Curvature and torsion

At a Glance

• Frequency: 6 sub-parts across 6 of 13 years (2014, 2016, 2017, 2018, 2019, 2024)
• Priority tier: T2
• Marks (count): 10 (1), 12 (1), 15 (2), 16 (1), 5 (1)
• Average solve time: ~12 min
• Difficulty mix: medium 4, easy 1, hard 1
• Section: B | Dominant type: computation

Why This Chapter Matters

This atom lands in Section B at 12–16 marks, making it one of the highest-value vector-analysis questions in Paper 1. UPSC has set it in six of the last eleven years, always asking for a direct computation — curvature vector, torsion, or polar radius of curvature — with no open-ended theory. Three formulas (the cross-product formula for $\kappa$, the triple-product formula for $\tau$, and the polar $\rho$ formula) handle every variant that has appeared; mastering the algebra around these three formulas is sufficient to score full marks.

Minimum Theory

Frenet–Serret vocabulary. For a curve $\mathbf{r}(t)$, let primes denote derivatives with respect to $t$ and $s$ denote arc length. The unit tangent is $\hat{T} = \mathbf{r}'/|\mathbf{r}'|$. The curvature vector is $\boldsymbol{\kappa} = d\hat{T}/ds$; its magnitude $\kappa = |\boldsymbol{\kappa}|$ is the curvature. The principal normal $\hat{N}$ points in the direction of $d\hat{T}/ds$. The binormal $\hat{B} = \hat{T}\times\hat{N}$ completes the right-handed Frenet frame. The torsion $\tau$ measures how fast $\hat{B}$ rotates out of the osculating plane: $d\hat{B}/ds = -\tau\hat{N}$. The figure below shows the three frame vectors at a point on a space curve.

Working formulas (all derivatives w.r.t.\ $t$).

$\kappa = \frac{|\mathbf{r}'\times\mathbf{r}''|}{|\mathbf{r}'|^3}, \qquad \tau = \frac{(\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}'''}{|\mathbf{r}'\times\mathbf{r}''|^2}.$

The curvature vector itself is $\boldsymbol{\kappa} = d\hat{T}/ds = \bigl[(\mathbf{r}'\times\mathbf{r}'')\times\mathbf{r}'\bigr]/|\mathbf{r}'|^4$.

Planar curve $\mathbf{r}=(f,g,0)$. $\kappa = \frac{|f'g''-g'f''|}{(f'^{\,2}+g'^{\,2})^{3/2}}, \qquad \tau = 0.$

Polar curve $r = r(\theta)$. Let $r_1 = dr/d\theta$, $r_2 = d^2r/d\theta^2$. Then $\rho = \frac{(r^2+r_1^2)^{3/2}}{|r^2+2r_1^2-r\,r_2|}.$

The radii $\rho = 1/\kappa$ and $\sigma = 1/|\tau|$ are called the radius of curvature and radius of torsion respectively.

Question Archetypes

curvature-computation (3 question(s); 2014, 2017, 2024)

Compute the curvature vector and/or magnitude of a parametric curve (and possibly a related locus)

Recognition Cues

The question asks for the “curvature vector” or “curvature” of a curve given by explicit parametric equations in two or three components. In 2017 a secondary sub-task asks you to show that the locus of feet of perpendiculars from the origin to the tangent lines lies on a specific surface — that is an additional step beyond computing $\kappa$.

Solution Template

1. Differentiate $\mathbf{r}(t)$ to get $\mathbf{r}'$ and $\mathbf{r}''$; compute $|\mathbf{r}'|$.
2. Compute $\mathbf{r}'\times\mathbf{r}''$; find $|\mathbf{r}'\times\mathbf{r}''|$.
3. Curvature magnitude: $\kappa = |\mathbf{r}'\times\mathbf{r}''|/|\mathbf{r}'|^3$.
4. Curvature vector: $\boldsymbol{\kappa} = \bigl[(\mathbf{r}'\times\mathbf{r}'')\times\mathbf{r}'\bigr]/|\mathbf{r}'|^4$.
5. (If a locus is required) Write the general point on the tangent as $\mathbf{r}(t) + \lambda\mathbf{r}'(t)$; impose the foot-of-perpendicular condition $\mathbf{OQ}\cdot\mathbf{r}'=0$ to find $\lambda$; then verify the claimed surface equation by direct substitution.

