This deck covers the unit tangent and principal unit normal vectors, the arc length of a space curve as the integral of speed, and the arc-length parameter. It then gives three curvature formulas, the radius of curvature, the osculating circle, and the split of acceleration into its tangential and normal components. It targets the classic errors: skipping normalization before finding N, integrating the wrong quantity for arc length, and reaching for the wrong curvature formula.
Subject: Calculus III · 114 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
This week we turn a moving-point description of a curve into its shape: which way it points, how far it runs, and how sharply it bends.
1. Build the unit tangent vector and the principal unit normal vector at any point.
2. Set up and evaluate the arc length of a space curve as an integral of speed.
3. Compute curvature with whichever of the three formulas fits the information you are given, and find the radius of curvature and osculating circle.
4. Split acceleration into a tangential part (changing speed) and a normal part (changing direction).
Warm-up
Discussion prompt
Before we open Week 4 - Unit Tangent, Arc Length & Curvature: without looking back, what was the main idea of Week 3 - Vector-Valued Functions, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck covers vector-valued functions and the space curves they trace: domain, limits and continuity taken componentwise, and the derivative read as a tangent and velocity vector. It then gives the product rules for dot and cross products, defines smooth curves, integrates with a vector constant, works projectile motion, and shows why a curve of constant magnitude has its position vector orthogonal to its velocity. It targets three classic errors: differentiating a dot or cross product without the product rule, treating the constant of integration as a scalar, and forgetting that the derivative is taken component by component.
Concept
From last week: differentiating a vector-valued function gives a vector that is tangent to the curve and points in the direction of motion.
\[ \mathbf{r}'(t) = \langle x'(t),\, y'(t),\, z'(t)\rangle \]
It carries two pieces of information at once: a direction and a length. This week we want to separate those two ideas.
\[ \text{speed} = |\mathbf{r}'(t)|, \qquad \text{direction} = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} \]
Counterexample
Discussion prompt
From last week: differentiating a vector-valued function gives a vector that is tangent to the curve and points in the direction of motion.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Concept
Everything this week assumes the curve is smooth: the derivative exists, is continuous, and is never the zero vector on the interval.
smooth curve — A curve traced by r(t) whose derivative r'(t) is continuous and nonzero. Nonzero means the point never stops, so a genuine direction of travel always exists.
That nonzero requirement matters: dividing by the length of the derivative is only legal when that length is not zero.
Analogy
Discussion prompt
Explain Smooth curves by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Everything this week assumes the curve is smooth: the derivative exists, is continuous, and is never the zero vector on the interval.
Intuition
Picture driving along the curve. The velocity vector is an arrow that points where you are headed and is longer when you go faster.
If all you care about is which way you face, shrink that arrow down to length one. Speed drops out; pure heading remains.
That length-one heading arrow is the unit tangent vector, and it is the building block for everything else this week.
Explain it
Discussion prompt
Explain Direction is an arrow of length one to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Picture driving along the curve. The velocity vector is an arrow that points where you are headed and is longer when you go faster.
Concept
The unit tangent vector is the velocity vector rescaled to length one.
\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} \]
unit tangent vector — T(t) is the direction of motion as a unit vector. By construction its length is always exactly 1, so it records heading only, not speed.
To build it: differentiate, take the magnitude, then divide. The divide step is the one people skip.
Definition probe
Sort into buckets
Every line below is part of the definition of smooth curve or of unit tangent vector — one or the other, never both. Put each where it belongs.
Ranking
Put in order
Put the moves of Worked example: T(t) for a circle into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The derivative of a vector function is taken in each slot separately.
Worked example
Find the unit tangent vector for the circle of radius three.
\[ \mathbf{r}(t) = \langle 3\cos t,\, 3\sin t\rangle \]
Differentiate component by component
Why: The derivative of a vector function is taken in each slot separately.
\[ \mathbf{r}'(t) = \langle -3\sin t,\, 3\cos t\rangle \]
Take the magnitude of the derivative
Why: Factor out 3 and use the Pythagorean identity for sine and cosine.
\[ |\mathbf{r}'(t)| = \sqrt{9\sin^2 t + 9\cos^2 t} = \sqrt{9} = 3 \]
Divide the derivative by its magnitude
Why: This rescales the velocity vector to length one.
\[ \mathbf{T}(t) = \frac{1}{3}\langle -3\sin t,\, 3\cos t\rangle = \langle -\sin t,\, \cos t\rangle \]
Verify the result has length one
Why: A unit tangent must satisfy |T| = 1; here sin squared plus cos squared is 1.
\[ |\mathbf{T}(t)| = \sqrt{\sin^2 t + \cos^2 t} = 1 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "T(t) for a circle", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A unit tangent must satisfy |T| = 1; here sin squared plus cos squared is 1.
Step zero
Discussion prompt
Worked example: T(t) for a helix — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Differentiate each component
Answer:
Worked example
Find the unit tangent vector for the circular helix.
\[ \mathbf{r}(t) = \langle \cos t,\, \sin t,\, t\rangle \]
Differentiate each component
Why: The z-component t differentiates to 1, adding a constant vertical piece to the velocity.
\[ \mathbf{r}'(t) = \langle -\sin t,\, \cos t,\, 1\rangle \]
Compute the magnitude
Why: The sine and cosine squares add to 1, then the constant 1 from the z-slot.
\[ |\mathbf{r}'(t)| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2} \]
Divide to normalize
Why: Every component is scaled by one over root two; the length is constant here, which is special to the helix.
\[ \mathbf{T}(t) = \frac{1}{\sqrt{2}}\langle -\sin t,\, \cos t,\, 1\rangle \]
Verify length one
Why: The squared components are one-half of (sin squared plus cos squared plus 1), which is one-half of 2.
\[ |\mathbf{T}(t)| = \sqrt{\tfrac{1}{2}(\sin^2 t + \cos^2 t + 1)} = \sqrt{\tfrac{1}{2}\cdot 2} = 1 \checkmark \]
Concept
Here is the fact that makes the normal vector work. If a vector always has constant length, then its derivative is perpendicular to it.
\[ |\mathbf{T}(t)| = 1 \;\Longrightarrow\; \mathbf{T}(t)\cdot \mathbf{T}(t) = 1 \]
Differentiate both sides using the dot-product rule
Why: The derivative of a constant is zero, and the dot product of T with itself differentiates to twice T dotted with T prime.
\[ 2\,\mathbf{T}(t)\cdot \mathbf{T}'(t) = 0 \;\Longrightarrow\; \mathbf{T}\cdot \mathbf{T}' = 0 \]
So the derivative of the unit tangent is perpendicular to the unit tangent. That perpendicular direction is exactly where the curve is turning.
