Orthogonal Friends

Parth Patel

Introduction:
Most calculus students at some point in their education will study differential equations
and their many applications, including modeling many natural phenomena.
We explore a certain application of the differential equation, namely finding theorthogonal trajectories of a curve. The curves explored may not fit the formal definition for
an orthogonal trajectory, so we will define some new terms to fit our needs.
This study relies heavily on the use of software to depict the graphs of various intricate
functions and is experienced best when read with an open copy of a Desmos file used to create
the provided images throughout the paper. The sliders can be used to change the variables asdesired. The file is linked here: https://rb.gy/fn72i.
Throughout this paper I will address the following topics:•

the traditional solutions of standard “Orthogonal Trajectory Problems”;•

different variations on the classical orthogonal trajectory;•

why those variations are not true orthogonal trajectories; and•further insights on the intersections of graphs.

Traditional Orthogonal Trajectories:
Let’s begin with a definition.

Definition: Orthogonal Trajectory
In mathematics, an orthogonal trajectory is a curve which intersects any
curve of a given pencil of planar curves orthogonally. Here a pencil is a set offunctions defined by a parameter such as k, and orthogonal simply means “ata right angle”.

Here, we can consider the tangent lines to the functions at the point of intersection. They
should be perpendicular.
These orthogonal trajectories are the traditional solutions to the problems from a first-year
calculus course and in many other classrooms. The question may read:Find the orthogonal trajectories of the family of functions 𝑦 = 𝑘𝑥 2 .
We know that the intersection should occur at a right angle, so the first derivatives of thefunctions should be negative reciprocals:𝑑𝑦𝑑𝑥

We also know that 𝑘 =𝑦𝑥2

= 2𝑘𝑥 thus

𝑑𝑦𝑑𝑥=−12𝑘𝑥.

, so we can substitute that and rearrange our differential equation

to prepare for integration:
𝑑𝑦𝑑𝑥

=−1𝑦𝑥

2𝑥 2

⟺ 2𝑦𝑑𝑦 = −𝑥𝑑𝑥 ⟺ ∫ 2𝑦𝑑𝑦 = ∫ −𝑥𝑑𝑥.

After integrating (making sure to add that pesky constant) and rearranging, we have thesolution.
2𝑦 2 + 𝑥 2 = 𝑐.
So, we say that the orthogonal trajectory of 𝑦 = 𝑘𝑥 2 is 2𝑦 2 + 𝑥 2 = 𝑐.

A picture of several functions of thepencil 𝑦 = 𝑘𝑥 2 (red) and its
orthogonal trajectory,
2𝑦 2 + 𝑥 2 = 𝑐 (blue). Note thateach member of one pencil
intersects every member of theother pencil orthogonally.

This is the standard solution of the problem seen in most calculus texts. However, on the day
my teacher planned to introduce orthogonal trajectories, she ran out of class time, so she told us

to try the homework and that we’d have the lesson in class the next day. As instructed, I did thehomework. Here was my solution, with the same start:𝑑𝑦𝑑𝑥= 2𝑘𝑥 thus

𝑑𝑦𝑑𝑥

=−1
2𝑘𝑥

Let’s separate the differential equation:
𝑑𝑦 = −𝑑𝑥⟺ ∫ 𝑑𝑦 = − ∫2𝑘𝑥

𝑑𝑥
2𝑘𝑥

.

No substitution is necessary to obtain the solution𝑦=−ln|𝑥|2𝑘+ 𝑐.

The knee-jerk response may be that it is wrong because it is not a true orthogonal trajectory of𝑦 = 𝑘𝑥 2 . However, it still turns out to be interesting.
The graphs are orthogonal, and that night, as I checked my calculus homework, I thought I
had correctly solved the problem. But my solution has the connection between 𝑦 = 𝑘𝑥 2 and𝑦=−ln|𝑥|2𝑘+ 𝑐. The parameter k remains in both equations.

A picture of 𝑦 =𝑘=1
10−ln|𝑥|2𝑘+ 𝑐 (green) and 𝑦 = 𝑘𝑥 2 (red) withand 𝑐 = 0. The graphs remain mutually orthogonal

for all 𝑘, 𝑐 ∈ ℝ, as can be seen in the Desmos project.

I was intrigued. This is not an orthogonal trajectory (recall that a member of one pencil must
intersect every member of the other pencil orthogonally). But it’s too special not to merit adefinition.

Definition: Orthogonal Friend
Two graphs are orthogonal friends if they are related by some parameter kand intersect orthogonally for all 𝑘 ∈ ℝ for which both functions are defined.

