Where things are: curves, frames, and the frame that lies to you

Idea. A partial derivative sees one line. A function can behave perfectly on every line through a point and still be wrong on a curve through it, so "differentiable" has to be defined by a statement about all directions at once.

1. \(f(x,y) = xy/(x^2+y^2)\). On the \(x\)-axis \(f(x,0) = 0\) for all \(x\), so \(\partial_1 f(0,0) = 0\); by symmetry \(\partial_2 f(0,0) = 0\). But along \(y = x\), \[ f(x,x) = \frac{x^2}{2x^2} = \tfrac12 \qquad (x \ne 0), \] while along \(y = 0\) the value is \(0\). Two paths to the origin give two limits, so \(\lim_{(x,y)\to 0}f\) does not exist and \(f\) is not continuous at the origin. In particular it is not differentiable there.

2. \(g(x,y) = x^3/(x^2+y^2)\). Continuity: \(|g| = |x|\cdot x^2/(x^2+y^2) \le |x| \to 0\), and \(g(0,0) = 0\). Directional derivatives: for a unit vector \(u = (u_1,u_2)\), \[ \frac{g(tu)-g(0)}{t} = \frac1t\cdot\frac{t^3u_1^3}{t^2(u_1^2+u_2^2)} = u_1^3 , \] so \(D_ug(0,0) = u_1^3\) exists for every \(u\). If \(g\) were differentiable at the origin with derivative \(\Lambda\), then \(D_ug(0,0) = \Lambda u\) would be linear in \(u\). Taking \(u = (1,0)\) gives \(\Lambda u = 1\), so \(\Lambda = (1,0)\) and \(\Lambda u = u_1\); but at \(u = (1/\sqrt2, 1/\sqrt2)\), \(u_1^3 = 0.3536 \ne 0.7071 = u_1\). Contradiction.

3. \(f\) fails the definition at the first hurdle: differentiability implies continuity, and \(f\) is not continuous. \(g\) fails the clause that the error be \(o(\lVert h\rVert)\) as \(h \to 0\) in any manner — it is \(o(|t|)\) along each fixed ray, at a rate that depends on the ray, and the uniformity over directions that a single linear map would give is exactly what is missing.

Check. Both examples are homogeneous of degree \(0\) and \(1\) respectively: \(f(tx,ty) = f(x,y)\) and \(g(tx,ty) = t\,g(x,y)\). A degree-0 function that is not constant on rays cannot be continuous at the origin, and a degree-1 function is differentiable at the origin exactly when its restriction to the unit circle is the restriction of a linear map. Both failures were predictable from the scaling alone.

Idea. Compute the two mixed partials at the origin from the definition — not from a formula valid only off it — and then show the formula off the origin has no limit, so the theorem's hypothesis, not its conclusion, is what fails.

1. For \(y \ne 0\), \[ \partial_1 f(0,y) = \lim_{h\to0}\frac{f(h,y)-f(0,y)}{h} = \lim_{h\to0}\frac1h\cdot\frac{hy(h^2-y^2)}{h^2+y^2} = \frac{y(-y^2)}{y^2} = -y , \] and \(\partial_1 f(0,0) = 0\) too, so \(\partial_1 f(0,y) = -y\) for every \(y\) and \[ \partial_2\partial_1 f(0,0) = \frac{\mathrm d}{\mathrm dy}(-y) = -1 . \] Since \(f(y,x) = -f(x,y)\), the same computation with the roles swapped gives \(\partial_2 f(x,0) = +x\) and \(\partial_1\partial_2 f(0,0) = +1\).

2. Away from the origin, writing \(f = (x^3y - xy^3)/(x^2+y^2)\) and differentiating twice, \[ \partial_1 f = \frac{y(x^4 + 4x^2y^2 - y^4)}{(x^2+y^2)^2}, \qquad \partial_2\partial_1 f = \frac{x^6 + 9x^4y^2 - 9x^2y^4 - y^6}{(x^2+y^2)^3} . \] Along the \(x\)-axis this is \(x^6/x^6 = +1\); along the \(y\)-axis it is \(-y^6/y^6 = -1\). So \(\partial_2\partial_1 f\) has no limit at the origin and is not continuous there. Clairaut's hypothesis fails, and nothing has been contradicted.

3. Any function assembled from polynomials, \(\exp\), \(\sin\) and \(\cos\) by composition, addition and multiplication is \(C^\infty\) wherever it is defined, so all its mixed partials are continuous and Clairaut always applies. The counterexample needs a quotient with a vanishing denominator — a point where smoothness genuinely breaks down.

Check. The signs are forced by antisymmetry: \(f(y,x) = -f(x,y)\) makes \(\partial_1\partial_2 f(0,0) = -\partial_2\partial_1 f(0,0)\), so the two mixed partials at the origin are negatives of each other and could only be equal if both were \(0\). They are \(\pm1\). The formula in part 2 is likewise antisymmetric under \(x \leftrightarrow y\) ✓.