Worked Example 1

2014 Paper 1, 2014-P1-Q5e (10 marks)

Find the curvature vector and its magnitude for $\mathbf{r}(t)=(t\cos t,\;t\sin t,\;0)$.

Step 1 — derivatives. $\mathbf{r}' = (\cos t - t\sin t,\; \sin t + t\cos t,\; 0),$

$\mathbf{r}'' = (-2\sin t - t\cos t,\; 2\cos t - t\sin t,\; 0).$

$|\mathbf{r}'|^2 = \cos^2 t - 2t\sin t\cos t + t^2\sin^2 t + \sin^2 t + 2t\sin t\cos t + t^2\cos^2 t = 1 + t^2.$

Step 2 — cross product. Both vectors have zero $z$-component, so $\mathbf{r}'\times\mathbf{r}'' = \bigl(0,\;0,\;({\cos t}-t\sin t)(2\cos t-t\sin t) - (\sin t+t\cos t)(-2\sin t-t\cos t)\bigr).$ Expanding the $z$-component: $= 2\cos^2 t - t\sin t\cos t - 2t\sin t\cos t + t^2\sin^2 t + 2\sin^2 t + t\sin t\cos t + 2t\sin t\cos t + t^2\cos^2 t$

$= 2(\cos^2 t+\sin^2 t) + t^2(\sin^2 t+\cos^2 t) = 2 + t^2.$ So $\mathbf{r}'\times\mathbf{r}'' = (0,0,2+t^2)$ and $|\mathbf{r}'\times\mathbf{r}''| = 2+t^2$.

Step 3 — curvature magnitude. $\boxed{\kappa = \frac{2+t^2}{(1+t^2)^{3/2}}.}$

Step 4 — curvature vector. Compute $(\mathbf{r}'\times\mathbf{r}'')\times\mathbf{r}'$: $\bigl(0,0,2+t^2\bigr)\times(\cos t-t\sin t,\;\sin t+t\cos t,\;0)$

$= \bigl(-(2+t^2)(\sin t+t\cos t),\;(2+t^2)(\cos t-t\sin t),\;0\bigr).$ Divide by $|\mathbf{r}'|^4 = (1+t^2)^2$: $\boxed{\boldsymbol{\kappa} = \frac{2+t^2}{(1+t^2)^2}\bigl(-(\sin t+t\cos t)\,\hat{\imath} + (\cos t-t\sin t)\,\hat{\jmath}\bigr).}$

Worked Example 2

2017 Paper 1, 2017-P1-Q7a (16 marks)

Find the curvature vector and its magnitude for the helix $\mathbf{r}=(\,a\cos\theta,\;a\sin\theta,\;a\theta\,)$. Show that the locus of the feet of perpendiculars from the origin to the tangent lines lies on the hyperboloid $x^2+y^2-z^2=a^2$.

Part 1 — curvature vector.

$\mathbf{r}' = (-a\sin\theta,\; a\cos\theta,\; a)$, so $|\mathbf{r}'| = a\sqrt{2}$.

Unit tangent: $\hat{T} = (-\sin\theta,\;\cos\theta,\;1)/\sqrt{2}$.

Differentiate w.r.t.\ arc length $s$ using $ds/d\theta = a\sqrt{2}$: $\boldsymbol{\kappa} = \frac{d\hat{T}}{ds} = \frac{1}{a\sqrt{2}}\frac{d\hat{T}}{d\theta} = \frac{1}{a\sqrt{2}}\cdot\frac{(-\cos\theta,-\sin\theta,0)}{\sqrt{2}} = \frac{(-\cos\theta,-\sin\theta,0)}{2a}.$

$\boxed{|\boldsymbol{\kappa}| = \kappa = \frac{1}{2a}.}$

Part 2 — locus of feet of perpendiculars.

The tangent line at parameter $\theta$ is $\mathbf{r}(\theta)+\lambda\mathbf{r}'(\theta)$. For the foot $Q$ of the perpendicular from the origin: $\mathbf{OQ}\cdot\mathbf{r}' = 0 \implies \bigl(\mathbf{r}+\lambda\mathbf{r}'\bigr)\cdot\mathbf{r}' = 0.$

$\mathbf{r}\cdot\mathbf{r}' = a^2(-\cos\theta\sin\theta + \sin\theta\cos\theta + \theta) = a^2\theta.$

$|\mathbf{r}'|^2 = 2a^2.$ So $\lambda = -\mathbf{r}\cdot\mathbf{r}'/|\mathbf{r}'|^2 = -\theta/2$.