To summarize so far, now we know that 𝑦 = 𝑘𝑥 2 has both a proper orthogonal trajectory andanother related graph which is not truly an orthogonal trajectory but is maybe close enough for
us to affectionately label it an orthogonal friend.
Remember that the orthogonal friend was obtained by not substituting for the parameter kduring the integration step. So, the orthogonal friend still contains a parameter:𝑦=−ln|𝑥|2𝑘+ 𝑐.

Let’s try substituting for k now and see what happens:
𝑦=−ln|𝑥|𝑦2 2𝑥+𝑐=−𝑥 2 ln|𝑥|2𝑦+ 𝑐 ⟺ 𝑦2 =−𝑥 2 ln|𝑥|
2+ 𝑐𝑦.

Let’s rewrite the equation as
𝑦 2 − 𝑐𝑦 +𝑥 2 ln|𝑥|2= 0.

And now we can solve for y:𝑦=𝑐± √𝑐 2 −2𝑥 2 ln|𝑥|2.

Let’s take the positive root because 𝑥 2 𝜖 ℝ+ . There are now two variables to manipulate, both cand k. Take a look at the graph for 𝑐 = 𝑘.

A picture of 𝑦 =𝑐+ √𝑐 2 −2𝑥 2 ln|𝑥|2(purple). It’s orthogonal

to 𝑦 = 𝑥 2 (red) as well, but there’s a catch. The graphs
are orthogonal only when 𝑐 = 𝑘, as can be seen using theDesmos project.

What a beautiful result! Not only have we produced the traditional orthogonal trajectory, but
we have also produced an additional graph, what we now call our orthogonal friend, and wederived a different equation from it. Let’s enjoy all thegraphs together:•The original function: 𝑦 = 𝑥 2 (red),•The orthogonal trajectory: 2𝑦 2 + 𝑥 2 = 𝑐 (blue),•The orthogonal friend: 𝑦 =•This new function: 𝑦 =−ln|𝑥|
2𝑘+ 𝑐 (green),𝑐+ √𝑐 2 −2𝑥 2 ln|𝑥|2(purple).

Consider another example, 𝑦 = 𝑘𝑥 3 , for which𝑑𝑦𝑑𝑥= 3𝑘𝑥 2 .

We will generate two differential equations, both by taking the reciprocal of the right-hand side,
but for second, substitute 𝑘 =𝑑𝑦 =−𝑑𝑥
3𝑘𝑥 2𝑦𝑥3:

and 3𝑦𝑑𝑦 = −𝑥𝑑𝑥.

Let’s integrate to find the equations:
𝑦=

13𝑘𝑥

+ 𝑐 and𝑥 2 + 3𝑦 2 = 𝑐.

Graphing them in Desmos, we see the following:•The original function: 𝑦 = 𝑘𝑥 3 (red),•The orthogonal trajectory: 𝑥 2 + 3𝑦 2 = 𝑐 (blue), and•The orthogonal friend: 𝑦 =13𝑘𝑥+ 𝑐 (green).

They all are orthogonal to 𝑦 = 𝑥 3 . Let’s now find the final graph by substituting 𝑘 =
𝑦=

13𝑘𝑥

+ 𝑐.

𝑦𝑥3

into

𝑥2

After substitution, we find that 𝑦 =3𝑦+ 𝑐 so 𝑦 2 − 𝑐𝑦 −𝑥2
3= 0. Since the original function

has a range ℝ, let’s leave this final equation in implicit form.
Graphing, we get the following:

The graphs are not orthogonal at all, and, in fact, no value of c makes the graphs orthogonal.
Substituting k after integration fails. Perhaps for 𝑦 = 𝑘𝑥 2 , we simply had a coincidence.
We can try to generalize this result for any 𝑦 = 𝑥 𝑛 for 𝑛 ∈ ℕ. And while we could put acoefficient on the x term, working without one will make the entire process easier to understand.
Let’s see if we can derive any expressions for various orthogonal graphs of powers of x, whichshould include one true orthogonal trajectory. Differentiating 𝑦 = 𝑥 𝑛 , we find𝑑𝑦𝑑𝑥= 𝑛𝑥 𝑛−1 .

Taking the negative reciprocal, we will find two solutions, both of which will be dependent on 𝑛:𝑑𝑦𝑑𝑥

1𝑥 𝑛−1𝑥𝑦

Substituting ==−1
𝑛𝑥 𝑛−1⇔𝑦𝑑𝑦𝑑𝑥

=−1
𝑥𝑛−1
𝑛 𝑦.

, we get∫ 𝑛𝑦𝑑𝑦 = ∫ −𝑥𝑑𝑥.

Hence,𝑛𝑦 2
2=−𝑥 22+ 𝑐.