Idea. Compose the linear maps; the subscripts are the matrix product and are never needed as a separate rule.

1. \(Df(u,v) = \begin{pmatrix} 2u & -2v \\ 2v & 2u\end{pmatrix}\), so \(Df(1,2) = \begin{pmatrix} 2 & -4 \\ 4 & 2\end{pmatrix}\). Also \(f(1,2) = (1-4,\ 4) = (-3,4)\), and \(Dg(x,y) = (2x,\ 2y)\), so \(Dg(-3,4) = (-6,\ 8)\). Composing, \[ D(g\circ f)(1,2) = (-6,\ 8)\begin{pmatrix}2 & -4\\ 4 & 2\end{pmatrix} = (-12+32,\ 24+16) = (20,\ 40). \] Directly: \(g(f(u,v)) = (u^2-v^2)^2 + 4u^2v^2 = (u^2+v^2)^2\), whose derivative is \(\big(4u(u^2+v^2),\ 4v(u^2+v^2)\big)\); at \((1,2)\) that is \((4\cdot5,\ 8\cdot5) = (20,\ 40)\) ✓.

2. \(D\Phi(r,\theta) = \begin{pmatrix}\cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta\end{pmatrix}\), \(\det D\Phi = r\cos^2\theta + r\sin^2\theta = r\). It is singular exactly at \(r = 0\). Geometrically the whole line \(\{r = 0\}\) is crushed to the single point \((0,0)\): \(\Phi\) is not injective on any neighbourhood of such a point, so it has no local inverse there, and \(\theta\) is undefined at the origin however the map is patched.

3. The columns are \((\cos\theta,\sin\theta)\) and \(r(-\sin\theta,\cos\theta)\); their inner product is \(-r\cos\theta\sin\theta + r\sin\theta\cos\theta = 0\). Dividing by the lengths \(1\) and \(r\) leaves \(\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix}\), which is orthogonal with determinant \(+1\).

Check. \(\det D\Phi = r\) has dimensions of length, as a map from \((\mathrm L, 1)\) to \((\mathrm L,\mathrm L)\) must ✓. It is the factor that will reappear in unit 05 as the area element \(r\,\mathrm dr\,\mathrm d\theta\), and it vanishing at \(r=0\) is why polar coordinates have a coordinate singularity there.

Idea. \((\hat e_r,\hat e_\theta)\) is a frame that turns at \(\dot\theta\). The familiar polar acceleration formula and the rotating-frame equation are the same statement.

1. \(\hat e_r = (\cos\theta,\sin\theta)\) gives \(\dot{\hat e}_r = \dot\theta(-\sin\theta,\cos\theta) = \dot\theta\hat e_\theta\), and \(\dot{\hat e}_\theta = \dot\theta(-\cos\theta,-\sin\theta) = -\dot\theta\hat e_r\). Then \[ \vec v = \frac{\mathrm d}{\mathrm dt}(r\hat e_r) = \dot r\hat e_r + r\dot\theta\hat e_\theta, \] \[ \vec a = \ddot r\hat e_r + \dot r\dot\theta\hat e_\theta +(\dot r\dot\theta + r\ddot\theta)\hat e_\theta + r\dot\theta(-\dot\theta\hat e_r) = (\ddot r - r\dot\theta^2)\hat e_r + (r\ddot\theta + 2\dot r\dot\theta)\hat e_\theta . \]

2. In the frame carried by \((\hat e_r,\hat e_\theta)\) the particle sits at \(\vec r_{\text{rot}} = (r,0)\), with \(\vec v_{\text{rot}} = (\dot r, 0)\) and \(\vec a_{\text{rot}} = (\ddot r, 0)\), and the frame turns at \(\vec\omega = \dot\theta\hat z\). The inertial acceleration is \[ \vec a_{\text{in}} = \vec a_{\text{rot}} + 2\vec\omega\times\vec v_{\text{rot}} +\vec\omega\times(\vec\omega\times\vec r) + \dot{\vec\omega}\times\vec r , \] and term by term: \(\vec a_{\text{rot}} = \ddot r\hat e_r\); \(2\dot\theta\hat z\times\dot r\hat e_r = 2\dot r\dot\theta\hat e_\theta\) (Coriolis); \(\dot\theta\hat z\times(\dot\theta\hat z\times r\hat e_r) = -r\dot\theta^2\hat e_r\) (centrifugal); \(\ddot\theta\hat z\times r\hat e_r = r\ddot\theta\hat e_\theta\) (Euler). Adding gives part 1 exactly.

