Mathematics SL Internal Assessment

Does my dog walk more than me?

Table of Contents
Introduction .......................................................................................................................................... 3
Scenario 1 .............................................................................................................................................. 3
Step 1: Finding the distance between my dog and myself ......................................................... 3
Step 2: Finding an equation for my walk ................................................................................... 3
Step 3: Finding an equation for my dog’s walk ......................................................................... 4
Step 4: Calculating the distance (in meters) walked .................................................................. 7
Scenario 2 .............................................................................................................................................. 8
Step 1: Finding coordinates for dog A ....................................................................................... 8
Step 2: Setting up an Excel model to find dog A coordinates ................................................. 10
Step 3: Graphing the Excel model for dog A and dog B ......................................................... 12
Step 4: Finding and manipulating a function ........................................................................... 12
Conclusion .......................................................................................................................................... 13
Works Cited ........................................................................................................................................ 14
Appendix ............................................................................................................................................. 15

2
Introduction
Last year, my pet dachshund had to undergo surgery to correct his hernia. The probability of
whether he would be able to walk again depended on how well his rehabilitation progressed. The
final stages of rehabilitation involved rebuilding his leg muscle by taking him on short walks. I
interpreted “short” as a 15-minute walk once around the block. However, I was later advised by the
vet to shorten this distance. The vet claimed that dogs can walk up to double the distance that we do
because they often walk circles around us. The aim of this investigation is to examine the validity of
the vet’s claim by creating a scenario in which I can graph the equation of my dog’s walk and my
own. In doing so, I can explore the relationship between the distance that dogs walk in comparison to
their owners.
Scenario 1
When developing this scenario, I recognized that I would not be able to find an equation for
my dog or my own path without establishing rules. Hence, the first challenge I faced was
determining what would be an appropriate value for each rule. Since I want my investigation to be
authentic, I heavily prioritized realism to base my choices on. Moreover, to ensure my process will
be easily understood, I chose more simple numbers for my speeds. Hence, based on these factors, I
decided to implement the following rules in my investigation:
1. A 2.5m leash is extended to its maximum length throughout the walk.
2. I walk at a constant speed of 1m/s.
3. My dog walks at a constant speed of 2m/s relative to a stationary point.
Step 1: Finding the distance between my dog and myself
The following diagram models the distance between my dog and myself given that his leash
is extended to maximum length (2.5m):
0.5m

𝑥m

From this, we can apply Pythagoras Theorem to find 𝑥:

𝑎2 + 𝑏 2 = 𝑐 2
0.52 + 𝑥 2 = 2.52
𝑥 = √6
Step 2: Finding an equation for my walk
Since my dog’s range of motion is restricted by his leash, his path revolves around me as the
central point. Hence, step 2 of this investigation is to graph an equation for my own walk. Because
the variable of time is involved, I chose to use parametric equations to model this scenario rather
than a rectangular equation (an equation composed of variables 𝑥 and 𝑦 which can be graphed on a
regular Cartesian plane). A curve in the plane is parameterized if the set of coordinates, (𝑥, 𝑦), on the
curve are represented as functions of a variable 𝑡.

3
 𝑥 = 𝑓 (𝑡 )
𝑦 = 𝑔(𝑡)
These functions describe the 𝑥 and 𝑦 position of an object moving in a plane. The advantages of
using parametric equations include being able to graph curves that are not functions and being able to
represent a third variable. The disadvantage when sketching parametric curves is that we must pick a
set range for 𝑡 based on intuition. Hence, there is always the possibility that the entire curve is not
modeled. Moreover, since parametric equations are not part of the syllabus, this approach would
require that I learn the basics of the topic myself.
The third variable in this scenario, called a parameter, is time (𝑡) in seconds. The purpose of
this investigation is to compare the distance that dogs walk to that of their owners over a set amount
of time. I chose to represent time between 0 ≤ 𝑡 ≤ 30 for this investigation. This range ensures
enough distance is covered for comparison and prevented the model from being overly complicated.
Given that I am walking in a straight line towards my destination, we could graph my walk as
a rectangular equation where 𝑥 = ℝ; where ℝ represents the set of all real numbers (real numbers are
values that can be represented as a quantity along a number line). We use parametric equations
instead because there is an advantage of being able to incorporate the variable of time (𝑡) and
demonstrate a direction of motion in the model. Let’s say that the road is represented by 𝑥 = 0. In
this case, I would be walking in one direction along 𝑥 = 0 at 1m/s. In other words, the y coordinate,
representing meters, will increase proportionally to 𝑡. From this information, we can base the
following parametric equations:
𝑥=0
𝑦=𝑡
0 ≤ 𝑡 ≤ 30