So, the solution is the ellipse given by 𝑛𝑦 2 + 𝑥 2 = 𝐶. No surprise there.

Now, we restart:
𝑑𝑦= 𝑛𝑥 𝑛−1 .

𝑑𝑥Taking the negative reciprocal to find the new solution, we have𝑑𝑦𝑑𝑥=−1𝑛𝑥 𝑛−1.

Thus,−𝑑𝑥∫ 𝑑𝑦 = ∫ 𝑛𝑥 𝑛−1.
Hence, as long as 𝑛 ≠ 2,𝑦=1𝑛(𝑛−2)𝑥 𝑛−2+ 𝐶.

The derivations are complete, and our results are as follows:•The original function: 𝑦 = 𝑘𝑥 𝑛 .

•The standard orthogonal trajectory: 𝑛𝑦 2 + 𝑥 2 = 𝐶.

•The new and interesting orthogonal friend: 𝑦 =1𝑛(𝑛−2)𝑥 𝑛−2+ 𝐶.

Recall our study of 𝑦 = 𝑥 2 , specifically when we derived the formula for its orthogonal friend
and then substituted 𝑘 =𝑦𝑥2

into the equation. The resulting function was orthogonal to 𝑦 = 𝑥 2

only when the integration constant c was equal to k but had quickly realized that it was acoincidence when the same procedure failed for 𝑦 = 𝑥 3 . Let’s see if we can derive anotherorthogonal friend for 𝑦 = 𝑥 𝑛 from the results of 𝑦 = 𝑥 2 .
The orthogonal friend of 𝑦 = 𝑥 2 is 𝑦 =−ln|𝑥|
2𝑘+ 𝑐. Instead of 𝑘 =𝑦𝑥2, Let’s substitute 𝑘 =𝑦𝑥𝑛

and see what happens. We find the following:𝑦=−𝑥 𝑛 ln|𝑥|2𝑦+ 𝑐.

We also need the graph to contain (1,1), so 𝑐 = 1. Also, when we take the derivative of 𝑦 = 𝑥 𝑛
and then the negative reciprocal, we will end up with an n in the denominator. So, while it may
seem strange now, substitute 𝑛 in the denominator for 2𝑦. We now have𝑦 =1−𝑥 𝑛 ln|𝑥|𝑛;. |𝑥| allows the largest possible domain. We now have𝑦 =1−|𝑥|𝑛 ln|𝑥|𝑛.

So, just like 𝑦 = 𝑥 𝑛 , this new graph always contains (1,1). Let’s now set the derivative of ournew function equal to the negative reciprocal of the derivative of 𝑦 = 𝑥 𝑛 , with careful use of theproduct rule:−1𝑛𝑥 𝑛−1=−𝑛|𝑥|𝑛−1 ln|𝑥|𝑛−𝑥 𝑛−1𝑛.

With some cancellation and rearrangement, we find:
1= 𝑛|𝑥|𝑛−1 ln|𝑥| − 𝑥 𝑛−1 .

𝑥 𝑛−1Plugging in our point of intersection, (1,1):
11𝑛−1= 𝑛 ∗ 1𝑛−1 ln|1| + 1𝑛−1 ,which is true for any 𝑛 ∈ ℕ, so our two graphs contain (1,1). Furthermore, their derivatives will
always be negative reciprocals at that point, and thus the two functions are orthogonal friends.
The exponent of x can be any positive real number, 𝑐. So, we have shown that 𝑦 = 1 −𝑥 𝑐 ln|𝑥|𝑛isa pencil of orthogonal friends of 𝑦 = 𝑥 𝑛 .

Definition: Orthogonal Friend GroupA graph has an orthogonal friend group if there is a pencil of functions definedby parameters, with at least one unique parameter, such that each member ofthat pencil is an orthogonal friend of the graph.
Since 𝑦 = 1 −𝑥 𝑐 ln|𝑥|𝑛contains a unique parameter c, it is an orthogonal friend group of 𝑦 = 𝑥 𝑛 .

To recap, we explicated the function, and our results are as follows:•The original function: 𝑦 = 𝑥 𝑛 (red)•The standard orthogonal trajectory: 𝑛𝑦 2 + 𝑥 2 = 𝐶 (blue).

•

The new and interesting orthogonal friend: 𝑦 =1𝑛(𝑛−2)𝑥 𝑛−2+𝐶 (green).
•

The new and even more interesting orthogonal friend group:•𝑦 =1−𝑥 𝑐 ln|𝑥|𝑛(purple).

Here is a picture of the function and its orthogonal trajectory, its orthogonal friend, and a single
member of its orthogonal friend group.