3. The Coriolis term \(2\dot r\dot\theta\). With no tangential force, \(r\ddot\theta + 2\dot r\dot\theta = 0\), i.e. \(\frac{\mathrm d}{\mathrm dt}(r^2\dot\theta) = 0\): pulling the arms in (\(\dot r < 0\)) forces \(\ddot\theta > 0\). It is the same conserved quantity that unit 08 will call angular momentum.

Check. Dimensions: \(r\dot\theta^2\) is \(\mathrm{L\,T^{-2}}\) ✓, \(\dot r\dot\theta\) is \(\mathrm{L\,T^{-2}}\) ✓. Limit \(\dot\theta = 0\): \(\vec a = \ddot r\hat e_r\), straight-line motion along a fixed ray ✓. Limit \(r\) constant, \(\dot\theta\) constant: \(\vec a = -r\dot\theta^2\hat e_r\), which is \(-v^2/r\) towards the centre with \(v = r\dot\theta\) ✓.

Idea. One lemma about constant-norm vectors does all the work, twice: once for \(\vec v\), once for \(\hat T\).

1. \(\frac{\mathrm d}{\mathrm dt}v^2 = \frac{\mathrm d}{\mathrm dt}(\vec v\cdot\vec v) = 2\vec v\cdot\vec a\) by the product rule for the dot product. If \(v\) is constant then \(v^2\) is, so the derivative vanishes and \(\vec a\cdot\vec v = 0\). Conversely if \(\vec a\cdot\vec v = 0\) on an interval then \(v^2\) has zero derivative there, hence is constant, hence so is \(v = \sqrt{v^2} \ge 0\). \(\square\)

2. Write \(\vec v = v\hat T\) with \(\hat T = \vec v/v\), which is defined because \(v > 0\), and is a unit vector. Differentiating, \[ \vec a = \dot v\,\hat T + v\,\dot{\hat T} . \] By part 1 applied to \(\hat T\) (constant norm \(1\)), \(\dot{\hat T}\cdot\hat T = 0\): \(\dot{\hat T}\) is perpendicular to \(\hat T\). Where \(\dot{\hat T} \ne 0\) define \[ \hat N = \frac{\dot{\hat T}}{\lVert\dot{\hat T}\rVert}, \qquad \kappa = \frac{\lVert\dot{\hat T}\rVert}{v} = \frac1\rho, \] the division by \(v\) turning "per unit time" into "per unit arc length", since \(\mathrm ds = v\,\mathrm dt\). Then \(\dot{\hat T} = \kappa v\hat N\) and \[ \vec a = \dot v\hat T + \kappa v^2\hat N = \dot v\hat T + \frac{v^2}{\rho}\hat N . \] There is no third term, and \(\vec a\cdot\hat B = 0\) follows because \(\vec a\) lies in \(\mathrm{span}(\hat T,\hat N)\) by construction. \(\square\)

3. On a circle of radius \(r\) at constant speed, \(\dot v = 0\) kills the first term, and \(\rho = r\) — a circle is its own osculating circle — leaves \(\lVert\vec a\rVert = v^2/r\) along \(\hat N\), which for a circle points at the centre. "Circle" was used in exactly one place: to evaluate \(\rho\). Parts 1 and 2 are statements about an arbitrary trajectory, and the familiar formula is what they become when the radius of curvature happens to be constant and equal to the radius of the path.

Check. Dimensions \(\mathrm{L\,T^{-2}}\) on both terms ✓. A straight line has \(\dot{\hat T} = 0\), so \(\rho = \infty\) and only the tangential term survives, recovering unit 00's one-dimensional problem ✓. A number: the helix of P6 has \(\dot v = 0\) and \(v^2/\rho = 36\ \mathrm{m\,s^{-2}}\), which is \(\lVert\ddot\gamma\rVert = R\omega^2 = 4\times 9 = 36\) ✓.

Idea. Everything follows from \(\dot\gamma\) and \(\ddot\gamma\); the only subtlety is remembering that primes are \(\mathrm d/\mathrm ds\) and dots are \(\mathrm d/\mathrm dt\), differing by a factor \(v\).

1. \(\dot\gamma = (-12\sin3t,\ 12\cos3t,\ 2)\), so \[ v = \sqrt{144 + 4} = \sqrt{148} = 12.1655\ \mathrm{m\,s^{-1}}, \] constant. A full turn takes \(2\pi/3 = 2.0944\) s, in which the arc length is \(v\cdot 2\pi/3 = 25.479\) m and the rise is \(2\cdot2\pi/3 = 4.1888\) m.