Intuition
Drive around a bend. Your heading is the tangent. But there is also a sideways pull toward the inside of the curve; that is the direction the tangent is rotating toward.
Figure (svg): A circular arc with a green tangent arrow along it and a red normal arrow pointing inward toward the center
The green arrow is the tangent; the red arrow, pointing into the turn, is the principal unit normal.
Concept
The principal unit normal is the unit tangent's rate of change, rescaled to length one.
\[ \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|} \]
principal unit normal — N(t) is the unit vector in the direction of T'(t). It is perpendicular to T and points toward the concave side of the curve, the side it is bending into.
Notice you differentiate T, the already-normalized tangent, not the raw velocity. That order is the whole trap on the next slide.
Estimation
Predict first
Continue the radius-three circle, where we already found the unit tangent.
Commit before you compute: what does Worked example: N(t) for the circle come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify N is a unit vector pointing to the center
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Its length is 1, and at every t it points from the circle straight toward the origin, the concave side.
Worked example
Continue the radius-three circle, where we already found the unit tangent.
\[ \mathbf{T}(t) = \langle -\sin t,\, \cos t\rangle \]
Differentiate the unit tangent
Why: Work with T, which is already length one, not with the raw velocity vector.
\[ \mathbf{T}'(t) = \langle -\cos t,\, -\sin t\rangle \]
Take the magnitude of T prime
Why: Again the Pythagorean identity collapses the squares to 1.
\[ |\mathbf{T}'(t)| = \sqrt{\cos^2 t + \sin^2 t} = 1 \]
Divide to get N
Why: Since the magnitude is 1, N equals T prime here.
\[ \mathbf{N}(t) = \langle -\cos t,\, -\sin t\rangle \]
Verify N is a unit vector pointing to the center
Why: Its length is 1, and at every t it points from the circle straight toward the origin, the concave side.
\[ |\mathbf{N}(t)| = 1, \qquad \mathbf{T}\cdot\mathbf{N} = \sin t\cos t - \cos t\sin t = 0 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "N(t) for the circle", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Its length is 1, and at every t it points from the circle straight toward the origin, the concave side.
Fill the middle
Fill in the blanks
From Worked example: N(t) for the helix — finish the line. Write what belongs on the right of the equals sign before you look.
\mathbf\frac{1}{\sqrt{2}}\langle -\sin t,\, \cos t,\, 1\rangle(t) = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The one over root two is a constant multiplier; the constant 1 in the last slot differentiates to 0.
Worked example
Use the helix unit tangent from earlier.
\[ \mathbf{T}(t) = \frac{1}{\sqrt{2}}\langle -\sin t,\, \cos t,\, 1\rangle \]
Differentiate T
Why: The one over root two is a constant multiplier; the constant 1 in the last slot differentiates to 0.
\[ \mathbf{T}'(t) = \frac{1}{\sqrt{2}}\langle -\cos t,\, -\sin t,\, 0\rangle \]
Compute the magnitude
Why: The one over root two comes out; the remaining squares add to 1.
\[ |\mathbf{T}'(t)| = \frac{1}{\sqrt{2}}\sqrt{\cos^2 t + \sin^2 t} = \frac{1}{\sqrt{2}} \]
Divide T prime by its magnitude
Why: Multiplying by root two cancels the one over root two, leaving a clean unit vector.
\[ \mathbf{N}(t) = \langle -\cos t,\, -\sin t,\, 0\rangle \]
Verify N is a unit vector
Why: The z-component is 0, and the horizontal squares add to 1.
\[ |\mathbf{N}(t)| = \sqrt{\cos^2 t + \sin^2 t + 0} = 1 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "N(t) for the helix", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The z-component is 0, and the horizontal squares add to 1.
Trap
Tempting shortcut: differentiate the raw velocity and normalize that to get N.
\[ \text{(wrong)}\quad \mathbf{N} \stackrel{?}{=} \frac{\mathbf{r}''(t)}{|\mathbf{r}''(t)|} \]
For the helix that gives the right answer by luck, but in general the second derivative is not perpendicular to the tangent, so this is not the normal.
\[ \mathbf{r}''\ \text{need not be}\ \perp\ \mathbf{r}' \]
Normalize first to get T, then differentiate T, then normalize again.
\[ \mathbf{T} = \frac{\mathbf{r}'}{|\mathbf{r}'|}, \qquad \mathbf{N} = \frac{\mathbf{T}'}{|\mathbf{T}'|} \]
Because T has constant length one, its derivative is guaranteed perpendicular to T. That guarantee is exactly what makes N a genuine normal direction.
Concept
At each point of a space curve, T and N span the plane the curve is momentarily turning in. A third unit vector completes a right-handed frame.
\[ \mathbf{B}(t) = \mathbf{T}(t)\times\mathbf{N}(t) \]
binormal vector — B = T cross N. Together T, N, B form the moving TNB frame that travels with the point. B is normal to the plane containing T and N.
We will meet B again briefly with torsion. For now, keep T and N front of mind.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of smooth curve, unit tangent vector, principal unit normal, binormal vector as Week 4 - Unit Tangent, Arc Length & Curvature uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Concept
Next question about a curve's shape: how far does the point travel from one time to another? That distance is the arc length.
The trick is the same one from single-variable calculus: chop the path into tiny straight pieces, add their lengths, and take a limit.
Each tiny piece is traveled in a tiny time, and its length is speed multiplied by that tiny time.
Intuition
If you drive at a known speed and record it every instant, the total distance is the speed accumulated over the whole trip.
\[ ds = |\mathbf{r}'(t)|\,dt \]
Here the speed is the magnitude of the velocity vector. Summing these tiny distances is exactly an integral of speed.