The foot is: $Q = \mathbf{r} - \frac{\theta}{2}\mathbf{r}' = \frac{a}{2}(2\cos\theta+\theta\sin\theta,\; 2\sin\theta-\theta\cos\theta,\; \theta).$

Let $(x,y,z) = Q$. Then $z = a\theta/2$, so $\theta = 2z/a$. Also: $x^2+y^2 = \frac{a^2}{4}\bigl[(2\cos\theta+\theta\sin\theta)^2+(2\sin\theta-\theta\cos\theta)^2\bigr]$

$= \frac{a^2}{4}\bigl[4+\theta^2\bigr] = a^2+\frac{a^2\theta^2}{4} = a^2+z^2.$

$\boxed{x^2+y^2-z^2 = a^2.}$

Worked Example 3

2024 Paper 1, 2024-P1-Q5e-i (5 marks)

For a plane curve $\mathbf{r}=(f(t),g(t),0)$, prove that $\kappa = |f'g''-g'f''|/(f'^{\,2}+g'^{\,2})^{3/2}$. Deduce the torsion $\tau$.

Cross product. $\mathbf{r}' = (f',g',0)$, $\mathbf{r}'' = (f'',g'',0)$. Then $\mathbf{r}'\times\mathbf{r}'' = (g'\cdot 0-0\cdot g'',\; 0\cdot f''-f'\cdot 0,\; f'g''-g'f'') = (0,0,f'g''-g'f'').$ So $|\mathbf{r}'\times\mathbf{r}''| = |f'g''-g'f''|$ and $|\mathbf{r}'|^3 = (f'^{\,2}+g'^{\,2})^{3/2}$. Dividing: $\boxed{\kappa = \frac{|f'g''-g'f''|}{(f'^{\,2}+g'^{\,2})^{3/2}}.}$

Torsion. $\mathbf{r}''' = (f''',g''',0)$, which has zero $z$-component. Therefore $(\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}''' = (0,0,f'g''-g'f'')\cdot(f''',g''',0) = 0,$ giving $\boxed{\tau = 0}$. A plane curve has zero torsion — it does not twist out of its own plane.

Common Traps

• Curvature vector vs.\ magnitude. UPSC often asks for both $\boldsymbol{\kappa}$ and $\kappa$; omitting the vector expression costs marks. The curvature vector is $\bigl[(\mathbf{r}'\times\mathbf{r}'')\times\mathbf{r}'\bigr]/|\mathbf{r}'|^4$, not merely $|\mathbf{r}'\times\mathbf{r}''|/|\mathbf{r}'|^3$.
• Sign of the cross product. In the 2014 problem, expanding $\mathbf{r}'\times\mathbf{r}''$ carefully shows the $z$-component is $2+t^2 > 0$; a sign slip yields $-(2+t^2)$ and then a negative “magnitude,” which is impossible.
• Foot-of-perpendicular condition. The condition $\mathbf{OQ}\cdot\mathbf{r}'=0$ involves the full vector $\mathbf{OQ} = \mathbf{r}+\lambda\mathbf{r}'$, not just $\mathbf{r}$. Missing the $\lambda|\mathbf{r}'|^2$ term gives the wrong $\lambda$.
• Planar $\tau = 0$ proof. Simply noting ”$\mathbf{r}'''$ has no $z$-component” is the complete argument; there is no need to expand $f'''$ or $g'''$.

curvature-torsion (2 question(s); 2018, 2019)

Compute both curvature $\kappa$ and torsion $\tau$ (or their reciprocals $\rho$, $\sigma$) for a space curve

Recognition Cues

The question gives a parametric space curve and asks for “curvature and torsion” or “radius of curvature and radius of torsion.” Both the cross-product formula (for $\kappa$) and the triple-product formula (for $\tau$) are needed. The 2019 question uses the helix $z = au\tan\alpha$, making it slightly different from the standard $z = au$ helix of other questions.

