curve is concave down at 𝑡 = −13
𝑑𝑦
6. Given that 𝑑𝑥 = 0 at 𝑡 = 2. Is the particle is at rest at this given time?
Justify your answer
The particle CANNOT be at rest because
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
= 0 at 𝑡 = 2. Wait a second…..isn’t
the particle at rest if = 0 or undefined? Well, remember that parametric
equations have different guidelines than regular functions. For a particle to be at
𝑑𝑦
rest in parametric equations 𝑑𝑥 must be indeterminate at the given time. Because
𝑑𝑦 describes the vertical movements and 𝑑𝑥 describes the horizontal movements,
0
both have to equal 0 for a particle to be at rest (remember that is indeterminate).
Did you answer
correctly? You
really should
have since you
have a 50%
chance of
getting it right!
0
𝑑𝑦
In this problem, 𝑑𝑥 = 0, thus there must be some sort of horizontal movements at 𝑡 = 2.
So who has worked with Parametric Equations?
Helaman Ferguson, a recipient of a PhD in Mathematics, has worked extensively with
parametric equations…but not in the way you might expect. Ferguson has takes various
mathematical concepts and applies them to his work as an artist. Surprisingly, the two seemingly
9
dissimilar fields come together flawlessly, as illustrated by Ferguson’s sculptures; here is a
sample of his work:
How does math relate to this type of sculpting? When creating such a sculpture, Ferguson
uses parametric equations and vector fields to craft the tori and double tori shapes which are
associated with his work. Since multiple equations need to be related in order to develop such a
complicated shape, parametric equations are essential in order to ensure that the sculpture ends
up just right.
* This piece features information from outside sources (see works cited)
* A calculator is necessary to solve this problem*
7. Pooh and piglet are attempting to move Pooh’s heavy pot of honey. They both decide to
pull on two ropes, which are conveniently attached to the honey pot. Pooh applies a force
of 17 Newtons, 45 degrees above the horizontal axis. Piglet exerts 25 Newtons, 20
degrees below the horizontal axis. Find the vector that describes the total force being
used to pull the honey pot.
To solve the problem, start by drawing a diagram describing the situation:
𝐹1 = 17𝑁
𝑦
𝜃1 =45⁰ 𝑥1
Type equation
here.
10
𝜃2 =-20⁰ 𝑥2
𝑦2
30202020⁰
𝐹Type
=
25𝑁
2
equation here.
Now that we have a picture, we can solve for the components of each vector using trigonometry:
𝐹1
sin 45 =
𝐹2
𝑦1
17
sin −20 =
𝑦1 = 17 sin 45
𝑦2 = 25 sin −20
y1 = 12.02081528 N
cos 45 =
𝑦2
25
𝑦2 = −8.550503583 N
𝑥1
17
cos −20 =
𝑥1 = 17cos 45
𝑥2
25
𝑥2 = 25 cos −20
𝑥1 = 12.02081528 N
𝑥2 = 23.49231552 N
Now that we have the individual horizontal and vertical components, we must add the
components together to find the total vertical and horizontal forces acting on the honey pot:
𝑥𝑡𝑜𝑡𝑎𝑙 = 𝑥1 + 𝑥2
𝑥𝑡𝑜𝑡𝑎𝑙 = 12.02081528 N + 23.49231552 N
𝑥𝑡𝑜𝑡𝑎𝑙 = 35.5131308 N
𝑦𝑡𝑜𝑡𝑎𝑙 = y1 + y2
𝑦𝑡𝑜𝑡𝑎𝑙 = 12.02081528 N + −8.550503583 N
𝑦𝑡𝑜𝑡𝑎𝑙 = 3.470311697 N
Since we have our total horizontal and vertical components, it may be helpful to draw a diagram
with these forces:
𝐹𝑡𝑜𝑡𝑎𝑙 = ?
𝑦𝑡𝑜𝑡𝑎𝑙 =3.470311697 N
𝜃 =?
𝑥𝑡𝑜𝑡𝑎𝑙 =35.5131308 N
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Now we must use the Pythagorean Theorem to solve for the total force (which is evident from
the diagram):
𝐹𝑡𝑜𝑡𝑎𝑙 = √xtotal 2 + ytotal 2
𝐹𝑡𝑜𝑡𝑎𝑙 = √35.51313082 + 3.4703116972
𝐹𝑡𝑜𝑡𝑎𝑙 = 35.6823 N
To find the direction of the force, the use of right triangle trigonometry is necessary:
tan−1 θ =
tan−1 θ =
𝑦𝑡𝑜𝑡𝑎𝑙
𝑥𝑡𝑜𝑡𝑎𝑙
3.470311697
35.5131308
𝜃 = 5.5812° above the horizontal
Works Cited
"The Artist." Helaman Ferguson Sculpture:. N.p., n.d. Web. 17 May 2012.
<http://www.helasculpt.com/theartist/index.html>.
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Larson. "Mathematical Sculpture: Parametric Surfaces." College.cengage.com. N.p., n.d. Web. 17 May
2012.
<http://college.cengage.com/mathematics/larson/calculus_analytic/7e/instructors/downloads/lab1
9der.pdf>.
Larson, Ron, Robert P. Hostetler, and Bruce H. Edwards. Calculus with Analytic Geometry. 8th ed.