Concept
The arc length of a smooth curve from one time value to another is the integral of the speed.
\[ L = \int_a^b |\mathbf{r}'(t)|\,dt = \int_a^b \sqrt{[x'(t)]^2 + [y'(t)]^2 + [z'(t)]^2}\,dt \]
The integrand is the magnitude of the derivative, a single positive number at each time, not the vector itself and not its components.
Missing information
Discussion prompt
Find the length of one full turn of the helix, for t from 0 to two pi.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
We already found this speed when building the unit tangent.
Worked example
Find the length of one full turn of the helix, for t from 0 to two pi.
\[ \mathbf{r}(t) = \langle \cos t,\, \sin t,\, t\rangle, \quad 0 \le t \le 2\pi \]
Differentiate and take the magnitude
Why: We already found this speed when building the unit tangent.
\[ |\mathbf{r}'(t)| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2} \]
Set up the arc-length integral
Why: Integrate the speed over the given time interval.
\[ L = \int_0^{2\pi} \sqrt{2}\,dt \]
Evaluate the integral
Why: The integrand is constant, so this is just the constant times the length of the interval.
\[ L = \sqrt{2}\,(2\pi - 0) = 2\pi\sqrt{2} \]
Verify the size is reasonable
Why: One loop rises a height of two pi while circling a radius-one circle of circumference two pi, so the slanted length should exceed two pi; and 2 pi root 2 is about 8.9, comfortably more than 2 pi which is about 6.3.
\[ 2\pi\sqrt{2} \approx 8.89 > 2\pi \approx 6.28 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "arc length of a helix", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: One loop rises a height of two pi while circling a radius-one circle of circumference two pi, so the slanted length should exceed two pi; and 2 pi root 2 is about 8.9, comfortably more than 2 pi which is about 6.3.
Step zero
Discussion prompt
Worked example: arc length with a square root that simplifies — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Differentiate each component
Answer:
Worked example
Find the length of this plane curve for t from 0 to 3.
\[ \mathbf{r}(t) = \left\langle t,\, \tfrac{2}{3}t^{3/2}\right\rangle, \quad 0 \le t \le 3 \]
Differentiate each component
Why: The three-halves power drops to a one-half power, and the fractions cancel.
\[ \mathbf{r}'(t) = \langle 1,\, t^{1/2}\rangle \]
Take the magnitude
Why: Square each component and add under the root; the t to the one-half squared is just t.
\[ |\mathbf{r}'(t)| = \sqrt{1 + t} \]
Integrate the speed
Why: This is a basic power-rule integral after a shift; the antiderivative of the square root of one plus t is two-thirds times one plus t to the three-halves.
\[ L = \int_0^3 \sqrt{1+t}\,dt = \left[\tfrac{2}{3}(1+t)^{3/2}\right]_0^3 \]
Evaluate at the limits
Why: At t equals 3 the inside is 4, whose three-halves power is 8; at t equals 0 the inside is 1.
\[ L = \tfrac{2}{3}(8) - \tfrac{2}{3}(1) = \tfrac{16}{3} - \tfrac{2}{3} = \tfrac{14}{3} \]
Verify by a rough bound
Why: The speed root of one plus t runs from 1 up to 2 over the interval of length 3, so the length lies between 3 and 6; and 14 thirds is about 4.67, right in that range.
\[ 3 < \tfrac{14}{3} \approx 4.67 < 6 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "arc length with a square root that simplifies", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The speed root of one plus t runs from 1 up to 2 over the interval of length 3, so the length lies between 3 and 6; and 14 thirds is about 4.67, right in that range.
Trap
A very common slip is to integrate the velocity components, or to forget the square root and integrate the sum of squares.
\[ \text{(wrong)}\quad L \stackrel{?}{=} \int_a^b \big([x']^2+[y']^2+[z']^2\big)\,dt \]
For the helix that would give the integral of 2, namely four pi, which is not a length at all.
\[ \int_0^{2\pi} 2\,dt = 4\pi \ne 2\pi\sqrt{2} \]
Arc length integrates the speed: the square root of the sum of the squared components.
\[ L = \int_a^b \sqrt{[x']^2+[y']^2+[z']^2}\,dt \]
The square root is essential. Speed is one positive number per instant; without the root you are adding up squared speeds, which has the wrong units and the wrong value.
\[ \int_0^{2\pi} \sqrt{2}\,dt = 2\pi\sqrt{2} \checkmark \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
Arc length integrates the speed: the square root of the sum of the squared components.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Concept
If we let the upper limit float, arc length becomes a function of time: the distance traveled so far.
\[ s(t) = \int_a^t |\mathbf{r}'(u)|\,du \]
Differentiate by the Fundamental Theorem of Calculus
Why: The derivative of an accumulated integral is its integrand, evaluated at the moving limit.
\[ \frac{ds}{dt} = |\mathbf{r}'(t)| \]
In words: the rate at which distance piles up is the speed. That single equation links time and distance along the curve.
Translation
\( \frac{ds}{dt} = |\mathbf{r}'(t)| \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Concept
Sometimes we want the curve labeled by distance traveled instead of by time. Solve for t in terms of s and substitute.
arc-length parameter — The variable s that measures distance along the curve from a fixed starting point. When a curve is written in terms of s, moving one unit of s means moving one unit of distance.
The payoff: a curve parametrized by arc length is traversed at unit speed, so its velocity vector is already the unit tangent.
\[ \left|\frac{d\mathbf{r}}{ds}\right| = 1 \]
Fill the middle
Fill in the blanks
From Worked example: reparametrize the helix by arc length — finish the line. Write what belongs on the right of the equals sign before you look.
\mathbf\langle \cos t,\, \sin t,\, t\rangle(t) = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The speed is constant at root two, so accumulated distance is root two times t.
