Added 2021-09-11, Modified 2022-08-31

Deriving cosh and sinh using implicit differentiation

Hyperbolas are great, they're used to describe gravity assists, the shape power lines or chains make when hanging off two points is a Catenary.

This post is about the hyperbolic functions cosh(t),sinh(t)\cosh(t), \sinh(t) that describe hyperbolics parametrically. In particular I show how you might have discovered them yourself!

This is the unit hyperbola, described by x2y2=1x^2 - y^2 = 1. We want to find x(t),y(t)x(t), y(t) that describe this curve. Since they are always on the curve they must satisfy

x(t)2y(t)2=1x(t)^2 - y(t)^2 = 1

Knowing how x(t),y(t)x(t),y(t) change will help us find x(t)x(t) and y(t)y(t), so let's differentiate using implicit differentiation

2x(t)x(t)2y(t)y(t)=0    x(t)x(t)=y(t)y(t)    xy=yx2x(t)x'(t) - 2y(t)y'(t) = 0 \implies x(t)x'(t) = y(t)y'(t) \implies \frac{x}{y} = \frac{y'}{x'}

Hey! That means x=Cyx' = Cy and y=Cxy' = Cx for some constant CC (since rational numbers are unique up to a constant multiple) This reminds me of eCte^{Ct}, except there are two functions in a cycle instead of one.

We can use this to find all the derivatives in terms of xx and yy

x=Cy    x=Cy=C2xx' = Cy \implies x'' = Cy' = C^2 x

Hey! If differentiating xx gives us CyCy and vise versa, then differentiating x(t)+y(t)x(t) + y(t) should stay the same up to a multiple of CC, Let's do this

ddt[x(t)+y(t)]=Cy(t)+Cx(t)=C(x(t)+y(t))\frac{d}{dt}\left[x(t) + y(t)\right] = Cy(t) + Cx(t) = C(x(t) + y(t))

If we denote s(t)s(t) to be the sum x(t)+y(t)x(t) + y(t) then we have s(t)=Cs(t)s'(t) = Cs(t).

This reminds me of eCte^{Ct} since it's derivative is CC times itself, so I'm going to write s(t)s(t) as a Maclaurin Series and try to find s(t)s(t) in terms of eCte^{Ct}, but first we need to decide what s(0)s(0) is, let's look at the graph again

Where should our parametric equations start? All points work fine, but we'll pick (1,0)(1, 0) for two reasons

  1. It's a natural seeming starting point
  2. Having x(0)=1,y(0)=0x(0) = 1, y(0) = 0 will drastically simplify our maclaurin series

Now that we have our starting point, we can find the maclaurin series of s(t)s(t)

s(t)=k=0s(n)(0)tkk!=k=0(Ct)kk!\begin{aligned} s(t) &= \sum_{k=0}^{\infty} \frac{s^{(n)}(0)t^k}{k!} \\ &= \sum_{k=0}^{\infty} \frac{(Ct)^k}{k!} \\ \end{aligned}

You might recognize this series, it's the series for eCte^{Ct}!, therefor s(t)=eCts(t) = e^{Ct}

Now I want to find x(t)x(t) and y(t)y(t) individually, going back to the original equation I notice it's a difference of squares, meaning we can factor it as

(xy)(x+y)=1    (xy)eCt=1    xy=eCt(x - y)(x + y) = 1 \implies (x - y)e^{Ct} = 1 \implies x-y = e^{-Ct}

Adding the sum and difference we can recover the individual equations!

x(t)=(x+y)+(xy)2=eCt+eCt2y(t)=(x+y)(xy)2=eCteCt2\begin{aligned} x(t) &= \frac{(x + y) + (x - y)}{2} = \frac{e^{Ct} + e^{-Ct}}{2} \\ y(t) &= \frac{(x + y) - (x - y)}{2} = \frac{e^{Ct} - e^{-Ct}}{2} \end{aligned}

Let's check our work, our functions should satisfy x(t)2y(t)2=1x(t)^2 - y(t)^2 = 1

14((eCt+eCt)2(eCteCt)2)=14((e2Ct+2eCteCt+e2Ct)(e2Ct2eCteCt+e2Ct))=14((e2Ct+2+e2Ct)(e2Ct2+e2Ct))=14(4)=1\begin{aligned} &\frac{1}{4}\left( (e^{Ct} + e^{-Ct})^2 - (e^{Ct} - e^{-Ct})^2 \right) \\ = &\frac 14\left( (e^{2Ct} + 2\bcancel{e^{Ct}e^{-Ct}} + e^{-2Ct}) - (e^{2Ct} - 2\bcancel{e^{Ct}e^{-Ct}} + e^{-2Ct}) \right) \\ = &\frac 14\left( (e^{2Ct} + 2 + e^{-2Ct}) - (e^{2Ct} - 2 + e^{-2Ct}) \right) \\ = &\frac 14(4) = 1 \end{aligned}

Great! our x(t)x(t) and y(t)y(t) work, and they are cosh(t)\cosh(t) and sinh(t)\sinh(t) when C=1C=1.

Arc length reparameterization

Picking C=1C=1 seems like an arbitrary choice, let's try to reparameterize in terms of arc length to give CC geometric meaning

s=x(t)2+y(t)2dt=Cy(t)2+x(t)2dt=Ccosh(t)2+sinh(t)2dt\begin{aligned} s &= \int \sqrt{x'(t)^2 + y'(t)^2}dt \\ &= C \int \sqrt{y(t)^2 + x(t)^2}dt \\ &= C \int \sqrt{\cosh(t)^2 + \sinh(t)^2}dt \end{aligned}

You can try to evaluate this integral, but you can't! this is a Nonelementary integral (see wolframalpha).

Relation to other conics

If you noticed the similarity between cosh,sinh\cosh, \sinh (for describing hyperbolas) and cos,sin\cos, \sin (for describing circles and ellipses) You might be wondering about expressing other conics parametrically.

The only other conic is a parabola (as circles are a special case of an ellipse). parabolas are described by y=ax2y = ax^2, which is already a function!

Visually parabolas (3) look different from the other conics, they just barely hug the cone

I like to think of this in terms of their Eccentricity

Parabolas really are the special child in the family! They are the only conic with a single eccentricity. Here's a desmos graph showing this off

I plan to write about parabolas some time, since they show up everywhere in algebra I want to become more acquainted with their amazing properties