2. \(\hat T = \dot\gamma/v = (-0.98639\sin3t,\ 0.98639\cos3t,\ 0.16440)\). Next \(\ddot\gamma = (-36\cos3t,\ -36\sin3t,\ 0)\), of norm \(36\), so \(\hat N = (-\cos3t,\ -\sin3t,\ 0)\) — horizontal, pointing at the axis — and \[ \kappa = \frac{\lVert\ddot\gamma\rVert}{v^2} = \frac{36}{148} = 0.24324\ \mathrm{m^{-1}}, \qquad \rho = \frac{148}{36} = 4.1111\ \mathrm m. \] (The middle step uses \(\dot{\hat T} = \kappa v\hat N\) with \(\dot v = 0\), so \(\ddot\gamma = \kappa v^2\hat N\).) Then \(\hat B = \hat T\times\hat N = (0.16440\sin3t,\ -0.16440\cos3t,\ 0.98639)\), and \(\dot{\hat B} = (3\cdot0.16440)(\cos3t,\sin3t,0) = -(0.49320)\hat N\), so \[ \tau = -\hat B'\cdot\hat N = -\frac{\dot{\hat B}\cdot\hat N}{v} = \frac{0.49320}{12.1655} = 0.040541\ \mathrm{m^{-1}} = \frac{6}{148}. \]

3. \(\dot v = 0\) and \(v^2/\rho = 148/4.1111 = 36.000\ \mathrm{m\,s^{-2}} = \lVert\ddot\gamma\rVert\) ✓, and the direction matches \(\hat N\) ✓. The shadow circle has radius \(4\) m, and \[ \frac{\rho}{4} = \frac{148}{144} = 1.02778 = \frac{v^2}{(R\omega)^2}: \] the factor is exactly the squared ratio of the true speed to the speed of the shadow. Stretching a circle into a helix straightens it, so \(\rho > R\); in the limit of a very steep helix \(\rho \to \infty\) and the curve is nearly a straight line.

Check. Dimensions: \(\kappa, \tau\) in \(\mathrm m^{-1}\) ✓. Limit: with the rise set to zero the curve is the circle of radius 4, and indeed \(\kappa \to R\omega^2/(R\omega)^2 = 1/R = 0.25\) and \(\tau \to 0\) ✓. Orthonormality: \(\hat T\cdot\hat N = 0.98639\sin3t\cos3t - 0.98639\cos3t\sin3t = 0\) ✓, and \(\lVert\hat B\rVert^2 = 0.16440^2 + 0.98639^2 = 1.0000\) ✓.

Idea. A graph is the curve \(t \mapsto (t, h(t))\); push it through the definition of \(\kappa\) once and the formula is yours for good.

1. Parametrise by \(x\): \(\gamma(x) = (x, h(x))\), \(\gamma' = (1, h')\), \(v = \sqrt{1+h'^2}\), \(\gamma'' = (0, h'')\). Since \(\vec a = \dot v\hat T + \kappa v^2\hat N\) and \(\hat N \perp \hat T\), take the component of \(\gamma''\) perpendicular to \(\gamma'\): the unit normal is \((-h', 1)/\sqrt{1+h'^2}\), so \[ \kappa v^2 = \gamma''\cdot\hat N = \frac{h''}{\sqrt{1+h'^2}} \quad\Longrightarrow\quad \kappa = \frac{|h''|}{(1+h'^2)^{3/2}} . \]

2. Here \(h = x^2/(2L)\), \(h' = x/L\), \(h'' = 1/L\) with \(L = 400\) m.

• At \(x = 0\): \(h' = 0\), so \(\kappa = 1/L = 2.500\times10^{-3}\ \mathrm{m^{-1}}\), \(\rho = 400\) m, and \(a_N = v^2/\rho = 900/400 = 2.250\ \mathrm{m\,s^{-2}} = 0.2294\,g\).
• At \(x = 200\) m: \(h' = 0.5\), \((1+0.25)^{3/2} = 1.39754\), so \(\kappa = 1.7889\times10^{-3}\ \mathrm{m^{-1}}\), \(\rho = 559.0\) m, and \(a_N = 900/559.0 = 1.6100\ \mathrm{m\,s^{-2}} = 0.1641\,g\).

3. At the bottom the road is concave up, so \(\hat N\) points upwards and the seat must supply \(m(g + a_N)\): the driver feels heavier, by \(2.250/9.81 = 22.9\%\).

Check. Dimensions: \([\kappa] = \mathrm L^{-1}\) and the formula's numerator \(h''\) is \(\mathrm L^{-1}\) with a dimensionless denominator ✓. Limit: a straight road, \(h'' = 0\), gives \(\kappa = 0\) and no normal acceleration ✓. Sanity: \(\kappa\) decreases away from the bottom, as it must — the parabola is flattest, hence most curved, at its vertex, and asymptotically straight far out ✓.

Idea. The cross product is tied to the determinant; the determinant is invariant under \(SO(3)\); everything else is bookkeeping.