Worked example
Rewrite the helix using arc length measured from t equals 0.
\[ \mathbf{r}(t) = \langle \cos t,\, \sin t,\, t\rangle \]
Compute the arc-length function
Why: The speed is constant at root two, so accumulated distance is root two times t.
\[ s(t) = \int_0^t \sqrt{2}\,du = \sqrt{2}\,t \]
Solve for t in terms of s
Why: Invert the relationship so time is expressed through distance.
\[ t = \frac{s}{\sqrt{2}} \]
Substitute back into r
Why: Replace every t by s over root two to label the curve by distance.
\[ \mathbf{r}(s) = \left\langle \cos\tfrac{s}{\sqrt{2}},\, \sin\tfrac{s}{\sqrt{2}},\, \tfrac{s}{\sqrt{2}}\right\rangle \]
Verify the speed in s is one
Why: Differentiate with respect to s; each derivative carries a factor one over root two, and the magnitude works out to exactly 1.
\[ \left|\frac{d\mathbf{r}}{ds}\right| = \sqrt{\tfrac{1}{2}\sin^2 + \tfrac{1}{2}\cos^2 + \tfrac{1}{2}} = \sqrt{\tfrac{1}{2}+\tfrac{1}{2}} = 1 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "reparametrize the helix by arc length", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Differentiate with respect to s; each derivative carries a factor one over root two, and the magnitude works out to exactly 1.
Pattern
1. Differentiate the position vector
Why: This gives the velocity r prime, tangent to the curve.
2. Divide velocity by its magnitude to get T
Why: Normalizing strips out speed and leaves pure direction.
3. Differentiate T, then divide by its magnitude to get N
Why: Because T has constant length, T prime is perpendicular to T, so N is a true normal.
4. Sanity-check both are unit and perpendicular
Why: Confirm each magnitude is 1 and that T dotted with N is 0.
Check
Apply the recipe carefully. Watch the normalization step.
\[ \mathbf{r}(t) = \langle 3\cos t,\, 3\sin t\rangle \]
Check your understanding
What is the unit tangent vector T(t)?
Answer: A
Why: Differentiate to get velocity with entries negative three sine t and three cosine t, whose magnitude is 3. Dividing by 3 gives entries negative sine t and cosine t, a unit vector. Verified: its magnitude is 1.
Concept
Now the headline idea of the week: curvature measures how sharply a curve bends, independent of how fast you drive along it.
A straight road has zero curvature. A tight hairpin has large curvature. A gentle bend has small curvature.
The key phrase is per unit distance: curvature is how fast the direction turns as you move one unit of arc length, not one unit of time.
Intuition
Two drivers take the same curve, one fast and one slow. They turn the steering wheel at different rates in time, yet the road bends the same amount. Curvature captures the road, not the driver.
That is why the definition uses distance, not time: it belongs to the curve itself.
\[ \kappa = \left|\frac{d\mathbf{T}}{ds}\right| \]
Concept
Curvature is the magnitude of the rate of change of the unit tangent with respect to arc length.
\[ \kappa = \left|\frac{d\mathbf{T}}{ds}\right| \]
curvature — The rate at which the unit tangent direction changes per unit of arc length. Written with the Greek letter kappa, it is always nonnegative and has units of one over length.
Differentiating with respect to s directly is awkward, so we convert to a formula in t using the chain rule on the next slide.
Concept
By the chain rule, the derivative with respect to s equals the derivative with respect to t divided by the speed.
\[ \frac{d\mathbf{T}}{ds} = \frac{d\mathbf{T}/dt}{ds/dt} = \frac{\mathbf{T}'(t)}{|\mathbf{r}'(t)|} \]
Taking magnitudes gives the first working formula.
\[ \kappa = \frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|} \]
Best when you already have T and its derivative in hand, or when T is simple to compute.
Estimation
Predict first
Find the curvature of a circle of radius a, and confirm it matches intuition.
Commit before you compute: what does Worked example: curvature of a circle come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify against intuition
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. A big circle bends gently and a small circle bends sharply; curvature one over a matches, since larger radius gives smaller curvature.
Worked example
Find the curvature of a circle of radius a, and confirm it matches intuition.
\[ \mathbf{r}(t) = \langle a\cos t,\, a\sin t\rangle,\quad a>0 \]
Find the speed
Why: Differentiate and take the magnitude; the radius a factors out.
\[ \mathbf{r}'(t) = \langle -a\sin t,\, a\cos t\rangle, \quad |\mathbf{r}'(t)| = a \]
Find the unit tangent and its derivative
Why: Divide velocity by a, then differentiate the result.
\[ \mathbf{T}(t) = \langle -\sin t,\, \cos t\rangle, \quad \mathbf{T}'(t) = \langle -\cos t,\, -\sin t\rangle \]
Apply formula one
Why: The magnitude of T prime is 1; divide by the speed a.
\[ \kappa = \frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|} = \frac{1}{a} \]
Verify against intuition
Why: A big circle bends gently and a small circle bends sharply; curvature one over a matches, since larger radius gives smaller curvature.
\[ \kappa = \tfrac{1}{a} \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "curvature of a circle", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A big circle bends gently and a small circle bends sharply; curvature one over a matches, since larger radius gives smaller curvature.
Concept
Computing T and then differentiating is often messy. This second formula uses only the first and second derivatives of the position vector.
\[ \kappa = \frac{|\mathbf{r}'(t)\times\mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3} \]
This is usually the most efficient choice for a space curve given as r of t, because you avoid normalizing and differentiating a messy unit tangent.
Note the cube on the speed in the denominator. Forgetting the third power is a frequent error.
Ranking
Put in order
Put the moves of Worked example: curvature of the helix (formula two) into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Differentiate twice, component by component.
Worked example
Find the curvature of the helix using the cross-product formula.
\[ \mathbf{r}(t) = \langle \cos t,\, \sin t,\, t\rangle \]
Compute the first two derivatives
Why: Differentiate twice, component by component.
\[ \mathbf{r}'(t) = \langle -\sin t,\, \cos t,\, 1\rangle, \quad \mathbf{r}''(t) = \langle -\cos t,\, -\sin t,\, 0\rangle \]
Take the cross product
Why: Expand the determinant; the identity sine squared plus cosine squared equals one cleans up the last component.
\[ \mathbf{r}'\times\mathbf{r}'' = \langle \sin t,\, -\cos t,\, 1\rangle \]
Take magnitudes
Why: Both the cross product and the velocity have the same tidy magnitude root two.
\[ |\mathbf{r}'\times\mathbf{r}''| = \sqrt{2}, \qquad |\mathbf{r}'| = \sqrt{2} \]
Apply formula two
Why: Cube the speed: root two cubed is two root two. Divide.
\[ \kappa = \frac{\sqrt{2}}{(\sqrt{2})^3} = \frac{\sqrt{2}}{2\sqrt{2}} = \frac{1}{2} \]
Verify it is constant and positive
Why: The helix bends the same amount everywhere, so a constant curvature of one-half is exactly what we expect.
\[ \kappa = \tfrac{1}{2} \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "curvature of the helix (formula two)", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The helix bends the same amount everywhere, so a constant curvature of one-half is exactly what we expect.