Solution Template

1. Compute $\mathbf{r}'$, $\mathbf{r}''$, $\mathbf{r}'''$.
2. Find $|\mathbf{r}'|^2$; simplify using trigonometric identities where possible.
3. Compute $\mathbf{r}'\times\mathbf{r}''$ component by component.
4. Compute $|\mathbf{r}'\times\mathbf{r}''|^2$ and take the square root.
5. Curvature: $\kappa = |\mathbf{r}'\times\mathbf{r}''|/|\mathbf{r}'|^3$.
6. Evaluate the scalar triple product $(\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}'''$.
7. Torsion: $\tau = (\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}'''/|\mathbf{r}'\times\mathbf{r}''|^2$.
8. Invert to get $\rho = 1/\kappa$ and $\sigma = 1/|\tau|$ if requested.

Worked Example 1

2018 Paper 1, 2018-P1-Q7b (12 marks)

Find the curvature and torsion of $\mathbf{r}=\bigl(a(u-\sin u),\;a(1-\cos u),\;bu\bigr)$.

Derivatives. $\mathbf{r}' = \bigl(a(1-\cos u),\; a\sin u,\; b\bigr),$

$\mathbf{r}'' = (a\sin u,\; a\cos u,\; 0),$

$\mathbf{r}''' = (a\cos u,\; -a\sin u,\; 0).$

Speed squared. $|\mathbf{r}'|^2 = a^2(1-\cos u)^2 + a^2\sin^2 u + b^2 = 2a^2(1-\cos u)+b^2.$

Cross product. $\mathbf{r}'\times\mathbf{r}'' = \begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\ a(1-\cos u)&a\sin u&b\\ a\sin u&a\cos u&0\end{vmatrix}$

$= \hat{\imath}(0-ab\cos u) - \hat{\jmath}(0-ab\sin u) + \hat{k}(a^2(1-\cos u)\cos u - a^2\sin^2 u)$

$= (-ab\cos u,\; ab\sin u,\; a^2(\cos u - \cos^2 u - \sin^2 u))$

$= (-ab\cos u,\; ab\sin u,\; a^2(\cos u - 1)).$

Magnitude squared. $|\mathbf{r}'\times\mathbf{r}''|^2 = a^2b^2\cos^2 u + a^2b^2\sin^2 u + a^4(1-\cos u)^2 = a^2b^2 + a^4(1-\cos u)^2 = a^2\bigl[b^2 + a^2(1-\cos u)^2\bigr].$

Curvature. Since $|\mathbf{r}'|^3 = \bigl[2a^2(1-\cos u)+b^2\bigr]^{3/2}$: $\boxed{\kappa = \frac{a\sqrt{b^2+a^2(1-\cos u)^2}}{\bigl[2a^2(1-\cos u)+b^2\bigr]^{3/2}}.}$

Triple product. $(\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}''' = (-ab\cos u)(a\cos u) + (ab\sin u)(-a\sin u) + a^2(\cos u-1)\cdot 0$

$= -a^2b\cos^2 u - a^2b\sin^2 u = -a^2b.$

Torsion. $\boxed{\tau = \frac{-a^2b}{a^2\bigl[b^2+a^2(1-\cos u)^2\bigr]} = \frac{-b}{b^2+a^2(1-\cos u)^2}.}$

Worked Example 2

2019 Paper 1, 2019-P1-Q7b (15 marks)

Find the radius of curvature $\rho$ and radius of torsion $\sigma$ for the helix $x=a\cos u$, $y=a\sin u$, $z=au\tan\alpha$.

Derivatives. $\mathbf{r}' = (-a\sin u,\; a\cos u,\; a\tan\alpha),$

$\mathbf{r}'' = (-a\cos u,\; -a\sin u,\; 0),$

$\mathbf{r}''' = (a\sin u,\; -a\cos u,\; 0).$

Speed. $|\mathbf{r}'|^2 = a^2\sin^2 u + a^2\cos^2 u + a^2\tan^2\alpha = a^2(1+\tan^2\alpha) = a^2\sec^2\alpha$. So $|\mathbf{r}'| = a\sec\alpha$.