Boston: Houghton Mifflin, 2006. Print.
Exercises
13
𝑑𝑦
1. Evaluate 𝑑𝑥 for the following at t=2:
𝑥(𝑡) = 3𝑡 2
𝑦(𝑡) = 6𝑡 5
𝑑𝑦
2. Evaluate 𝑑𝑥 for the following at t=0 :
𝑥(𝑡) = sin 2𝑡 − 5𝑡 2
𝑦(𝑡) =
1
𝑡 − cos 𝑡
5
𝜋
3. Find the velocity vector of the following at t= :
2
< sin 5𝑡 − 1 , cos 2𝑡 − 7 >
4. Find the velocity vector of the following at t=3:
5 + 4𝑡 3 2𝑡 2
<
, 𝑒 (5𝑡 ) >
𝑡2
5. Find the acceleration vector of the following at t=2:
𝑡 3 + 6𝑡
<
, 6𝑡 3 + 𝑡 2 >
𝑡4
6. Find the speed of an object at t=5 using the following information:
𝑥(𝑡) = 4𝑡 2
𝑦(𝑡) = 𝑡 3
7. Find the speed of an object at g =1.5 using the following information (a calculator is
needed to solve this problem):
𝑥(𝑔) = 𝑠𝑖𝑛 𝑔 +
1
𝑔2
𝑦(𝑔) = 𝑡𝑎𝑛 4𝑔 + 2𝑔3
8. A curve can be described by the following equations:
𝑥(𝑡) = 2𝑡 2
𝑦(𝑡) = 5𝑡 4 + 6𝑡
Find the equation of the line tangent to the aforementioned curve at t=1
9. A curve can be described by the following equations:
14
𝑥(𝑡) = 5𝑡 3 + 6𝑡 +
1
𝑡2
𝑦(𝑡) = 4𝑡 2 + √𝑡
Find the line tangent to the aforementioned curve at t= 4
10. A curve can be described by the following equations:
3𝑡 4 5𝑡 3
+
+ 𝑡2
4
3
2𝑡 3 𝑡 2
𝑦(𝑡) =
+ − 3𝑡
3
2
Find the t-values of the points where the curve has vertical tangents and the t-values
where the curve has horizontal tangents.
11. An object is being pulled by both Tigger and Pooh. Pooh pulls with a force 22N 62
degrees above the horizontal axis. Tigger pulls with a force of 17N 42 degrees below the
horizontal axis. Find the magnitude and direction of the total force being exerted on the
object.
𝑔
12. A curve is given parametrically by the equations 𝑥(𝑔) = 𝑔2 − 1 and 𝑦(𝑔) = (𝑔−3) such
𝑥(𝑡) =
that g is all real numbers. At what value 𝑔 does the curve go through the line x= 8 ?
a) 𝑔 = −3
b) 𝑔 = −3, 3
c) 𝑔 = 0
d) 𝑔 = 3
e) No solution
13. If the path of Pooh Bear is defined by 𝑥(𝑡) = 𝑡 2 − 15 and 𝑦(𝑡) = 24 − 5𝑡 such that 𝑡 ≥
0. Find Pooh’s speed at time 𝑡 = 3.
a) 1.414
b) 7.810
c) 61
d) 18.065
e) 10.817
𝑦(𝑡)
14. Find lim 𝑥(𝑡) if 𝑦(𝑡) = 4𝑒 −𝑡+4 − 4 and 𝑥(𝑡) = 𝑡 − 4
𝑡→4
a) 4
b) Undefined
c) Indeterminate
d) - 4
e) 0
15. The position of Tigger in an x-y plane is given by 𝑥 = 2√𝑡 + 2 and 𝑦 = 2𝑡 3 + 3. What
is the acceleration vector of Tigger at = 4 ?
15
1
a) ⟨2 , 96⟩
1
b) ⟨16 , 48⟩
c) ⟨6, 128⟩
−1
d) ⟨ 16 , 48 ⟩
e) ⟨−16, 48 ⟩
16. Because Pooh is going to a party, he decided to lose some weight. The path in which
1
Pooh runs is modeled by 𝑥(𝑡) = 𝑡 2 and 𝑦(𝑡) = 3 𝑡 3 where t is measured in minutes and x
and y is measured in meters. If Pooh can burn 20 calories every 100 meters he run, how
much calories did Pooh burned if he runs for 20 minutes continuously.
(A calculator may be helpful :P )
a) 1154133.333 calories
b) 1787.715 calories
c) 71.508 calories
d) 40065.203 calories
e) 1602.608 calories
The balloon in which Pooh is holding is moving in the x-y plane at position(𝑥(𝑡), 𝑦(𝑡)), such
that t is measured in minutes. Given that
𝑑𝑦
𝑑𝑡
= 3𝑡 2 + 4𝑡 and
𝑑𝑥
𝑑𝑡
3
= 2 𝑡 − 5. At time 𝑡 = 0, the
position of the balloon is (1,2).
a) Find the position of the balloon at time t = 4
b) Is the balloon moving up or moving down at time t = 6? Explain your answer. Find the
slope of the path at time t = 6.
c) If Pooh reached his destination after 12 minutes, how long has Pooh traveled?
d) What is the balloon’s speed when Pooh reaches his destination?
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