Asymptotes and curvature

Two questions close out the qualitative study of a plane curve: where does it go when $x$ runs off to infinity or crashes into a forbidden point, and how sharply does it bend right here, at a single point? The first question is answered by asymptotes — straight lines the curve hugs ever more closely without touching. The second is answered by curvature — a single number that turns "sharply bent" into something measurable, built from the first and second derivatives already familiar from the concavity topic. Together they finish the toolkit MA10001's Unit 1 builds for reading the shape of $y=f(x)$ : limits at infinity and at discontinuities to spot asymptotes, and second-derivative arithmetic to quantify bending. Both ideas below extend the closing topic of the differential-calculus unit and stay strictly within it — no optimization, no tangent-line word problems.

The core insight

An asymptote is a straight line a curve approaches as $x$ (or $y$ ) tends to infinity or to a singular point, found by limit tests, while curvature $\kappa = |y''|/(1+y'^2)^{3/2}$ measures how fast the tangent direction turns per unit arc length at a point.

The core idea

Vertical asymptotes occur where the function blows up. At $x=a$ ,

\lim_{x\to a^{\pm}} f(x) = \pm\infty,

where: $a$ is typically a zero of the denominator that is not cancelled by the numerator. Check both one-sided limits, since the curve can shoot to $+\infty$ on one side and $-\infty$ on the other.

Horizontal and oblique asymptotes describe the far-away behaviour. If $\lim_{x\to\pm\infty} f(x) = L$ (finite), then $y=L$ is a horizontal asymptote. Otherwise, an oblique (slant) asymptote $y=mx+c$ may exist:

m = \lim_{x\to \pm\infty} \frac{f(x)}{x}, \qquad c = \lim_{x\to\pm\infty}\big(f(x) - mx\big),

where: $m$ is the slope and $c$ the intercept of the limiting line; both limits must exist and be finite, and $m\neq 0$ (an $m=0$ result just reduces to the horizontal case). The two directions $x\to+\infty$ and $x\to-\infty$ must be checked separately, since they can give different lines. For a rational function, a quick shortcut before doing any limit work is comparing degrees: numerator degree less than or equal to denominator degree suggests a horizontal asymptote, exactly one more suggests an oblique one, and two or more usually means neither — though the limits should still be computed to confirm, since the shortcut only predicts which case applies, not the actual values of $L$ , $m$ or $c$ .

Curvature turns bending into a number. For $y=f(x)$ twice differentiable,

\kappa = \frac{|y''|}{\left(1+y'^2\right)^{3/2}}, \qquad \rho = \frac{1}{\kappa} = \frac{(1+y'^2)^{3/2}}{|y''|},

where: $y'$ and $y''$ are the first and second derivatives at the point, $\kappa$ is the curvature (units of inverse length), and $\rho$ is the radius of curvature — the radius of the circle that best hugs the curve there. A straight line has $\kappa=0$ everywhere; a circle of radius $r$ has constant $\kappa=1/r$ ; a parabola bends most sharply at its vertex and flattens as $|x|$ grows.

Figure

A curve rising steeply and settling toward a straight oblique line at large x, with a small osculating circle drawn tangent to the curve near its bend to show radius of curvature

Derivation ACurvature from the turning tangent

The goal is to derive $\kappa = |y''|/(1+y'^2)^{3/2}$ from first principles by measuring how fast the tangent line's direction angle changes per unit of arc length travelled along the curve, using only the chain rule and the derivative of the arctangent — no new machinery beyond what the concavity topic already introduced.

Step 1. Let $\varphi$ be the angle the tangent line makes with the positive $x$ -axis at a point on $y=f(x)$ . Since the slope of the tangent is $y'$ , we have $\tan\varphi = y'$ , so $\varphi = \arctan(y')$ .

Step 2. Curvature is defined as the rate at which this direction angle turns as we move along the curve, per unit arc length $s$ : $\kappa = \left|\dfrac{d\varphi}{ds}\right|$ . To compute it, use the chain rule to write $\dfrac{d\varphi}{ds} = \dfrac{d\varphi}{dx}\cdot\dfrac{dx}{ds}$ .

Step 3. Differentiate $\varphi = \arctan(y')$ with respect to $x$ : $\dfrac{d\varphi}{dx} = \dfrac{y''}{1+y'^2}$ , using the standard derivative of arctangent with $y'$ in place of the variable and the chain rule to bring down $y''$ .

Step 4. The arc-length element along $y=f(x)$ is $ds = \sqrt{1+y'^2}\,dx$ , so $\dfrac{dx}{ds} = \dfrac{1}{\sqrt{1+y'^2}}$ . Multiplying the two pieces from Steps 3 and 4 gives $\dfrac{d\varphi}{ds} = \dfrac{y''}{1+y'^2}\cdot\dfrac{1}{\sqrt{1+y'^2}} = \dfrac{y''}{(1+y'^2)^{3/2}}$ , and taking the absolute value (since curvature is a magnitude, not a signed turning rate) yields exactly $\kappa = |y''|/(1+y'^2)^{3/2}$ .