Cross product. $\mathbf{r}'\times\mathbf{r}'' = \begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\ -a\sin u&a\cos u&a\tan\alpha\\ -a\cos u&-a\sin u&0\end{vmatrix}$

$= \hat{\imath}(0+a^2\sin u\tan\alpha) - \hat{\jmath}(0+a^2\cos u\tan\alpha) + \hat{k}(a^2\sin^2 u+a^2\cos^2 u)$

$= (a^2\sin u\tan\alpha,\; -a^2\cos u\tan\alpha,\; a^2).$

Magnitude. $|\mathbf{r}'\times\mathbf{r}''|^2 = a^4\tan^2\alpha(\sin^2 u+\cos^2 u)+a^4 = a^4(\tan^2\alpha+1) = a^4\sec^2\alpha.$ So $|\mathbf{r}'\times\mathbf{r}''| = a^2\sec\alpha$.

Radius of curvature. $\kappa = \frac{a^2\sec\alpha}{a^3\sec^3\alpha} = \frac{\cos^2\alpha}{a}, \qquad \boxed{\rho = \frac{a}{\cos^2\alpha} = a\sec^2\alpha.}$

Triple product. $(\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}''' = a^2\sin u\tan\alpha\cdot a\sin u + (-a^2\cos u\tan\alpha)(-a\cos u) + a^2\cdot 0$

$= a^3\tan\alpha\sin^2 u + a^3\tan\alpha\cos^2 u = a^3\tan\alpha.$

Torsion and radius of torsion. $|\tau| = \frac{a^3\tan\alpha}{a^4\sec^2\alpha} = \frac{\tan\alpha}{a\sec^2\alpha} = \frac{\sin\alpha\cos\alpha}{a} = \frac{\sin 2\alpha}{2a}.$

$\boxed{\sigma = \frac{1}{|\tau|} = \frac{2a}{\sin 2\alpha}.}$

Common Traps

• $k$-component of $\mathbf{r}'\times\mathbf{r}''$ for the cycloid. The $k$-component simplifies as $\cos u - \cos^2 u - \sin^2 u = \cos u - 1$, not $+1$. A sign error here flips the sign of the triple product and hence the sign of $\tau$.
• Triple product is constant for the cycloid. The $-a^2b$ result is independent of $u$; if your answer for the triple product still contains $u$, recheck the dot product.
• $1+\tan^2\alpha = \sec^2\alpha$ shortcut. This collapses both $|\mathbf{r}'|$ and $|\mathbf{r}'\times\mathbf{r}''|$ to simple forms for the tilted helix. Without it, the algebra is unwieldy.
• Sign of $\tau$. Torsion can be negative (the $-b$ in the cycloid answer reflects a left-handed twist); the radius of torsion $\sigma = 1/|\tau|$ is always positive.

radius-of-curvature-polar (1 question(s); 2016)

Find the polar radius of curvature of a curve (e.g.\ cardioid) and verify a proportionality relation

Recognition Cues

The curve is given in polar form $r = r(\theta)$. The question asks to “show $\rho^2$ is proportional to $r$” or to evaluate $\rho$ at specific angles. The keyword “cardioid” or “lemniscate” in combination with a polar equation signals this archetype.

Solution Template

1. Compute $r_1 = dr/d\theta$ and $r_2 = d^2r/d\theta^2$.
2. Evaluate $r^2 + r_1^2$ (numerator base) and $r^2 + 2r_1^2 - r\,r_2$ (denominator expression).
3. Simplify both using the original curve equation; factor out common factors.
4. Apply $\rho = (r^2+r_1^2)^{3/2}/|r^2+2r_1^2-r\,r_2|$.
5. To show proportionality, express $\rho^2$ in terms of $r$ only (eliminating $\theta$).
6. Substitute the specific $\theta$ values to get numerical answers.

Worked Example

2016 Paper 1, 2016-P1-Q8d (15 marks)

For the cardioid $r = a(1+\cos\theta)$, show that $\rho^2$ is proportional to $r$. Find $\rho$ at $\theta = 0$, $\theta = \pi/4$, and $\theta = \pi/2$.