Pattern
Three formulas, one curvature. Choose by what you are given.
| You are given | Best formula |
|---|---|
| A curve as y equals f of x | the plane-curve formula in y prime and y double prime |
| A space curve r of t, need speed anyway | the cross-product formula with the cube of the speed |
| T already computed or very simple | T prime over r prime |
Match the formula to the information
Why: All three give the same number; the fastest route depends on whether you have a function y of x, a parametrization r of t, or the unit tangent in hand.
Comparison
Comparison matrix
From Recipe: pick the curvature formula: refill the Best formula column from what you know. The rest of the table is as it appeared.
| You are given | Best formula |
|---|---|
| A curve as y equals f of x | the plane-curve formula in y prime and y double prime |
| A space curve r of t, need speed anyway | the cross-product formula with the cube of the speed |
| T already computed or very simple | T prime over r prime |
Elimination
Eliminate the wrong options
Which formula is the most direct route to curvature here?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: With a space curve given as r of t, the cross-product formula needs only r prime and r double prime, which are quick to compute, and reuses the speed you already need. It avoids normalizing a messy unit tangent.
Check
You are given a space curve as a position vector r of t and asked for its curvature. You will need the speed for other parts of the problem anyway.
Check your understanding
Which formula is the most direct route to curvature here?
Answer: A
Why: With a space curve given as r of t, the cross-product formula needs only r prime and r double prime, which are quick to compute, and reuses the speed you already need. It avoids normalizing a messy unit tangent.
Concept
When the curve is the graph of a function in the plane, there is a shortcut in terms of ordinary derivatives.
\[ \kappa = \frac{|y''|}{\big(1 + (y')^2\big)^{3/2}} \]
It comes from the cross-product formula applied to the parametrization with x as the parameter. Use it only for a graph y equals f of x, not for a general space curve.
The absolute value on the top keeps curvature nonnegative; the three-halves power on the bottom is easy to drop, so write it deliberately.
Hypothesis
Predict first
Worked example: curvature of a parabola is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Take the first and second derivatives
Why: Plain single-variable differentiation of x squared.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Find the curvature of this parabola, then evaluate it at the vertex.
\[ y = x^2 \]
Take the first and second derivatives
Why: Plain single-variable differentiation of x squared.
\[ y' = 2x, \qquad y'' = 2 \]
Substitute into the plane-curve formula
Why: Top is the absolute value of the second derivative; bottom is one plus the first derivative squared, raised to the three-halves.
\[ \kappa(x) = \frac{2}{\big(1 + 4x^2\big)^{3/2}} \]
Evaluate at the vertex
Why: At x equals 0 the denominator is one to the three-halves, which is 1.
\[ \kappa(0) = \frac{2}{(1+0)^{3/2}} = 2 \]
Verify the behavior away from the vertex
Why: As x grows, the denominator grows, so curvature decreases; the parabola is sharpest at the vertex, which matches a curvature that is largest there.
\[ \kappa(x) \to 0 \ \text{as}\ x \to \pm\infty \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "curvature of a parabola", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: As x grows, the denominator grows, so curvature decreases; the parabola is sharpest at the vertex, which matches a curvature that is largest there.
Worked example
Treat the same parabola as a parametrized plane curve and confirm the two formulas agree.
\[ \mathbf{r}(t) = \langle t,\, t^2\rangle \]
Differentiate twice
Why: The parameter t plays the role of x here.
\[ \mathbf{r}'(t) = \langle 1,\, 2t\rangle, \qquad \mathbf{r}''(t) = \langle 0,\, 2\rangle \]
Use the planar cross-product magnitude
Why: For plane vectors the cross product points out of the plane with magnitude the absolute value of x prime times y double prime minus y prime times x double prime.
\[ |\mathbf{r}'\times\mathbf{r}''| = |(1)(2) - (2t)(0)| = 2 \]
Divide by the cube of the speed
Why: The speed is the root of one plus four t squared; cube it.
\[ \kappa = \frac{2}{\big(1 + 4t^2\big)^{3/2}} \]
Verify it equals the plane-curve result
Why: Setting t equal to x reproduces the earlier expression exactly, so both formulas give the same curvature.
\[ \frac{2}{(1+4t^2)^{3/2}} = \frac{|y''|}{(1+(y')^2)^{3/2}} \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the plane formula matches the cross-product form", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Setting t equal to x reproduces the earlier expression exactly, so both formulas give the same curvature.
Trap
Reaching for the plane-curve formula on a genuine space curve, or forgetting the cube on the speed in the cross-product formula.
\[ \text{(wrong)}\quad \kappa \stackrel{?}{=} \frac{|\mathbf{r}'\times\mathbf{r}''|}{|\mathbf{r}'|^2} \]
For the helix that returns root two over two, that is one over root two, instead of the correct one-half. The wrong power of the speed silently changes the answer.
\[ \frac{\sqrt{2}}{(\sqrt{2})^2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \ne \frac{1}{2} \]
Cube the speed in the cross-product formula, and save the plane formula for graphs y equals f of x.
\[ \kappa = \frac{|\mathbf{r}'\times\mathbf{r}''|}{|\mathbf{r}'|^3} \]
For the helix the correct denominator is root two cubed, which is two root two, giving one-half.
\[ \frac{\sqrt{2}}{2\sqrt{2}} = \frac{1}{2} \checkmark \]
Check
For the helix you have the two magnitudes ready. Apply the cross-product formula with care about the power.
\[ |\mathbf{r}'\times\mathbf{r}''| = \sqrt{2}, \qquad |\mathbf{r}'| = \sqrt{2} \]
Check your understanding
What is the curvature of the helix?
Answer: A
Why: Curvature is the cross-product magnitude over the cube of the speed: root two divided by root two cubed. Root two cubed is two root two, so the result is one-half. This matches the constant bending of the helix.
Concept
Curvature is one over a length. Flip it and you get an actual length: the radius of curvature.
\[ \rho = \frac{1}{\kappa} \]
radius of curvature — The reciprocal of curvature. It is the radius of the circle that bends at the same rate as the curve at that point. Large radius means gentle bend; small radius means sharp bend.