1. Expanding, \[ (\vec u\times\vec v)\cdot\vec w = (u_2v_3-u_3v_2)w_1 + (u_3v_1-u_1v_3)w_2 +(u_1v_2-u_2v_1)w_3 , \] which is precisely the cofactor expansion of \(\det[\vec u\ \vec v\ \vec w]\) along its third column. \(\square\)

2. For \(R \in SO(3)\) and any \(\vec w\), writing \(\vec w = R(R^{\!\top}\vec w)\), \[ \big((R\vec u)\times(R\vec v)\big)\cdot\vec w = \det[R\vec u\ R\vec v\ RR^{\!\top}\vec w] = \det R\cdot\det[\vec u\ \vec v\ R^{\!\top}\vec w] = (\vec u\times\vec v)\cdot R^{\!\top}\vec w = \big(R(\vec u\times\vec v)\big)\cdot\vec w . \] True for every \(\vec w\), so \((R\vec u)\times(R\vec v) = R(\vec u\times\vec v)\). Then for any \(\vec w\), \[ R\,\hat a\,R^{\!\top}\vec w = R\big(\vec a\times(R^{\!\top}\vec w)\big) = (R\vec a)\times(RR^{\!\top}\vec w) = (R\vec a)\times\vec w = \widehat{R\vec a}\,\vec w , \] so \(R\hat aR^{\!\top} = \widehat{R\vec a}\). \(\square\)

\(\det R = +1\) is used exactly once, in the middle equality of the first display. For \(\det R = -1\) it gives \(-1\) instead, so \((R\vec u)\times(R\vec v) = -R(\vec u\times\vec v)\): under a reflection the cross product acquires a sign, which is the precise content of calling it a pseudovector.

3. From \(\Omega = R^{\!\top}\dot R\) we get \(\dot R = R\Omega\), hence \[ \tilde\Omega = \dot RR^{\!\top} = R\Omega R^{\!\top} = \widehat{R\vec\omega} \] by part 2. So the antisymmetric matrix built in the inertial frame corresponds to the vector \(R\vec\omega\) — and \(x_{\text{in}} = Rx_{\text{rot}}\) is exactly how the components of one vector in the two frames are related. The two frames report the same arrow. \(\square\)

4. Apply the transport theorem to \(\vec u = \vec\omega\): \[ \Big(\frac{\mathrm d\vec\omega}{\mathrm dt}\Big)_{\text{in}} = \Big(\frac{\mathrm d\vec\omega}{\mathrm dt}\Big)_{\text{rot}} + \vec\omega\times\vec\omega = \Big(\frac{\mathrm d\vec\omega}{\mathrm dt}\Big)_{\text{rot}} . \] Equivalently, in inertial components, \(\frac{\mathrm d}{\mathrm dt}(R\vec\omega) = \dot R\vec\omega + R\dot{\vec\omega} = R(\vec\omega\times\vec\omega) + R\dot{\vec\omega} = R\dot{\vec\omega}\). \(\square\)

Check. Part 2 with \(R\) a rotation by \(\pi/2\) about \(\hat z\): \(\hat e_1\times\hat e_2 = \hat e_3\) becomes \(\hat e_2\times(-\hat e_1) = \hat e_3\) ✓. Part 3's dimensions: both sides of \(\tilde\Omega = R\Omega R^{\!\top}\) are \(\mathrm T^{-1}\) ✓. Part 4 is why the Euler term needs no frame label in the rotating-frame equation — a fact used silently on the slides.

Idea. Apply \((\mathrm d/\mathrm dt)_{\text{in}} = (\mathrm d/\mathrm dt)_{\text{rot}} + \vec\omega\times\) to \(\vec r\), then to what comes out. Every term appears exactly once, which is why counting them is easy.

Working. First application, to the position vector: \[ \vec v_{\text{in}} = \dot{\vec r} + \vec\omega\times\vec r , \] where the dot is the rotating-frame derivative, i.e. \(\vec v_{\text{rot}}\). Second application, to \(\vec v_{\text{in}}\): \[ \vec a_{\text{in}} = \frac{\mathrm d}{\mathrm dt}\Big|_{\text{rot}}\!\big(\dot{\vec r} + \vec\omega\times\vec r\big) +\vec\omega\times\big(\dot{\vec r} + \vec\omega\times\vec r\big) . \] Expanding the first bracket with the product rule for \(\times\), and using \(\dot{\vec\omega}\) for the rotating-frame derivative of \(\vec\omega\) (which by P8.4 is unambiguous): \[ \vec a_{\text{in}} = \ddot{\vec r} + \dot{\vec\omega}\times\vec r +\vec\omega\times\dot{\vec r} + \vec\omega\times\dot{\vec r} +\vec\omega\times(\vec\omega\times\vec r) = \vec a_{\text{rot}} + 2\vec\omega\times\vec v_{\text{rot}} +\vec\omega\times(\vec\omega\times\vec r) + \dot{\vec\omega}\times\vec r . \] Multiply by \(m\) and substitute \(m\vec a_{\text{in}} = \vec F\), then solve for \(m\vec a_{\text{rot}}\): \[ m\vec a_{\text{rot}} = \vec F - 2m\vec\omega\times\vec v_{\text{rot}} -m\vec\omega\times(\vec\omega\times\vec r) - m\dot{\vec\omega}\times\vec r . \qquad\square \]

The factor 2. \(\vec\omega\times\dot{\vec r}\) is produced twice: once by differentiating \(\vec\omega\times\vec r\) inside the rotating frame (the \(\vec\omega\times\dot{\vec r}\) half of the product rule), and once by the \(\vec\omega\times\) that the second application of the theorem adds to \(\dot{\vec r}\). The centrifugal and Euler terms each arise from one source only.

Translating origins. If \(\vec x_{\text{in}} = \vec d(t) + R\vec r\), the extra term differentiates independently, giving \(\vec a_{\text{in}} = \ddot{\vec d} + (\text{everything above})\), so the rotating-frame equation acquires one more fictitious term, \(-m\ddot{\vec d}\). Nothing else changes, because translation and rotation act on separate parts of the expression.

Check. Dimensions: every term \(\mathrm{M\,L\,T^{-2}}\) ✓. Limit \(\vec\omega \to 0\): all three vanish and \(m\vec a = \vec F\) ✓. Special case \(\vec r\) fixed in the rotating frame, \(\vec\omega\) constant: only the centrifugal term survives, of magnitude \(m\omega^2r_\perp\) outward ✓. The matrix derivation on the slides gives the identical four terms, which is the strongest check available.

Idea. Solve to zeroth order, feed the answer back in once, and check that the next round would be negligible.

1. In the local east–north–up frame at latitude \(\lambda\), \(\vec\omega = \omega(0,\cos\lambda,\sin\lambda)\) and the equation, with centrifugal already absorbed into \(g\) and \(\dot{\vec\omega} = 0\), is \(\vec a = -g\hat z - 2\vec\omega\times\vec v\). To zeroth order \(\vec v = (0,0,-gt)\), so \[ -2\vec\omega\times\vec v = -2\omega(0,\cos\lambda,\sin\lambda)\times(0,0,-gt) = (2\omega gt\cos\lambda,\ 0,\ 0), \] due east. Integrating \(\ddot x = 2\omega g t\cos\lambda\) twice from rest gives \(x = \tfrac13\omega gt^3\cos\lambda\), and the drop time is \(t = \sqrt{2h/g}\).

The direction is east in both hemispheres because the deflection carries the factor \(\cos\lambda > 0\) for every \(\lambda \in (-90^\circ, 90^\circ)\) — unlike the horizontal Coriolis deflection, which carries \(\sin\lambda\) and does change sign at the equator.

2. \(t = \sqrt{160/9.81} = 4.03855\) s, so \[ x = \tfrac13(7.29212\times10^{-5})(9.81)(4.03855)^3\cos 51.0^\circ = 0.015706\times0.62932 = 9.88\times10^{-3}\ \mathrm m \approx 0.99\ \mathrm{cm}. \] The naive estimate: the mass begins with the eastward speed of a point \(h\) further from the axis, an excess \(\omega h\cos\lambda\), which over \(t\) seconds gives \(\omega h t\cos\lambda = 7.29212\times10^{-5}\times80\times4.03855 \times 0.62932 = 1.48\ \mathrm{cm}\). The ratio is exactly \(2/3\), since \(h = \tfrac12gt^2\) makes the naive answer \(\tfrac12\omega gt^3\cos\lambda\) against the true \(\tfrac13\omega gt^3\cos\lambda\). The naive estimate overshoots because the mass also falls inwards, shrinking its distance from the axis and so giving up part of its initial lead.

3. Feeding the first-order eastward velocity \(\dot x = \omega gt^2\cos\lambda\) back into \(-2\vec\omega\times\vec v\) produces a correction smaller by a factor of order \(\omega t = 7.29\times10^{-5}\times4.04 = 2.9\times10^{-4}\), i.e. about \(3\ \mu\mathrm m\). That is three and a half orders of magnitude below the answer and far below any achievable measurement, so one iteration is enough. (Unit 21 calls this a regular perturbation expansion in the small parameter \(\omega t\).)

Check. Dimensions: \(\omega gt^3 = \mathrm{T^{-1}\,L\,T^{-2}\,T^3} = \mathrm L\) ✓. Limit \(\omega \to 0\): no deflection ✓. Limit \(\lambda \to 90^\circ\): no deflection, correct, since at the pole the fall is along the rotation axis ✓. Cross-check against the slide's worked example: at the equator with \(h = 100\) m the same formula gives \(2.20\) cm, and Reich's 1831 measurements down a \(158\) m shaft at Freiberg agree with the predicted \(2.75\) cm ✓.

Idea. Resolve one cross product in the local frame, then maximise a product of a sine and a cosine.

1. Take the local frame \(\hat x\) east, \(\hat y\) north, \(\hat z\) up, so \(\vec\omega = \omega(0,\cos\lambda,\sin\lambda)\) and \(\vec r = R\hat z\). Then \(\vec\omega\times\vec r = \omega R(\cos\lambda, 0, 0)\) — pointing east — and \[ -\vec\omega\times(\vec\omega\times\vec r) = -\omega^2R\,(0,\cos\lambda,\sin\lambda)\times(\cos\lambda,0,0) = \omega^2R\,(0,\ -\sin\lambda\cos\lambda,\ \cos^2\lambda). \] So the centrifugal term is \(\omega^2R\cos^2\lambda\) upward and \(\omega^2R\sin\lambda\cos\lambda\) southward (towards the equator) in the northern hemisphere. Its magnitude is \(\omega^2R\cos\lambda\), directed away from the polar axis, as it must be.

2. The tilt angle satisfies \[ \tan\alpha = \frac{\omega^2R\sin\lambda\cos\lambda}{g - \omega^2R\cos^2\lambda} \approx \frac{\omega^2R\sin\lambda\cos\lambda}{g}, \] the approximation costing a relative \(\omega^2R\cos^2\lambda/g = 0.17\%\) at \(45^\circ\). Since \(\sin\lambda\cos\lambda = \tfrac12\sin2\lambda\), the tilt is largest at \(\lambda = 45^\circ\), where \[ \alpha \approx \frac{0.0339157\times0.5}{9.81} = 1.72863\times10^{-3}\ \text{rad} = 0.09904^\circ = 5.94' . \] (Here \(\omega^2R_{\mathrm E} = (7.29212\times10^{-5})^2\times6.378137\times10^6 = 0.0339157\ \mathrm{m\,s^{-2}}\).)

3. At the equator the whole centrifugal term is vertical and equal to \(0.0339157\ \mathrm{m\,s^{-2}}\), a reduction of \(0.0339157/9.81 = 0.346\%\). The measured equator-to-pole difference is \(9.832 - 9.780 = 0.052\ \mathrm{m\,s^{-2}}\), of which rotation explains \(0.0339\), i.e. \(65\%\). The remaining \(0.018\ \mathrm{m\,s^{-2}}\) is the equatorial bulge: the Earth's equatorial radius exceeds its polar radius by \(21\) km, so an equatorial observer is further from the bulk of the mass, which lowers \(g\) further — but the nearby bulge mass itself raises it, and the net effect of the shape is what the residual measures. Unit 10 computes the bulge's own field from its quadrupole moment.

Check. Dimensions: \(\omega^2R/g\) is dimensionless ✓. Limits: the tilt is zero at the pole (\(\cos\lambda = 0\)) and at the equator (\(\sin\lambda = 0\)), both correct by symmetry ✓. Size: \(6\) arcminutes is a tenth of a degree — a metre of horizontal error over \(580\) m of plumb line, invisible to a builder and unmissable to a geodesist ✓.

Idea. One formula: \(T = T_{\text{sid}}/\sin\lambda\), with \(T_{\text{sid}} = 86\,164.09\ \mathrm s = 23.93447\) h.

1.

2. \(1^\circ\) in \(15\) min is a rate of \(\tfrac{\pi}{180}/900 = 1.93925\times10^{-5}\ \mathrm{s^{-1}}\). Setting \(\omega\sin\lambda\) equal to it, \[ \sin\lambda = \frac{1.93925\times10^{-5}}{7.29212\times10^{-5}} = 0.265939 \quad\Longrightarrow\quad \lambda = 15.42^\circ , \] north or south. (Equivalently: \(1^\circ\) per \(15\) min is \(4^\circ\) per hour, so \(T = 90\) h, and \(\sin\lambda = 23.93447/90\) ✓.)

3. At \(48.85^\circ\) N the plane turns clockwise seen from above at \(\omega_z = \omega\sin\lambda = 5.4903\times10^{-5}\ \mathrm{s^{-1}}\); at \(48.85^\circ\) S it turns anticlockwise at the same rate, since \(\sin(-\lambda) = -\sin\lambda\). The relative rate is \(2\omega_z\), and a plane of swing is a line, so "perpendicular" means a relative turn of \(\pi/2\): \[ t = \frac{\pi/2}{2\omega_z} = \frac{1.570796}{1.098060\times10^{-4}} = 1.4304\times10^{4}\ \mathrm s = 3.97\ \mathrm h . \]

Check. Every period exceeds \(23.93\) h ✓, as \(\sin\lambda \le 1\) forces. The equatorial limit is infinite ✓ and the polar limit is one sidereal day ✓. Part 2's answer lies between the equator (infinitely slow) and \(45^\circ\) (\(33.8\) h, i.e. \(10.6^\circ\) per hour — faster than the \(4^\circ\) per hour asked for), so a latitude below \(45^\circ\) is right ✓. Part 3: a quarter of a relative turn at a combined \(11.33 + 11.33 = 22.66\) degrees per hour is \(90/22.66 = 3.97\) h ✓.

