Differentiation of a vector function of a scalar variable

At a Glance

• Frequency: 4 sub-parts across 4 of 13 years (2013, 2017, 2023, 2024)
• Priority tier: T3
• Marks (count): 10 (4)
• Average solve time: ~8 min
• Difficulty mix: easy 2, medium 2
• Section: B | Dominant type: computation

Why This Chapter Matters

Differentiation of vector functions appears in Section B compulsory (Q5) at 10 marks. The three archetypes are: (a) angle between tangent vectors at two points — straightforward dot product; (b) acceleration components along velocity and perpendicular to the plane of $(\mathbf{r},\mathbf{v})$ — projects onto $\mathbf{v}$ and $\mathbf{r}\times\mathbf{v}$; (c) differentiating a cross or triple product with respect to a parameter — use the product rule. All three reduce to differentiating component-wise and combining.

Minimum Theory

Vector function of a scalar. If $\mathbf{r}(t) = f(t)\hat{i}+g(t)\hat{j}+h(t)\hat{k}$, then $\mathbf{r}'(t) = f'(t)\hat{i}+g'(t)\hat{j}+h'(t)\hat{k}$ (differentiate each component).

Tangent and angle. The tangent vector at parameter $t$ is $\mathbf{r}'(t)$. The angle $\theta$ between tangents at $t_1$ and $t_2$ satisfies $\cos\theta = \frac{\mathbf{r}'(t_1)\cdot\mathbf{r}'(t_2)}{|\mathbf{r}'(t_1)|\,|\mathbf{r}'(t_2)|}.$

Acceleration components. For a particle with position $\mathbf{r}(t)$, velocity $\mathbf{v}=\dot{\mathbf{r}}$, and acceleration $\mathbf{a}=\ddot{\mathbf{r}}$:

• Along $\mathbf{v}$: scalar projection $\dfrac{\mathbf{a}\cdot\mathbf{v}}{|\mathbf{v}|}$.
• Perpendicular to the plane of $\mathbf{r}$ and $\mathbf{v}$: the plane’s normal is $\mathbf{r}\times\mathbf{v}$; the component is $\dfrac{\mathbf{a}\cdot(\mathbf{r}\times\mathbf{v})}{|\mathbf{r}\times\mathbf{v}|}$.

Product rule for vector products. $\dfrac{d}{dt}(\mathbf{a}\times\mathbf{b}) = \mathbf{a}'\times\mathbf{b}+\mathbf{a}\times\mathbf{b}'$. Similarly for dot and triple products. When the vectors depend on a parameter $\theta$, differentiate component-by-component.

Question Archetypes

tangent-angle (1 question(s); 2013)

Recognition Cues

• A space curve $\mathbf{r}(t)$ is given; asked for “angle between tangents” at two named values of $t$.

Solution Template

1. Compute $\mathbf{r}'(t)$; evaluate at $t_1$ and $t_2$.
2. Dot product $\mathbf{r}'(t_1)\cdot\mathbf{r}'(t_2)$; magnitudes $|\mathbf{r}'(t_1)|$ and $|\mathbf{r}'(t_2)|$.
3. $\cos\theta = \text{dot}/(\text{mag}_1\times\text{mag}_2)$; state $\theta=\cos^{-1}(\ldots)$.

Worked Example

2013 Paper 1, 2013-P1-Q8b (10 marks)

Angle between tangents to $\mathbf{r}=(t^2,2t,-t^3)$ at $t=+1$ and $t=-1$.

Step 1. $\mathbf{r}'=(2t,2,-3t^2)$. At $t=1$: $(2,2,-3)$. At $t=-1$: $(-2,2,-3)$.

Step 2. $(2)(-2)+(2)(2)+(-3)(-3) = -4+4+9 = 9$. $|\mathbf{r}'(\pm1)| = \sqrt{4+4+9}=\sqrt{17}$.