This is why a circle of radius a has curvature one over a: its radius of curvature is just a, everywhere.
Step zero
Discussion prompt
Worked example: radius of curvature at a vertex — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Recall the curvature there
Answer:
Worked example
Find the radius of curvature of the parabola at its vertex.
\[ y = x^2 \ \text{at}\ x = 0 \]
Recall the curvature there
Why: We found the curvature at the vertex earlier.
\[ \kappa(0) = 2 \]
Take the reciprocal
Why: Radius of curvature is one over curvature.
\[ \rho = \frac{1}{\kappa} = \frac{1}{2} \]
Verify the units and size make sense
Why: Curvature 2 is fairly sharp, so the matching circle should be small; a radius of one-half is indeed small, consistent with a tightly bending vertex.
\[ \rho = \tfrac{1}{2} \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "radius of curvature at a vertex", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Curvature 2 is fairly sharp, so the matching circle should be small; a radius of one-half is indeed small, consistent with a tightly bending vertex.
Concept
At a point, the circle that best hugs the curve is the osculating circle: it shares the curve's tangent and its curvature there.
\[ \text{radius} = \rho = \frac{1}{\kappa}, \qquad \text{center} = \mathbf{r} + \rho\,\mathbf{N} \]
osculating circle — The circle of radius one over kappa whose center lies a distance one over kappa from the point in the direction of the principal unit normal N. It touches the curve to second order.
The center sits on the concave side, along N, exactly the radius of curvature away.
Intuition
A tangent line matches a curve's direction. The osculating circle does one better: it matches the direction and the bending.
Think of it as the circle you would trace if the curve froze its steering at that instant and you kept driving.
Sharper bend, smaller osculating circle; straighter piece, enormous circle that flattens toward the tangent line.
Fill the middle
Fill in the blanks
From Worked example: osculating circle at a vertex — finish the line. Write what belongs on the right of the equals sign before you look.
x^2 + \left(y - \tfrac\tfrac{1}{4} \checkmark___\right)^2 = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Curvature at the vertex is 2, so the radius is one-half.
Worked example
Find the osculating circle of the parabola at the origin.
\[ y = x^2 \ \text{at}\ (0,0) \]
Get the radius from the curvature
Why: Curvature at the vertex is 2, so the radius is one-half.
\[ \rho = \frac{1}{\kappa} = \frac{1}{2} \]
Find the normal direction at the vertex
Why: At the vertex the curve is concave up, so the principal normal points straight up in the positive y-direction.
\[ \mathbf{N} = \langle 0,\, 1\rangle \]
Locate the center
Why: Move from the point a distance rho along N: from the origin, up one-half.
\[ \text{center} = (0,0) + \tfrac{1}{2}\langle 0,1\rangle = \left(0,\, \tfrac{1}{2}\right) \]
Verify the circle passes through the point with the right radius
Why: The distance from the center at height one-half to the origin is one-half, matching the radius.
\[ x^2 + \left(y - \tfrac{1}{2}\right)^2 = \tfrac{1}{4} \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "osculating circle at a vertex", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The distance from the center at height one-half to the origin is one-half, matching the radius.
Concept
Acceleration is the rate of change of velocity. Velocity can change in two independent ways: its length can change, or its direction can change.
Speeding up or slowing down changes the length. Turning changes the direction. Real motion usually does both at once.
So it is natural to split the acceleration vector into a piece along T and a piece along N.
\[ \mathbf{a} = a_{\mathbf{T}}\,\mathbf{T} + a_{\mathbf{N}}\,\mathbf{N} \]
Intuition
The tangential component is your gas pedal and brake: it changes how fast you go. The normal component is your steering wheel: it changes where you point.
Drive straight at changing speed and all acceleration is tangential. Drive a circle at steady speed and all acceleration is normal.
There is no acceleration piece along the binormal: a curve does not accelerate sideways out of its own turning plane.
Concept
The tangential component is the rate of change of speed, which works out to velocity dotted with acceleration over speed.
\[ a_{\mathbf{T}} = \frac{d}{dt}\,|\mathbf{v}| = \frac{\mathbf{v}\cdot\mathbf{a}}{|\mathbf{v}|} \]
It can be positive (speeding up), negative (slowing down), or zero (steady speed). It is the only component that changes speed.
Concept
The normal component measures acceleration due to turning. It is the magnitude of velocity cross acceleration, divided by speed.
\[ a_{\mathbf{N}} = \frac{|\mathbf{v}\times\mathbf{a}|}{|\mathbf{v}|} \]
It is always nonnegative, since the normal always points into the turn. A larger normal component means a sharper or faster turn.
Concept
Because T and N are perpendicular unit vectors, the two components are the legs of a right triangle whose hypotenuse is the full acceleration.
\[ a_{\mathbf{T}}^2 + a_{\mathbf{N}}^2 = |\mathbf{a}|^2 \]
This gives a free check on any decomposition: square the two components, add, and compare with the squared magnitude of the acceleration vector.
Explain it
Discussion prompt
Explain The two components and the total acceleration to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Because T and N are perpendicular unit vectors, the two components are the legs of a right triangle whose hypotenuse is the full acceleration.
Missing information
Discussion prompt
Find the tangential and normal components for this path at time t equals 1.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Differentiate once for velocity, again for acceleration.
Worked example
Find the tangential and normal components for this path at time t equals 1.
\[ \mathbf{r}(t) = \langle t,\, t^2\rangle \]
Find velocity and acceleration
Why: Differentiate once for velocity, again for acceleration.
\[ \mathbf{v} = \langle 1,\, 2t\rangle, \quad \mathbf{a} = \langle 0,\, 2\rangle; \quad \text{at } t=1:\ \mathbf{v}=\langle 1,2\rangle,\ \mathbf{a}=\langle 0,2\rangle \]
Compute the tangential component
Why: Dot velocity with acceleration and divide by speed; the speed is the root of one plus four, which is root five.
\[ a_{\mathbf{T}} = \frac{\mathbf{v}\cdot\mathbf{a}}{|\mathbf{v}|} = \frac{(1)(0)+(2)(2)}{\sqrt{5}} = \frac{4}{\sqrt{5}} \]
Compute the normal component
Why: The planar cross magnitude is the absolute value of one times two minus two times zero, which is 2; divide by the speed root five.
\[ a_{\mathbf{N}} = \frac{|\mathbf{v}\times\mathbf{a}|}{|\mathbf{v}|} = \frac{2}{\sqrt{5}} \]
Verify against the acceleration magnitude
Why: The squares should add to the squared acceleration magnitude, which is 4.
\[ a_{\mathbf{T}}^2 + a_{\mathbf{N}}^2 = \frac{16}{5} + \frac{4}{5} = \frac{20}{5} = 4 = |\mathbf{a}|^2 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "decomposing acceleration", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The squares should add to the squared acceleration magnitude, which is 4.