Derivatives. $r_1 = -a\sin\theta, \qquad r_2 = -a\cos\theta.$

Numerator base. $r^2+r_1^2 = a^2(1+\cos\theta)^2 + a^2\sin^2\theta = a^2(1+2\cos\theta+\cos^2\theta+\sin^2\theta) = 2a^2(1+\cos\theta) = 2ar.$

Denominator expression. $r^2+2r_1^2-r\,r_2 = a^2(1+\cos\theta)^2 + 2a^2\sin^2\theta - a(1+\cos\theta)(-a\cos\theta).$

$= a^2\bigl[(1+\cos\theta)^2 + 2\sin^2\theta + (1+\cos\theta)\cos\theta\bigr]$

$= a^2\bigl[1+2\cos\theta+\cos^2\theta + 2\sin^2\theta + \cos\theta+\cos^2\theta\bigr]$

$= a^2\bigl[3 + 3\cos\theta\bigr] = 3a^2(1+\cos\theta) = 3ar.$

Radius of curvature. $\rho = \frac{(2ar)^{3/2}}{3ar} = \frac{2\sqrt{2}\,a^{1/2}\,r^{3/2}}{3ar} = \frac{2\sqrt{2}}{3}\sqrt{\frac{r}{a}\cdot a} = \frac{2\sqrt{2}}{3}\sqrt{ar}.$

Wait — let us be careful: $\rho = \frac{(2ar)^{3/2}}{3ar} = \frac{2^{3/2}a^{3/2}r^{3/2}}{3ar} = \frac{2\sqrt{2}\,a^{1/2}\,r^{1/2}}{3}.$

Proportionality. $\rho^2 = \dfrac{8a\,r}{9}$, which is proportional to $r$. $\quad\checkmark$

Equivalently, $\rho = \dfrac{2\sqrt{2}}{3}\sqrt{ar} = \dfrac{2}{3}\sqrt{2ar}$.

Values at specific angles. Use $r = a(1+\cos\theta)$:

$\boxed{\rho\big|_{\theta=0} = \tfrac{4a}{3}, \quad \rho\big|_{\theta=\pi/4} = \tfrac{2a}{3}\sqrt{2+\sqrt{2}}, \quad \rho\big|_{\theta=\pi/2} = \tfrac{2a\sqrt{2}}{3}.}$

Common Traps

• Sign of $-r\,r_2$. For the cardioid, $r_2 = -a\cos\theta$, so $-r\,r_2 = +a(1+\cos\theta)\cdot a\cos\theta > 0$ for $\theta < \pi/2$. Students often drop this term or give it the wrong sign, wrecking the denominator.
• Numerator exponent. The formula has $(r^2+r_1^2)^{3/2}$ in the numerator; writing it as $(r^2+r_1^2)^{1/2}$ (forgetting the cube) is a common error that makes $\rho$ dimensionally wrong.
• Proportionality check. Once $\rho = (2\sqrt{2}/3)\sqrt{ar}$, compute $\rho^2 = 8ar/9$; the factor $8/9$ need not be simplified further — “proportional to $r$” is established.
• $\theta=0$ endpoint. At $\theta=0$, $r=2a$ (not $r=a$); substituting $r=a$ instead gives $\rho=2\sqrt{2}a/3$, which is the $\theta=\pi/2$ answer.

Marks-Aware Writing

For a 10–12 mark curvature question: Write $\mathbf{r}'$, $\mathbf{r}''$, and $|\mathbf{r}'|$ explicitly. Display $\mathbf{r}'\times\mathbf{r}''$ as a $3\times3$ determinant, even if two rows are obvious — the examiner awards a process mark here. State the formula $\kappa = |\mathbf{r}'\times\mathbf{r}''|/|\mathbf{r}'|^3$ before substituting. Box the final answer.

For a 15–16 mark question: The same requirements apply, plus you must show the locus derivation (2017) or both $\rho$ and $\sigma$ (2019) fully worked. For the locus problem, set up the foot-of-perpendicular condition in vector form, find $\lambda$ explicitly, then verify the surface identity by expanding and cancelling. Skipping the “verify by algebra” step loses 3–4 marks even if the setup is correct.

For the 5-mark proof (2024): A three-step proof (write $\mathbf{r}'\times\mathbf{r}''$ for a planar curve, note the $z$-component equals $f'g''-g'f''$, divide) is all that is needed. Do not pad with unnecessary theory; the mark scheme rewards conciseness.

General principle: Every intermediate vector — the cross product, its magnitude, the triple product — must appear on the page. Examiners mark the process as much as the result.

Practice Set

• 2024-P1-Q5e-ii (5 m) — — second part of the 2024 question; find the curvature of a specific planar curve using the formula just proved
• 2023-P1-Q7c (15 m) — — full curvature-torsion computation for a space curve; test whether the methods from 2018/2019 transfer directly