Idea. Coriolis alone does no work, so the speed is fixed; a force of fixed magnitude always perpendicular to a fixed-speed velocity gives fixed curvature; and fixed curvature in a plane means a circle.

1. \(\vec v\cdot\vec a = \vec v\cdot(-2\vec\omega\times\vec v) = 0\), because \(\vec\omega\times\vec v\) is perpendicular to \(\vec v\). By P5.1, \(\lVert\vec v\rVert\) is constant. \(\square\)

2. Since \(\vec v\) is horizontal and \(\vec\omega = \Omega\hat z\) is vertical, \(\lVert\vec\omega\times\vec v\rVert = \Omega v\), so \(\lVert\vec a\rVert = 2\Omega v\). But \(\vec a = \dot v\hat T + (v^2/\rho)\hat N\) with \(\dot v = 0\) by part 1, so \[ \frac{v^2}{\rho} = 2\Omega v \quad\Longrightarrow\quad \rho = \frac{v}{2\Omega}, \qquad \kappa = \frac{2\Omega}{v}, \] both constant, since \(v\) and \(\Omega\) are. The motion also stays horizontal: \(\vec a = -2\Omega\hat z\times\vec v\) has no vertical component when \(\vec v\) has none. \(\square\)

3. Take the curve unit-speed in the plane. \(\hat T\) is a unit vector so \(\hat T' = \kappa\hat N\) with \(\hat N \perp \hat T\); and \(\hat N\) is a unit vector so \(\hat N' \perp \hat N\), i.e. \(\hat N' = \alpha\hat T\) for some \(\alpha\). Differentiating \(\hat T\cdot\hat N = 0\) gives \(\kappa + \alpha = 0\), so \(\hat N' = -\kappa\hat T\). Now set \(\vec c = \vec r + \rho\hat N\) with \(\rho = 1/\kappa\) constant: \[ \vec c\,{}' = \hat T + \rho(-\kappa\hat T) = \hat T - \hat T = 0 . \] So \(\vec c\) is a fixed point, and \(\vec r - \vec c = -\rho\hat N\) has constant length \(\rho\), since \(\hat N\) is a unit vector: the curve lies on the circle of centre \(\vec c\) and radius \(\rho\). \(\square\)

4. \(\rho = 0.20/(2\times2) = 0.050\) m, and one circuit takes \[ T_{\text{circle}} = \frac{2\pi\rho}{v} = \frac{2\pi}{2\Omega} = \frac{\pi}{\Omega} = 1.5708\ \mathrm s, \] against the dish's own period \(2\pi/\Omega = 3.1416\) s. The ratio is exactly \(1/2\), and exactly, for every \(v\): the speed cancels between the circumference \(2\pi v/(2\Omega)\) and the speed \(v\). The circuit is traversed clockwise seen from above when \(\Omega > 0\) — opposite to the dish — because \(-2\Omega\hat z\times\vec v\) always points to the right of \(\vec v\).

Check. Dimensions: \(v/\Omega\) is \(\mathrm{L\,T^{-1}\,T} = \mathrm L\) ✓. Limit \(\Omega\to0\): \(\rho\to\infty\), a straight line, correct for a free puck on a non-rotating table ✓. The same formula with \(\Omega \to \omega_{\mathrm E}\sin\lambda\) is the atmospheric inertial circle of the Coriolis slide: \(30/(2\times5.157\times10^{-5}) = 291\) km at \(45^\circ\) N, with a period of \(\pi/(\omega\sin45^\circ) = 16.9\) h — the "inertial oscillation" that oceanographers see in drifter tracks ✓.

Another way (part 3)

Without any Frenet machinery: a unit-speed plane curve can be written \(\hat T(s) = (\cos\phi(s), \sin\phi(s))\) for a differentiable angle \(\phi\), and then \(\kappa = |\phi'|\). Constant \(\kappa\) gives \(\phi = \phi_0 \pm \kappa s\), and integrating \(\vec r\,{}' = \hat T\) gives \(\vec r(s) = \vec c \pm \rho(\sin\phi, -\cos\phi)\), a circle of radius \(\rho\) traversed at unit speed. The version on the slide generalises to space curves; this one does not, but it is the quickest route in the plane.