Worked example

Find all asymptotes of $f(x) = \dfrac{x^2+1}{x-2}$ , then find the curvature of $y=x^2$ at $(1,1)$ .

Step 1. Vertical asymptote: the denominator vanishes at $x=2$ , and the numerator there is $2^2+1=5\neq0$ , so $\lim_{x\to 2^+} f(x) = +\infty$ and $\lim_{x\to 2^-} f(x) = -\infty$ . Thus $x=2$ is a vertical asymptote.

Step 2. Horizontal asymptote: as $x\to\pm\infty$ , $f(x)\to\pm\infty$ since the numerator's degree exceeds the denominator's, so no horizontal asymptote exists — check for an oblique one instead.

Step 3. Oblique asymptote: $m = \lim_{x\to\infty} \dfrac{x^2+1}{x(x-2)} = \lim_{x\to\infty}\dfrac{1+1/x^2}{1-2/x} = 1$ , and then $c = \lim_{x\to\infty}\left(\dfrac{x^2+1}{x-2}-x\right) = \lim_{x\to\infty}\dfrac{2x+1}{x-2} = 2$ . So $y=x+2$ is the oblique asymptote (the same result holds as $x\to-\infty$ ). Summary: vertical asymptote $x=2$ , oblique asymptote $y=x+2$ , no horizontal asymptote.

Step 4. Now the curvature question. For $y=x^2$ : $y'=2x$ , so $y'(1)=2$ , and $y''=2$ everywhere. Substituting into $\kappa = |y''|/(1+y'^2)^{3/2}$ gives $\kappa = 2/(1+4)^{3/2} = 2/5^{3/2} = 2/(5\sqrt5) \approx 0.1789$ , and the radius of curvature is $\rho = 1/\kappa = 5\sqrt5/2 \approx 5.590$ . The point $(1,1)$ on $y=x^2$ bends like a circle of radius about $5.59$ units, far gentler than at the vertex $(0,0)$ , where $\rho=0.5$ .

Competitive stretch

At the vertex $(0,0)$ , $y'=0$ so $\kappa = |2|/(1+0)^{3/2} = 2$ and $\rho=0.5$ — a much tighter bend, confirming that this parabola is sharpest at its vertex and flattens on either side, exactly the trend the curvature formula predicts as $|y'|$ grows.

Common traps

Assuming "no horizontal asymptote" automatically means "there is an oblique one" — the oblique-asymptote limits for $m$ and $c$ must both exist and be finite; some functions have neither.
Checking only $x\to+\infty$ and forgetting $x\to-\infty$ (or vice versa) — the two directions can give different asymptotes.
Dropping the absolute value or the $3/2$ power in the curvature formula — both errors change the numerical answer, not just its sign.
Treating a vertical tangent ( $y'\to\infty$ ) as "no curvature" — the Cartesian formula breaks down there, but the curve can still bend with a finite, well-defined curvature; parametric or polar forms handle this case.
Confusing curvature with concavity — concavity (the sign of $y''$ ) only says whether the curve bends upward or downward, while curvature is always non-negative and measures how much bending happens, regardless of direction.

Ask the tutor: if you want to see the curvature formula rewritten for a curve given parametrically or in polar form, or want another asymptote-hunting example worked step by step, ask the AI tutor.

How do I know whether to look for a horizontal or an oblique asymptote first?

Always try the horizontal limit first, since it is simpler: compute $\lim_{x\to\pm\infty} f(x)$ . If it comes out finite, that finite value is the horizontal asymptote and you stop — an oblique asymptote cannot coexist with a horizontal one in the same direction. Only if the limit is infinite (typically when the numerator's degree exceeds the denominator's degree by exactly one, for rational functions) do you move on to computing $m$ and $c$ for a possible oblique asymptote.

Why does curvature use y double prime and not just y prime?

The first derivative $y'$ only tells you the slope of the tangent line at a point — it says nothing about how that slope is changing. Curvature is about the rate of change of the tangent's direction as you move along the curve, which is fundamentally a second-derivative idea, exactly like how acceleration (a second derivative of position) captures change in velocity, not velocity itself. That is why $y''$ sits in the numerator: no bending means $y''=0$ , which correctly gives $\kappa=0$ for a straight line.

Is the radius of curvature the same as the actual radius of the curve?

A general curve does not have one fixed radius the way a circle does — its bending changes from point to point. The radius of curvature $\rho$ at a specific point is the radius of the unique circle (called the osculating circle) that best approximates the curve right at that point, matching both its slope and its rate of bending there. Move to a different point on the curve and $\rho$ generally changes, as the worked example shows: $\rho=0.5$ at the vertex of $y=x^2$ but $\rho\approx5.59$ at $(1,1)$ .

Asymptotes and curvature

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads