MATH U101 · Doubt clinic · topic page

Curvature, unit tangent and normal, the osculating circle

Parametrise → v, |v| → T → dT/dt → κ, N — or the graph / parametric shortcut — then assemble the circle of curvature. Notes: B4; lesson: the recipe. Cards are collapsed — open one, or use the buttons. Newest at the bottom. Back to the clinic index.

Tutorial 4 Q3 · 13 SepOsculating circle of y = x² at x = 1What the circle is, why its radius is 1/κ and its centre P + ρN, which side, and how to write the parametrization.
Doubt 1 · Tutorial 4, Q3 · filed 13 Sep Find a parametrization of the osculating circle for the parabola y = x2 when x = 1.

Where this lives Notes B4 · κ and N · Lesson · the osculating-circle recipe · Companion T4 Q3 · Thomas §13.4 (circle of curvature) · kit: parametrising a circle, chain rule

The doubt, named

Three things usually stop this question, and all three are about what is being asked, not the algebra. (1) "Osculating circle" is a new name — what object is it? (2) "Parametrization" of a circle — what does the answer even look like? (3) Which side of the curve does the circle sit on, and how do you make the maths choose the side? Settle those three and the computation is a recipe you already own from the helix example.

Step 0What the osculating circle is, and what "parametrize it" means

At any point of a bending curve there is exactly one circle that hugs the curve there: it passes through the point, has the same tangent line, and bends exactly as much as the curve does. That is the osculating circle (Latin osculari, to kiss), also called the circle of curvature. It sits on the inside of the bend — the side the curve is turning toward.

"Bends exactly as much" pins down its size. A circle of radius ρ has curvature κ = 1/ρ everywhere (notes B4, anchor fact). So for the hugging circle to have the curve's curvature κ, its radius must be

ρ = 1κ

A circle is fixed by its centre and radius. The centre is at distance ρ from the point, perpendicular to the tangent, on the inside of the bend — and "perpendicular to the tangent, toward the bend" is exactly the direction of the principal unit normal N. Hence

centre C = P + ρ N

Finally, "a parametrization" of a circle with centre (h, k) and radius ρ is the pair of formulas

x(s) = h + ρ cos s, y(s) = k + ρ sin s, 0 ≤ s ≤ 2π
Unpack this step: why those two lines trace a circle

The point (h + ρ cos s, k + ρ sin s) sits at distance √(ρ²cos²s + ρ²sin²s) = ρ√(cos²s + sin²s) = ρ from (h, k) for every s, and as s runs from 0 to 2π it goes once round. (Kit item: parametrising a circle.)

So the whole question is: find κ, find N, assemble the centre, write the two lines. That is the plan for Steps 1–4.

Step 1Parametrise the parabola and run the §13.4 recipe

The recipe needs a vector function. A graph y = f(x) becomes one by using x itself as the clock: r(t) = ⟨t, t2⟩. The point x = 1 is then t = 1, i.e. P = (1, 1).

  1. Velocity and speed. v = ⟨1, 2t⟩, |v| = √(1 + 4t2).
  2. Unit tangent. T = v/|v| = ⟨1, 2t⟩ (1 + 4t2)−1/2.
  3. Turn rate dT/dt. Product rule on (vector) × (scalar):
    dTdt = ⟨0, 2⟩ (1 + 4t2)−1/2 + ⟨1, 2t⟩ · (−4t)(1 + 4t2)−3/2
    Unpack this step: the derivative of (1 + 4t²)−1/2

    Chain rule: outer power gives −½ (1 + 4t2)−3/2, inner derivative is 8t; product −4t(1 + 4t2)−3/2. (Kit: chain rule on a power; the daily drill has this shape.)

    Factor out (1 + 4t2)−3/2 (the smaller power) to combine:
    dTdt = [⟨0, 2⟩(1 + 4t2) − 4t⟨1, 2t⟩] (1 + 4t2)−3/2
    = ⟨−4t, 2 + 8t2 − 8t2⟩ (1 + 4t2)−3/2 = ⟨−4t, 2⟩(1 + 4t2)3/2
  4. Now put t = 1 (only now — differentiate first, substitute second). |v| = √5, and dT/dt = ⟨−4, 2⟩/53/2, whose magnitude is √(16 + 4)/53/2 = √20/(5√5) = 2√5/(5√5) = 2/5.
  5. Curvature. κ = |dT/dt| / |v| = (2/5)/√5 = 2/(5√5), so ρ = 1/κ = 5√5/2 ≈ 5.59.
  6. Normal. N = (dT/dt)/|dT/dt| is the direction of ⟨−4, 2⟩, i.e. N = ⟨−4, 2⟩/√20 = ⟨−2, 1⟩/√5. It points up-and-left: into the bowl of the parabola. ✓
Faster κ, same answer — the graph formula

Because the curve is a graph, the lesson's shortcut κ = |f″| / (1 + f′2)3/2 gives it in one line: f′(1) = 2, f″(1) = 2, κ = 2/53/2 = 2/(5√5) ✓. And N without differentiating T: rotate T = ⟨1, 2⟩/√5 by 90° to ⟨−2, 1⟩/√5 or ⟨2, −1⟩/√5, and keep the one pointing to the concave side (up, since f″ > 0). Use whichever route the question's wording allows; if it says "find T, N, κ", it wants the recipe above.