Estimation
Predict first
Split the acceleration of the helix into tangential and normal parts.
Commit before you compute: what does Worked example: acceleration on the helix come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with the curvature link
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The normal component should equal curvature times speed squared: one-half times two equals one, matching.
Worked example
Split the acceleration of the helix into tangential and normal parts.
\[ \mathbf{r}(t) = \langle \cos t,\, \sin t,\, t\rangle \]
Get velocity and acceleration
Why: First and second derivatives of the position vector.
\[ \mathbf{v} = \langle -\sin t,\, \cos t,\, 1\rangle, \quad \mathbf{a} = \langle -\cos t,\, -\sin t,\, 0\rangle \]
Compute the tangential component
Why: Velocity dotted with acceleration is negative sine cosine plus cosine times negative sine, which cancels to 0; the speed is constant.
\[ a_{\mathbf{T}} = \frac{\mathbf{v}\cdot\mathbf{a}}{|\mathbf{v}|} = \frac{0}{\sqrt{2}} = 0 \]
Compute the normal component
Why: We already found velocity cross acceleration has magnitude root two; divide by the speed root two.
\[ a_{\mathbf{N}} = \frac{|\mathbf{v}\times\mathbf{a}|}{|\mathbf{v}|} = \frac{\sqrt{2}}{\sqrt{2}} = 1 \]
Verify with the curvature link
Why: The normal component should equal curvature times speed squared: one-half times two equals one, matching.
\[ a_{\mathbf{N}} = \kappa\,|\mathbf{v}|^2 = \tfrac{1}{2}\cdot 2 = 1 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "acceleration on the helix", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The normal component should equal curvature times speed squared: one-half times two equals one, matching.
Trap
Assuming the normal component equals the length of the acceleration vector, or using the acceleration alone without dividing by speed.
\[ \text{(wrong)}\quad a_{\mathbf{N}} \stackrel{?}{=} |\mathbf{a}| \]
For the parabola at t equals 1 that would claim the normal component is 2, but the acceleration also has a tangential piece, so the normal part is smaller.
\[ |\mathbf{a}| = 2, \quad \text{but}\ a_{\mathbf{N}} = \tfrac{2}{\sqrt{5}} \approx 0.89 \]
Split the acceleration into perpendicular pieces; only when the tangential part is zero does the normal part equal the whole magnitude.
\[ a_{\mathbf{N}} = \frac{|\mathbf{v}\times\mathbf{a}|}{|\mathbf{v}|}, \qquad a_{\mathbf{T}}^2 + a_{\mathbf{N}}^2 = |\mathbf{a}|^2 \]
The two components are legs of a right triangle; the acceleration magnitude is the hypotenuse, generally larger than either leg.
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Concept
There is a clean link between turning acceleration and how sharply the path bends.
\[ a_{\mathbf{N}} = \kappa\,|\mathbf{v}|^2 \]
This is why fast, tight turns feel intense: the normal acceleration grows with the square of speed and with the curvature. Double the speed on the same curve and the sideways pull quadruples.
Prediction
Predict first
What is the normal component of acceleration a_N at t = 1?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: two over root five
Why: The planar cross magnitude of v and a is the absolute value of one times two minus two times zero, which is 2. Dividing by the speed root five gives two over root five, about 0.89.
Check
For the parabola path, at time t equals 1 you have velocity and acceleration ready.
\[ \mathbf{v} = \langle 1,\,2\rangle, \qquad \mathbf{a} = \langle 0,\,2\rangle \]
Check your understanding
What is the normal component of acceleration a_N at t = 1?
Answer: A
Why: The planar cross magnitude of v and a is the absolute value of one times two minus two times zero, which is 2. Dividing by the speed root five gives two over root five, about 0.89.
Concept
Curvature says how much a curve bends inside its turning plane. Torsion says how much the curve twists out of that plane.
torsion — A measure, written with the Greek letter tau, of how fast the osculating plane turns as you move along the curve. A flat plane curve has zero torsion; a helix has constant nonzero torsion.
A purely two-dimensional curve never leaves its plane, so its torsion is zero. The helix, which climbs while it circles, has constant torsion one-half.
\[ \tau = \frac{(\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}'''}{|\mathbf{r}'\times\mathbf{r}''|^2} \]
Analogy
Discussion prompt
Explain A brief look at torsion by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Curvature says how much a curve bends inside its turning plane. Torsion says how much the curve twists out of that plane.
Pattern
1. Differentiate for velocity and acceleration
Why: Velocity is r prime; acceleration is r double prime. Speed is the magnitude of velocity.
2. Get T from velocity, and N from T
Why: Normalize velocity for T, then differentiate and normalize for N.
3. Compute curvature by the fitting formula
Why: Cross-product form for a space curve, plane formula for a graph, T prime over r prime if T is easy.
4. Split acceleration into components
Why: Tangential is v dotted with a over speed; normal is the cross magnitude over speed, or curvature times speed squared.
5. Check the triangle relation
Why: Confirm the two squared components add to the squared acceleration magnitude.
Real world
Discussion prompt
Outside this lesson: where does Week 4 - Unit Tangent, Arc Length & Curvature actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Recipe: full motion analysis of r(t) is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck covers the unit tangent and principal unit normal vectors, the arc length of a space curve as the integral of speed, and the arc-length parameter. It then gives three curvature formulas, the radius of curvature, the osculating circle, and the split of acceleration into its tangential and normal components. It targets the classic errors: skipping normalization before finding N, integrating the wrong quantity for arc length, and reaching for the wrong curvature formula.