(Hohenwarter, Markus)

Step 3: Finding an equation for my dog’s walk

For this scenario, I am walking at a speed of 1m/s along 𝑥 = 0. My dog, who is always √6m from
my coordinates, walks in circles around me at a speed of 2m/s. To derive an equation for my dog’s
walk, I chose to build off the parametric equations for a unit circle (a circle centered at (0,0) with
radius 1). This would allow me to build my equation based on what I already know about
trigonometric identities. If A(𝑥, 𝑦) moves around the unit circle such that OA makes an angle of 𝑡
with a positive x-axis, then:
𝑥 = cos 𝑡
𝑦 = sin 𝑡

4
 (Hohenwarter, Markus)
The parametric equations for a unit circle would represent the following situation: I am
standing at the origin (0,0) and my dog is continuously walking in a circle, of radius 1m, around me.
To satisfy the requirements of this investigation, the equations must consider the distance between
my dog and I, as well as the speed that we are walking. We can base our changes using periodic
function formulas, 𝑦 = 𝐴 sin(𝐵𝑥 + 𝐶 ) + 𝐷 and 𝑦 = 𝐴 cos(𝐵𝑥 + 𝐶 ) + 𝐷. The variables in this
2𝜋
formula refer to amplitude (𝐴), period ( ), horizontal shift (𝐶), and vertical shift (𝐷).
𝑏

Amplitude
The first step is to incorporate the distance between my dog and myself into the equations.
This has to do with the amplitude (A) of the graph (the amplitude refers to the distance from the
midpoint to the highest or lowest point of the function). We found in step 1 that my dog is always
√6m from my coordinates. The radius of the circle that my dog walks and therefore the amplitude of
this graph is √6.
𝑥 = √6 cos 𝑡
𝑦 = √6 sin 𝑡
Period
The next step is to find and incorporate the period (the time required for the function to make one
full oscillation) into the equations. For this scenario, an oscillation is one full revolution walked
around me by my dog.
Distance required to make one full oscillation (where radius is √6):
Circumference = 2πr
Circumference = 2√6π
Time required for my dog to make one full oscillation (where my dog walks at 2m/s):
𝑑 =𝑠×𝑡
2√6π = 2 × 𝑡
𝑡 = √6π
Hence, we can find 𝑏:

5
 2𝜋
Period =
𝑏
2𝜋
√6π =
𝑏
2
𝑏=
√6
Then,
2𝑡
𝑥 = √6 cos
√6
2𝑡
𝑦 = √6 sin
√6
Vertical Shift
The final step is the addition of a vertical shift (D) to the parametric equation for 𝑦 (a vertical
shift is a measure of how far the graph has shifted vertically from its’ initial position). Because my
dog’s path revolves around me as the central point, the equations for my walk must be implemented
into his. This can be accomplished by adding the parametric equations from my path to his:
2𝑡
𝑥 = √6 cos +0
√6
2𝑡
= √6 cos
√6
2𝑡
𝑦 = √6 sin +𝑡
√6
0 ≤ 𝑡 ≤ 30
By adding the variable 𝑡 to the equation for 𝑦, we can sketch the effect time has on distance walked.
Graphing my dog’s walk

6
Step 4: Calculating the distance (in meters) walked
Distance I walked
The parameter, time (𝑡) in seconds, was represented between 0 ≤ 𝑡 ≤ 30 for this
investigation. Given that I am walking 1m/s in a straight line to my destination, we can find the
distance that I walked using the following equation:
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑠𝑝𝑒𝑒𝑑 =
𝑡𝑖𝑚𝑒
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
1=
30
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 30 meters
Distance my dog walked
I had difficulty with this step of my investigation because in SL Math, we are not taught
parametric equations in depth. Hence, I approached this challenge by researching online to see if
there was an existing method to derive the length of a parametric curve. I found the following
formula in a tutorial by Paul Dawkins from California State University Northridge:

𝛽
𝑑𝑥 2 𝑑𝑦 2
𝐿 = ∫ √( ) + ( ) 𝑑𝑡
𝛼 𝑑𝑡 𝑑𝑡

(Dawkins, Paul)
We can use the formula above to determine the length of the parametric curve for 0 ≤ 𝑡 ≤ 30
given by the following parametric equations:
2𝑡
𝑥 = √6 cos
√6
2𝑡
𝑦 = √6 sin +𝑡
√6
We will first need to find the derivatives of each equation. For my dog’s 𝑥 equation,
𝑑𝑥 2𝑡
(√6 cos )
𝑑𝑡 √6
𝑑𝑥 2𝑡
= √6 × (cos )
𝑑𝑡 √6
2𝑡 𝑑𝑥 2𝑡
= √6 (− sin ) × ( )
√6 𝑑𝑡 √6
2𝑡 𝑑𝑥 2𝑡
= −√6 × × (𝑡) × sin
√6 𝑑𝑡 √6
2𝑡
= −2 sin
√6
For my dog’s 𝑦 equation,
𝑑𝑦 2𝑡
(√6 sin +𝑡)
𝑑𝑡 √6
𝑑𝑦 2𝑡
= √6 × (sin +𝑡)
𝑑𝑡 √6

7
 2𝑡 𝑑𝑦 2𝑡
= √6 cos × ( + 𝑡)
√6 𝑑𝑡 √6
2𝑡 𝑑𝑦 𝑑𝑦 2𝑡
= √6 ( × (𝑡 ) + (𝑡)) cos ( + 𝑡)
√6 𝑑𝑡 𝑑𝑡 √6
2 2𝑡
= ( + 1) × √6cos ( + 𝑡)
√6 √6
2𝑡
= (√6 + 2) × cos ( + 𝑡)
√6
The length is then,

𝛽
𝑑𝑥 2 𝑑𝑦 2
𝐿 = ∫ √( ) + ( ) 𝑑𝑡
𝛼 𝑑𝑡 𝑑𝑡
30 2 2
2𝑡 2𝑡
𝐿 = ∫ √(−2 sin ) + ((√6 + 2) × cos ( + 𝑡)) 𝑑𝑡
0 √6 √6
Since this integration would be extremely difficult to do by hand, I chose to use the integrate
function on my graphing display calculator (GDC) to solve for 𝐿:

𝐿 = 97.78 meters

Scenario 2
When I take my dog for a walk, realistically it is never just a straight line to my destination.
Let’s say we walk towards an intersection and another dog (dog B) crosses in front of us. In this
scenario, my dog (dog A) would stop circling me and begin to pursue dog B. The resulting curve is
referred to as a pursuit curve, where dog A is always directed towards dog B. For this investigation,
dog A and B will move with uniform velocities (1m/s). To avoid unnecessary complication, if my
dog begins to curve, we will assume I follow his path without resistance. This scenario considers a
common obstacle in dog walks and therefore presents a more realistic conclusion to the aim outlined
in the introduction.
Step 1: Finding coordinates for dog A
Same as scenario 1, I am walking dog A from the origin (0,0) along 𝑥 = 0. Since dog B will
initially be walking perpendicular to dog A, we can represent dog B’s walk as 𝑦 = ℝ; where ℝ
represents the set of all real numbers. For simplicity, let’s say dog B is walking at 1m/s along 𝑦 = 3.
We can show this information with the following vector equation:

⃗B = ( 𝑡 )
3
My initial approach to model dog A was to find a standard formula for equations of pursuit. Research
𝑥
⃗ = (𝑡) .
⃗ = ( ) and B
online directed me to a pursuit formula, where I substituted A 𝑦 3
⃗ − ⃗A) × ⃗Ȧ
(B
=1
⃗ − ⃗A|
|B

8
 𝑡 − 𝑥 𝑥̇
(3 − 𝑦) ( )
𝑦̇
1= 𝑡−𝑥
| |
3−𝑦
(Weisstein, Eric W.)
Unfortunately, I cannot solve the derived equation because of the “overdot” (indicating
differentiation in respect to time). Instead of using a formula to find the pursuit curve, I changed my
approach and decided to use vectors to find coordinates for dog A. Coordinates will be found for
small time intervals of constant velocity: I chose to use 0.1 second time intervals. The two dogs will
be in eye range when dog A is at (0,0) and dog B is at (0,3). From this position onwards, dog A will
always move in the direction of dog B:

B (0,3)
⃗B = ( 𝑡 )
3

A (0,0)

First, we must find dog A’s position when 𝑡 = 0.1 (represented by P in the diagram below).