Step 2Assemble the centre

C = P + ρN = (1, 1) + 5√52 · ⟨−2, 1⟩√5 = (1, 1) + 52⟨−2, 1⟩
= (1 − 5, 1 + 52) = (−4, 72)

The √5 in ρ cancels the √5 in N — expect that; it happens because both came from the same |v|.

Step 3Write the parametrization, then audit it

x(s) = −4 + 5√52 cos s, y(s) = 72 + 5√52 sin s, 0 ≤ s ≤ 2π

Audit (one line, catches nearly every slip): the distance from the centre back to P must equal ρ. √((1 − (−4))2 + (1 − 7/2)2) = √(25 + 25/4) = √(125/4) = 5√5/2 ✓. If this fails, the error is in the sign of N or the arithmetic of Step 2.

The osculating circle (green) hugs the parabola at P from the inside of the bend. The dashed circle is what you get by taking −N: same radius, wrong side — it crosses the parabola instead of kissing it.
Classic traps · this exact question

1) Substituting t = 1 before differentiating T. T(1) is a constant vector; its derivative is zero and κ comes out 0. Differentiate the general T(t), then substitute. 2) Centre on the wrong side. C = P + ρN, never P − ρN and never along T. For a parabola opening upward the centre must have a larger y than P; here 7/2 > 1 ✓. 3) Stopping at the centre and radius. The question says parametrization: the two cos/sin lines with the range of s are the answer. 4) Radius = κ. ρ is 1/κ; a nearly flat curve (small κ) has a huge hugging circle, which is why ρ ≈ 5.6 is right for a point where the parabola is already fairly straight.

Try the twin: parametrize the osculating circle of y = x³ at x = 1. (Use the graph-formula route; then check the distance.)

P = (1, 1). f′(1) = 3, f″(1) = 6: κ = 6/(1 + 9)3/2 = 6/(10√10), so ρ = 10√10/6 = 5√10/3. T = ⟨1, 3⟩/√10; rotated candidates ⟨−3, 1⟩/√10 and ⟨3, −1⟩/√10; the curve is concave up at x = 1 (f″ > 0), so N = ⟨−3, 1⟩/√10. Centre: (1, 1) + (5√10/3)·⟨−3, 1⟩/√10 = (1, 1) + (5/3)⟨−3, 1⟩ = (−4, 8/3). Parametrization: x = −4 + (5√10/3) cos s, y = 8/3 + (5√10/3) sin s, 0 ≤ s ≤ 2π. Audit: distance from (−4, 8/3) to (1, 1) is √(25 + 25/9) = √(250/9) = 5√10/3 ✓.

Tutorial 4 Q4 · 13 SepWhere an ellipse bends mostShow κ is largest on the major axis and smallest on the minor: what a finished "show that" looks like.
Doubt 2 · Tutorial 4, Q4 · filed 13 Sep Show that the ellipse x = a cos t, y = b sin t, a > b > 0, has its largest curvature on its major axis and its smallest curvature on its minor axis.

Where this lives Notes B4 · κ · Lesson · κ of a parametric plane curve · Companion T4 Q4 · Thomas §13.4 exercises · kit: sin² + cos² = 1, derivatives of sin/cos

The doubt, named

This is a "show that", and the usual stall is not knowing what a finished answer looks like. Unpack the sentence: "on its major axis" means at the two points where the ellipse meets its major axis, (±a, 0); "on its minor axis" means at (0, ±b). So the job is: (1) get κ as a formula in t; (2) find which t makes it biggest and which makes it smallest; (3) name the points those t give. Nothing else is required — no calculus max/min test, as it turns out.

Step 1Which formula, and why the ellipse needs it

The ellipse is not a graph y = f(x) (it fails the vertical-line test), so the graph formula from Doubt 1 isn't available, and running the T-differentiation recipe with a non-constant speed |v| = √(a²sin²t + b²cos²t) is a page of algebra. The tool built for this — Thomas §13.4 exercises, derived in the lesson — is the parametric formula for a plane curve ⟨x(t), y(t)⟩:

κ(t) = |x′y″ − y′x″|(x′2 + y′2)3/2
Unpack this step: where the parametric formula comes from

Near any point where x′ ≠ 0 the curve is locally a graph, with slope dy/dx = y′/x′ and second derivative d²y/dx² = (x′y″ − y′x″)/x′3 (quotient rule, then divide by x′ again). Feed those into the graph formula |f″|/(1 + f′2)3/2 and the powers of x′ cancel to give the box. Check: a graph has x = t, so x′ = 1, x″ = 0, and the box collapses back to the graph formula.

It is worth a place on your formula sheet: sanity-check it once on the circle ⟨a cos t, a sin t⟩ — numerator a2, denominator a3, κ = 1/a ✓.

Step 2Compute κ(t) for the ellipse

  1. Derivatives. x = a cos t ⇒ x′ = −a sin t, x″ = −a cos t; y = b sin t ⇒ y′ = b cos t, y″ = −b sin t.
  2. Numerator. x′y″ − y′x″ = (−a sin t)(−b sin t) − (b cos t)(−a cos t) = ab sin2t + ab cos2t = ab. Watch the signs: minus times minus, then minus times a minus again — both terms come out positive. The identity kills the t: the top is a constant.
  3. Denominator. x′2 + y′2 = a2 sin2t + b2 cos2t.
κ(t) = ab(a2 sin2t + b2 cos2t)3/2

Now the one algebraic move the whole argument turns on: get the bracket in terms of a single trig function so you can see when it is big and when small. Replace cos2t = 1 − sin2t:

a2 sin2t + b2(1 − sin2t) = b2 + (a2 − b2) sin2t

Since a > b, the coefficient a2 − b2 is positive. So the bracket is b2 plus a positive number times sin2t: it is smallest when sin2t is smallest (0) and largest when sin2t is largest (1). That is the entire "max/min" analysis — no derivative test needed, because sin2t is known to live in [0, 1].

Step 3Read off the extremes and name the points

κ is a constant over the bracket, so κ is largest where the bracket is smallest, and vice versa (a fixed top divided by a smaller bottom is bigger).

sin²twherethe point (a cos t, b sin t)bracketκwhich axis
0t = 0, π(a, 0) and (−a, 0)b2 — smallestab/b3 = a/b2 — largestmajor axis (the long one, length 2a)
1t = π/2, 3π/2(0, b) and (0, −b)a2 — largestab/a3 = b/a2 — smallestminor axis (the short one, length 2b)

That sentence — "κ = ab/(b² + (a² − b²) sin²t)3/2 is maximised when sin²t = 0, i.e. at (±a, 0) on the major axis, and minimised when sin²t = 1, i.e. at (0, ±b) on the minor axis" — is the proof. Written out with the two values of κ, it earns full marks.

Sanity checks: put a = b and both extremes become 1/a, the circle ✓. With a = 2, b = 1: κmax = 2 (ρ = ½, a tight circle at the pointy ends) and κmin = ¼ (ρ = 4, a huge gentle circle at the flat top) — the figure below.

Where it bends: tight circle (ρ = ½) at the end of the major axis, gentle circle (ρ = 4, cropped) at the top of the minor axis.
κ(t) computed from the formula: peaks of 2 at t = 0, π (major axis), troughs of ¼ at π/2, 3π/2 (minor axis).
Classic traps · this exact question

1) "Long axis, so it bends less." The ellipse bends most at the ends of its long axis (the pointy ends) — think of a running track: the tightest turns are at the far ends. Largest κ ↔ smallest ρ. 2) Maximising the denominator by mistake. With a constant on top, κ is largest where the bottom is smallest. Write the "constant over bracket" sentence explicitly so the examiner sees you know which way it goes. 3) Rewriting with cos² instead. Also fine — you get a2 − (a2 − b2) cos2t, and now the bracket is smallest when cos²t = 1 — the same points, (±a, 0). Either single-function form works; a mixture of both doesn't. 4) Dropping the sign condition. The step "bracket grows with sin²t" needs a² − b² > 0, which is exactly why the question says a > b. Quote it.

Try the twin: for x = 3 cos t, y = 2 sin t, find the largest and smallest curvature and the points where each occurs.

a = 3, b = 2. Numerator ab = 6; bracket 4 + 5 sin2t. Largest κ at sin²t = 0, the points (±3, 0): κmax = 6/43/2 = 6/8 = 3/4 (matches a/b² = 3/4 ✓), ρ = 4/3. Smallest κ at sin²t = 1, the points (0, ±2): κmin = 6/93/2 = 6/27 = 2/9 (matches b/a² = 2/9 ✓), ρ = 9/2.