Curves & Motion

Why one coordinate at a time: the difference quotient $\dfrac{\mathbf r(t + h) - \mathbf r(t)}{h}$ is a vector whose components are the ordinary difference quotients of $x$, $y$ and $z$, and a vector converges exactly when each component does. The velocity is tangent because the chord $\mathbf r(t + h) - \mathbf r(t)$ joins two nearby points of the path, and as $h \to 0$ its direction turns into the direction of the path itself. For the helix, $\mathbf r'(t) = (-\sin t, \cos t, 1)$: the horizontal part always has length $1$ and the vertical part is always $1$, so the speed is $\sqrt{1 + 1} = \sqrt2$ at every instant.

Curvature is the rate at which the tangent direction turns per unit of distance travelled (the formula is derived below). A circle of radius $R$ turns through $2\pi$ radians over a length $2\pi R$, so its curvature is $\tfrac1R$. That is why $\tfrac1\kappa$ is the radius of the osculating circle: the circle that matches the curve's position, slope and bending at the point. At the vertex of $y = x^2$, $y' = 0$ and $y'' = 2$, so $\kappa = 2$ and the osculating circle has radius $\tfrac12$, centred at $\left(0, \tfrac12\right)$.

Read each coordinate as its own story. The $x$-velocity stays $3$ because nothing pushes sideways; the $y$-velocity $4 - 10t$ falls by $10$ every second because gravity is $(0, -10)$. At the top the vertical velocity is zero, $4 - 10t = 0$, so $t = \tfrac25$ — the same condition as $\mathbf v\cdot\mathbf a = (3, 4 - 10t)\cdot(0, -10) = -40 + 100t = 0$. The top is at $x = \tfrac65$, height $4\cdot\tfrac25 - 5\cdot\tfrac{4}{25} = \tfrac45$. The projectile lands when $4t - 5t^2 = 0$, at $t = \tfrac45$ and $x = \tfrac{12}{5}$, and its velocity there is $(3, -4)$: the launch velocity reflected.

Why divide by the speed: $\mathbf T'(t)$ measures how fast the direction turns per unit of time, but curvature should belong to the path, not to how fast it is traced. The speed $|\mathbf r'| = \tfrac{ds}{dt}$ converts per-time into per-length. The cross-product form skips computing $\mathbf T$: writing $\mathbf r' = v\mathbf T$ with $v = |\mathbf r'|$ gives $\mathbf r'' = v'\mathbf T + v^2\kappa\,\mathbf N$, and crossing with $\mathbf r'$ removes the $\mathbf T$ part, leaving $|\mathbf r'\times\mathbf r''| = v^3\kappa$.

The cross product is $\mathbf r'\times\mathbf r'' = (12\sin t,\ -12\cos t,\ 9)$, whose length is $\sqrt{144 + 81} = 15$ for every $t$. So $\kappa = \tfrac{15}{5^3} = \tfrac{3}{25}$, the same at every point, as the helix's symmetry demands. The flat circle of radius $3$ has $\kappa = \tfrac13$; climbing spreads each turn over more length, so the helix bends less. In general the helix $(a\cos t, a\sin t, bt)$ has $\kappa = \dfrac{a}{a^2 + b^2}$, here $\tfrac{3}{9 + 16}$.

So $\kappa(t) = \dfrac{8}{(16\sin^2t + 4\cos^2t)^{3/2}}$. At $t = 0$, the point $(4, 0)$: $\kappa = \tfrac{8}{4^{3/2}} = 1$, osculating radius $1$. At $t = \tfrac\pi2$, the point $(0, 2)$: $\kappa = \tfrac{8}{16^{3/2}} = \tfrac18$, radius $8$. The ellipse is sharpest at the ends of its long axis and flattest at the ends of its short axis, and the two circles are centred at $(3, 0)$ and $(0, -6)$ on the inner side of the curve.