B (0,3) ⃗B = ( 𝑡 )
3
P (𝑥, 𝑦)
A (0,0)

Before we can find P (𝑥, 𝑦), we must find ⃗⃗⃗⃗⃗ ̂ (unit vector).
AB (direction vector) and AB

Finding ⃗⃗⃗⃗⃗
AB:
⃗⃗⃗⃗⃗ = B
AB ⃗ −A⃗
⃗⃗⃗⃗⃗ = ( ) − (0)
AB
0
3 0
0
⃗⃗⃗⃗⃗ = ( )
AB
3
̂:
Finding AB
⃗⃗⃗⃗⃗
AB
̂ =
AB
⃗⃗⃗⃗⃗ |
|AB
0
( )
̂ =
AB 3
√02 + 32
̂ = (0)
AB
1

9
Coordinate P(𝑥, 𝑦) can thus be found by,
𝑥
⃗ + 𝑡AB
(𝑦 ) = A ̂
𝑥 0 0
(𝑦 ) = ( ) + 𝑡 ( )
0 1
𝑥=0
𝑦=𝑡
= 0.1
Therefore, after 0.1 seconds, dog A has moved to P(0,0.1). Since calculating each coordinate by
hand would be inefficient, I chose to set up a model on Excel.
Step 2: Setting up an Excel model to find dog A coordinates

Time
As you go down a row, the value for time will increase by 0.1 intervals. This column represents the
overall time which has passed.
⃗⃗⃗⃗⃗ )
Direction Vector (AB

To find any direction vector,

⃗⃗⃗⃗⃗ = B
AB ⃗ −A⃗
B𝑥 A𝑥
⃗⃗⃗⃗⃗
AB = (B ) − (A )
𝑦 𝑦
AB𝑥 = B𝑥 − A𝑥
AB𝑦 = B𝑦 − A𝑦

We can use the information stored in excel cells to calculate the direction vector ⃗⃗⃗⃗⃗
AB. When 𝑡 = 0.1,
we use cells C4 (stores A𝑥 ), D4 (stores A𝑦 ), E4 (stores B𝑥 ), and F4 (stores B𝑦 ) as following:
AB𝑥 = B𝑥 − A𝑥
= E4 − C4
AB𝑦 = B𝑦 − A𝑦
= F4 − D4

10
Dog B
Given that dog B is walking along 𝑦 = 3, we can deduce that his 𝑦-coordinate will not change with
time. Since he walks at a speed of 1m/s, his 𝑥-coordinate will increase proportionally to time: each
0.1 second increase, dog B will move an additional 0.1m from the previous 𝑥-coordinate.
𝑑
𝑠=
𝑡
𝑑
1=
0.1
𝑑 = 0.1
Dog A
𝑥
⃗ + 𝑡AB
(𝑦 ) = A ̂
𝑥 ⃗⃗⃗⃗⃗
AB
(𝑦) = ⃗A + 𝑡
⃗⃗⃗⃗⃗ |
|AB
(Note: For this equation, 𝑡 is always substituted by 0.1 regardless of how much time has passed
overall. This is because we are finding the coordinates for P(𝑥, 𝑦) with different direction vectors
each calculation. If at any other point in these report 𝑡 is referred to, it should be defined as the
overall time which has passed.)
Hence,
⃗⃗⃗⃗⃗ 𝑥
AB
⃗ 𝑥 + 0.1
𝑥=A
⃗⃗⃗⃗⃗ |
|AB
⃗⃗⃗⃗⃗ 𝑦
AB
𝑦 = ⃗A𝑦 + 0.1
⃗⃗⃗⃗⃗ |
|AB
As an example, consider finding the 𝑦-coordinate for dog A when 𝑡 = 0.2. To find the dog’s new
position, we must know the current 𝑦-coordinate for dog A (stored in cell D4), and the corresponding
direction vector ⃗⃗⃗⃗⃗
AB (stored in cells G4 and H4).

⃗⃗⃗⃗⃗
AB𝑦
𝑦 = ⃗A𝑦 + 0.1
⃗⃗⃗⃗⃗ |
|AB
H4
𝑦 = D4 + 0.1
√G42 + H42
The same process is repeated for the 𝑥-coordinate. We can use the fill handle to copy the formula
into other cells and then proceed to graph these coordinates.

