# Linear approximation of hyperbola vs of circle?

I was trying to do a linear approximation using hyperbolic trig functions and comparing that with a linear approximation to a circle at any given angle $a$. So I found that for the unit circle:

$$f(x)=-\cot(a)(x-\cos(a)) + \sin(a)$$

But for a hyperbola $x^2-y^2=1$: $$g(x)=\coth(a)(x-\cosh(a)) + \sinh(a)$$

If my equations are correct, I’m wondering why they’re the only difference between the two is the negative term before cotangent in the circle linearization but no negative term in the hyperbola linearization. Any ideas as to why this is?

First, notice that the cosine and sine terms in your linear approximation don't matter. They only reflect the fact that your linear approximation is made to pass through the point $(x,y)$, which is defined in terms of the angle $a$ as $(\cos a, \sin a)$ and $(\cosh a, \sinh a)$ respectively.
Your question then essentially boils down to "why does the slopes of the circle and hyperbola have different signs". Geometrically, it is because the angle is always defined from the origin, relative to the $+x$ axis. However, the circle is "concave down" as viewed from the origin (it curves towards the origin), while the hyperbola is "concave up" (it curves away from the origin). This means that the slope is necessarily an increasing function of the angle for a circle, and an decreasing function for the hyperbola. This is essentially the geometric reason for the sign change.
On a more algebraic level, the slope can be directly computed from the defining equations. Collectively, we can write the defining parametrizations of the unit circle/hyperbola as $$x^2 \pm y^2 = 1.$$
Implicitly differentiating with respect to $x$, we get $$2x \pm 2yy' = 0 \iff y' = \mp \frac{x}{y},$$ with the top sign for the circle and the bottom sign for the hyperbola. This shows you that the sign directly comes from the differences in the defining equation.
• Thank you so much! Can you please explain more about how the concavity works towards the origin? I specifically don't understand "This means that the slope is necessarily a decreasing function of the angle for a circle, and an increasing function for the hyperbola." As theta increases for a circle, the slope of the linear approx goes from more negative (tangent to $\frac{\pi}{4}$ has a derivative of $-1$), but $\frac{3\pi}{4}$ has a derivative of $1$. But the hyperbolic cotangent starts, when theta is 0, at infinity, and then decreases, converging to 1, as theta increases. – rb612 Oct 24 '17 at 22:56
• Imagine you have a regular concave or convex function. This is a function defined on $x$ such that $f''(x)$ is positive or negative, which means that the slope $f'(x)$ is either increasing or decreasing. Now, imagine you look at your circle from the origin. The circle is parametrized as a function of the angle $\theta$ but the idea is very much the same. As $\theta$ increases, the slope is increasing (I mixed up the circle and the hyperbola in my original answer, as you rightly pointed out) as a result of the concavity. – EuYu Oct 25 '17 at 21:06