Why: differentiate $\mathbf v = v\mathbf T$ with the product rule, $\mathbf a = v'\mathbf T + v\mathbf T'$. Because $\mathbf T$ always has length $1$, differentiating $\mathbf T\cdot\mathbf T = 1$ gives $2\,\mathbf T\cdot\mathbf T' = 0$, so $\mathbf T'$ is perpendicular to $\mathbf T$. Its length is $|\mathbf T'| = \kappa v$ by the definition of curvature, so $v\mathbf T' = \kappa v^2\mathbf N$. A driver feels exactly these two parts: the accelerator changes $a_T$, and taking a curve of radius $\tfrac1\kappa$ at speed $v$ produces $a_N = \kappa v^2$.

The tangential part is $a_T = \dfrac{\mathbf v\cdot\mathbf a}{|\mathbf v|} = \dfrac{4}{\sqrt5}$, and the normal part is $a_N = \dfrac{|x'y'' - y'x''|}{|\mathbf v|} = \dfrac{|1\cdot2 - 2\cdot0|}{\sqrt5} = \dfrac{2}{\sqrt5}$. Check: $\tfrac{16}{5} + \tfrac45 = 4 = |\mathbf a|^2$. The curvature follows from $a_N = \kappa v^2$: $\kappa = \dfrac{2/\sqrt5}{5} = \dfrac{2}{5\sqrt5}$, the same value the worked example below finds for $y = x^2$ at $x = 1$ with the graph formula.

The two parts are perpendicular, so $|\mathbf a| = \sqrt{9 + 64} = \sqrt{73} \approx 8.54$ m/s$^2$. The turning part dominates. Doubling the speed to $40$ m/s would quadruple $a_N$ to $\tfrac{1600}{50} = 32$ m/s$^2$, more than three times the acceleration of gravity, which is why curves carry speed limits.

Summing the segments gives a Riemann sum $\sum |\mathbf r'(t_i)|\,\Delta t$, and the limit as $\Delta t \to 0$ is by definition $\int_a^b |\mathbf r'(t)|\,dt$. $\blacksquare$

Divide: $\kappa = \left|\tfrac{d\phi/dx}{ds/dx}\right| = \dfrac{|y''|}{(1 + y'^2)^{3/2}}$. $\blacksquare$

The SPEED $|\mathbf r'(t)|$ is a number — the length of that vector. Distance traveled is speed accumulated over time, which is why arc length is $L = \int_a^b|\mathbf r'(t)|\,dt$ and not anything simpler: over a tiny interval $\Delta t$ the point moves approximately $\mathbf r'(t)\Delta t$, a straight step of length $|\mathbf r'(t)|\Delta t$, and adding infinitely many tiny straight steps IS the integral. Note what this is not: it is not $|\mathbf r(b) - \mathbf r(a)|$, the straight-line displacement — a runner on a circular track covers a full lap of distance with zero displacement.

Step 1 — differentiate coordinate-wise: $\mathbf r'(t) = (-\sin t, \cos t, 1)$.

Step 2 — speed: $|\mathbf r'(t)| = \sqrt{\sin^2t + \cos^2t + 1} = \sqrt{1 + 1} = \sqrt2$, using the Pythagorean identity to collapse the first two squares.

Step 3 — notice the speed is CONSTANT: the integral needs no antiderivative work, $L = \int_0^{2\pi}\sqrt2\,dt = 2\sqrt2\,\pi$.

Step 4 — sanity check by unrolling: the helix wraps around a cylinder of circumference $2\pi$ while climbing $2\pi$; unrolled flat it is the hypotenuse of a right triangle with both legs $2\pi$, length $\sqrt{(2\pi)^2 + (2\pi)^2} = 2\sqrt2\,\pi$ ✓ — the same answer with no calculus at all.

Step 3 — compare with the vertex, where $y' = 0$ gives $\kappa = 2$: at $x = 1$ the curvature is $\tfrac{2}{5\sqrt5} \approx 0.18$, roughly a tenth of the vertex value.

Step 4 — interpret via the osculating circle: the best-fitting circle at $x=1$ has radius $\tfrac1\kappa = \tfrac{5\sqrt5}{2} \approx 5.6$, versus radius $\tfrac12$ at the vertex — the parabola straightens out fast as you leave the bottom, and the formula quantifies exactly how fast.