11
Step 3: Graphing the Excel model for dog A and dog B

A Graph to Model the Pursuit Curve for Dog A and Dog B

y

0 x
0 1 2 3 4 5 6 7 8 9 10 11

After graphing dog A’s path, I recognized that the intervals between points were not consistent as
they should have been. While this may become a source of error, I chose to continue as is after
confirming my formulas were correct. While I had originally planned to rely on excel to find a line
of best fit for the pursuit curve, I discovered that this would not be possible given the complexity of
dog A’s path. Instead, I will need to find a function which has similar characteristics to the pursuit
curve graphed in Fig. 11.
Step 4: Finding and manipulating a function
Finding a function
My approach to this task reminded me of how I found my dog’s equation in scenario 1. Hence, like I
did for that task, I will find a similar function to manipulate. Here, I noticed that 𝑦 = tan−1 𝑥 and
1
𝑦 = − 𝑥 have similar visual shapes. The defining feature of my pursuit curve is that it has
1
asymptotes: the curve is bound by 0 ≤ y ≤ 3 and 𝑥 ≥ 0. Therefore, I chose to manipulate 𝑦 = − 𝑥
because I know this function has two asymptotes: 𝑦 = 0 and 𝑥 = 0. (Note: I will only consider the
positive side of the 𝑥-axis from this point onwards.)
Manipulating a function
Instead of the asymptote being 𝑦 = 0, I will vertically shift the equation by 3 so that the asymptote is
𝑦 = 3. While the asymptote should really be 𝑦 = 3. 0̇1, this is not possible to input on GeoGebra.
Since this will not result in any statistically significant problems, I decided to leave it as 3. This
equation will be graphed with 15 randomly chosen coordinates from the excel spreadsheet to show
how it relates to the ideal pursuit curve.
1
𝑦 = − +3
𝑥

12
 (Hohenwarter, Markus)
Point Q, where the curve intersects the 𝑥-axis, is (0.33,0). For the curve to intersect the origin, we
must horizontally shift it right by 0.33. The resulting curve and its’ graph is,
1
𝑦=− +3
𝑥 + 0.33

(Hohenwarter, Markus)
Conclusion
In the modelled 30 seconds for scenario 1, I walked 30 meters. In the same duration, my dog walked
97.78 meters. Hence, under the variables given for this scenario, we can deduce that my dog walked
3.26 times the distance I did. Therefore, the vet’s claim is only partially valid because dogs walk
triple, instead of double, the distance we do: a ratio that results from dogs circling their owners
during walks. This discovery is of significance because it means owners are misunderstanding the
intensity of exercise that their dogs are doing. If your pet is hurt, intense exercise can cause
additional stress to the injury and hinder the rehabilitation progress. Therefore, it becomes very
important for owners to recognize how much more their dog could be walking and adjust based on
this recognition. The major limitation to this scenario is that it is heavily based on assumptions. If
you’ve ever walked a dog, you would know that they don’t continuously circle you. Therefore, in
scenario 2, I investigated how dog A’s walk changes based on the presence of dog B. This made me
aware of how obstacles in a walk can decrease the dog’s total walking distance. Instead of circling
1
me, my dog would change his path and walk along a pursuit curve: given by 𝑦 = − 𝑥+0.33 + 3. This
suggests that the obstacles result in the total distance being walked decreasing. Since we did not
consider obstacles when finding that my dog walks 3.26 times the distance I do, we can assume that
this is unrealistic. Considering how obstacles decrease the total walking distance, it becomes more
legitimate for the vet to claim a dog walks double the distance I do. However, it should be noted that
the extent to which a dog’s paths is affected by obstacles, whether that be other dogs or fire hydrants,
is dependent on the dog. Moreover, the extent to which a dog circles their own is also subjective.
These assumptions and generalizations are the biggest limitations to this investigation. Additionally,
the inability to find an exact equation for the pursuit curve (having to rely on approximation) also
becomes a source of error. The investigation could have been improved by combining what was

13
found in both scenarios to determine what distance my dog and I respectively would have walked if
we encountered a certain number of dogs. This would have allowed for a more realistic conclusion to
be drawn and therefore should be considered as a possible extension. My methodology was a
strength considering how I adapted my process to find a pragmatic solution to problems: such as
using excel after my initial pursuit curve formula was too complex to calculate. This investigation
taught me how to use digital skills to solve real life application problems: for instance, using a
variety of softwares to draw a conclusion for scenario 2. The investigation could be improved greatly
if I could evaluate distance walked based on different types of dogs. A possible extension would
therefore be to find the distance walked for three different situations: where dog A circles his owner
without obstacle, dog B pursuits another dog, and dog C walks towards each fire hydrant.

Works Cited
Dawkins, Paul. “Calculus II.” Parametric Equations and Polar Coordinates, 2007, pp. 26-29.
California State University Northridge: College of Engineering and Computer Science,
www.csun.edu/matabots/resources/CalcII_ParametricEqns_PolarCoords.pdf.
Hohenwarter, Markus. “Discover Math with GeoGebra.” GeoGebra - Dynamic Mathematics, IGI,
Dec. 2013, www.geogebra.org/.
Weisstein, Eric W. “Pursuit Curve.” Wolfram MathWorld, Wolfram Research, Inc.,
mathworld.wolfram.com/PursuitCurve.html.

14
Appendix

15