Check
Set up and evaluate the length of one full turn of the helix.
\[ \mathbf{r}(t) = \langle \cos t,\, \sin t,\, t\rangle, \quad 0 \le t \le 2\pi \]
Check your understanding
What is the arc length over one full turn?
Answer: A
Why: The speed is the root of sine squared plus cosine squared plus one, which is root two, a constant. Integrating root two from 0 to two pi gives root two times two pi, that is two pi root two.
Step zero
Discussion prompt
Worked example: putting it together — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Velocity and its value at t equals 0
Answer:
Worked example
For this plane curve, find the unit tangent at t equals 0 and the curvature there.
\[ \mathbf{r}(t) = \langle t,\, t^2\rangle \]
Velocity and its value at t equals 0
Why: Differentiate, then substitute the time.
\[ \mathbf{r}'(t) = \langle 1,\, 2t\rangle, \quad \mathbf{r}'(0) = \langle 1,\, 0\rangle \]
Unit tangent at t equals 0
Why: The velocity already has length 1 there, so it is its own unit vector.
\[ \mathbf{T}(0) = \frac{\langle 1,0\rangle}{1} = \langle 1,\, 0\rangle \]
Curvature at t equals 0 by the cross-product formula
Why: The cross magnitude is 2 and the speed at 0 is 1, cubed is 1.
\[ \kappa(0) = \frac{2}{1^3} = 2 \]
Verify against the plane-curve formula
Why: This is the graph y equals x squared at its vertex, where we already found curvature 2; the two routes agree.
\[ \kappa(0) = \frac{|y''|}{(1+(y')^2)^{3/2}} = \frac{2}{1} = 2 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "putting it together", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: This is the graph y equals x squared at its vertex, where we already found curvature 2; the two routes agree.
Concept
All three compute the same curvature; they differ only in what you start from.
| Formula | Use when |
|---|---|
| kappa equals magnitude of T prime over magnitude of r prime | the unit tangent is already known or simple |
| kappa equals magnitude of r prime cross r double prime over speed cubed | a space curve is given as r of t |
| kappa equals absolute y double prime over one plus y prime squared to the three-halves | the curve is a graph y equals f of x |
When more than one applies, pick the one that reuses work you have already done.
Comparison
Comparison matrix
From The three curvature formulas at a glance: refill the Use when column from what you know. The rest of the table is as it appeared.
| Formula | Use when |
|---|---|
| kappa equals magnitude of T prime over magnitude of r prime | the unit tangent is already known or simple |
| kappa equals magnitude of r prime cross r double prime over speed cubed | a space curve is given as r of t |
| kappa equals absolute y double prime over one plus y prime squared to the three-halves | the curve is a graph y equals f of x |
Elimination
Eliminate the wrong options
What can you conclude about the acceleration components?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The tangential component is the rate of change of speed. Constant speed makes that rate zero, so the tangential component vanishes. The path still curves, so the normal component carries all of the acceleration.
Check
A particle moves along a curved path at a constant speed, never speeding up or slowing down.
Check your understanding
What can you conclude about the acceleration components?
Answer: A
Why: The tangential component is the rate of change of speed. Constant speed makes that rate zero, so the tangential component vanishes. The path still curves, so the normal component carries all of the acceleration.
Concept
The extreme cases anchor the idea. A straight line never changes direction, so its unit tangent is constant and its curvature is zero.
\[ \mathbf{T} = \text{constant} \;\Longrightarrow\; \mathbf{T}' = \mathbf{0} \;\Longrightarrow\; \kappa = 0 \]
At the other extreme, a tiny circle turns quickly and has large curvature. Every real curve lives between these, bending by an amount its curvature records at each point.
Counterexample
Discussion prompt
The extreme cases anchor the idea. A straight line never changes direction, so its unit tangent is constant and its curvature is zero.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
At the other extreme, a tiny circle turns quickly and has large curvature. Every real curve lives between these, bending by an amount its curvature records at each point.
Ranking
Put in order
Put the moves of Worked example: helix curvature by formula one into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The constant one over root two stays out front; the last component is constant and dies.
Worked example
Recompute the helix curvature using T prime over r prime, and confirm it agrees with the cross-product answer.
\[ \mathbf{T}(t) = \frac{1}{\sqrt{2}}\langle -\sin t,\, \cos t,\, 1\rangle \]
Differentiate the unit tangent
Why: The constant one over root two stays out front; the last component is constant and dies.
\[ \mathbf{T}'(t) = \frac{1}{\sqrt{2}}\langle -\cos t,\, -\sin t,\, 0\rangle \]
Take the magnitude of T prime
Why: The one over root two comes out; the remaining squares add to one.
\[ |\mathbf{T}'(t)| = \frac{1}{\sqrt{2}} \]
Divide by the speed
Why: The speed of the helix is root two; apply formula one.
\[ \kappa = \frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|} = \frac{1/\sqrt{2}}{\sqrt{2}} = \frac{1}{2} \]
Verify it matches formula two
Why: The cross-product route earlier also gave one-half, so both formulas agree, as they must.
\[ \kappa = \tfrac{1}{2} \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "helix curvature by formula one", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The cross-product route earlier also gave one-half, so both formulas agree, as they must.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Recipe: unit tangent and unit normal · Recipe: pick the curvature formula · Recipe: full motion analysis of r(t) · The derivative points along the curve · Smooth curves. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
You can now describe a curve's shape independently of how it is traced: heading, length, bending, and how a moving object accelerates along it.
Unit tangent and normal: normalize the velocity for T, then differentiate and normalize for N. Always normalize before differentiating for N.
Arc length: integrate the speed, the magnitude of the derivative, never the components or the squared speed.
| Quantity | How to get it |
|---|---|
| Unit tangent T | velocity divided by its magnitude |
| Principal normal N | T prime divided by its magnitude |
| Arc length | integral of the speed over the interval |
| Curvature kappa | one of three formulas, chosen by what is given |
| Radius of curvature | one over curvature |
| Acceleration split | tangential from v dot a over speed, normal from the cross magnitude over speed |
Curvature: three formulas, one number. Match the formula to what you are given, and mind the cube on the speed. Radius of curvature is its reciprocal, and the osculating circle sits a radius away along N.
Want this taught 1-on-1? Alexander tutors Calculus III — $55/session, free consultation.