Step 3. $\cos\theta = 9/17$. $\boxed{\theta = \cos^{-1}(9/17).}$

Common Traps

• Tangents at $t$ and $-t$ are not negatives. Here the $\hat{i}$-component $2t$ is odd but $\hat{j}=2$ (even) and $\hat{k}=-3t^2$ (even), so the two tangent vectors are not antiparallel.

motion-components (1 question(s); 2017)

Recognition Cues

• A position vector $\mathbf{r}(t)$ is given.
• Asked for the component of acceleration (a) “parallel to velocity” and (b) “perpendicular to the plane of $\mathbf{r}$ and $\mathbf{v}$.”
• “Perpendicular to the plane of $\mathbf{r}$ and $\mathbf{v}$” means project onto $\mathbf{r}\times\mathbf{v}$, not onto the principal normal.

Solution Template

1. Compute $\mathbf{v}=\dot{\mathbf{r}}$ and $\mathbf{a}=\ddot{\mathbf{r}}$; evaluate at the given $t$.
2. Component along $\mathbf{v}$: $(\mathbf{a}\cdot\mathbf{v})/|\mathbf{v}|$.
3. Compute $\mathbf{r}\times\mathbf{v}$ at the given $t$; its magnitude $|\mathbf{r}\times\mathbf{v}|$.
4. Component along $\mathbf{r}\times\mathbf{v}$: $(\mathbf{a}\cdot(\mathbf{r}\times\mathbf{v}))/|\mathbf{r}\times\mathbf{v}|$.

Worked Example

2017 Paper 1, 2017-P1-Q5e (10 marks)

$\mathbf{r}=(\sin t,\cos2t,t^2+2t)$. Find components of $\mathbf{a}$ parallel to $\mathbf{v}$ and perpendicular to the plane $(\mathbf{r},\mathbf{v})$ at $t=0$.

Step 1. $\mathbf{v}=(\cos t,-2\sin2t,2t+2)$, $\mathbf{a}=(-\sin t,-4\cos2t,2)$. At $t=0$: $\mathbf{r}_0=(0,1,0),\quad\mathbf{v}_0=(1,0,2),\quad\mathbf{a}_0=(0,-4,2).$

Step 2. $\mathbf{a}_0\cdot\mathbf{v}_0 = 0+0+4=4$. $|\mathbf{v}_0|=\sqrt{5}$. Component along $\mathbf{v}$: $\boxed{4/\sqrt{5}=4\sqrt{5}/5.}$

Step 3. $\mathbf{r}_0\times\mathbf{v}_0 = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\0&1&0\\1&0&2\end{vmatrix} = (2,0,-1)$. $|\mathbf{r}_0\times\mathbf{v}_0|=\sqrt{5}$.

Step 4. $\mathbf{a}_0\cdot(2,0,-1)=0+0-2=-2$. Component: $\boxed{-2/\sqrt{5}=-2\sqrt{5}/5.}$

Common Traps

• “Perpendicular to the plane of $\mathbf{r}$ and $\mathbf{v}$” ≠ “principal normal.” The principal normal is $d\hat{T}/ds$, which is in the osculating plane. The direction perpendicular to the plane of $\mathbf{r}$ and $\mathbf{v}$ is $\mathbf{r}\times\mathbf{v}$.
• Differentiate $\cos2t$ with the chain rule: derivative is $-2\sin2t$, so at $t=0$ it is $0$, and the double derivative $-4\cos2t$ at $t=0$ is $-4$.
• Report the signed component. The problem asks for the component in the direction of $\mathbf{r}\times\mathbf{v}$; the signed value $-2/\sqrt{5}$ indicates $\mathbf{a}$ points opposite to $\mathbf{r}\times\mathbf{v}$.

vector-derivative (2 question(s); 2023, 2024)

Recognition Cues

• A vector expression $\mathbf{a}\times(\mathbf{b}\times\mathbf{c})$ or a scalar triple product depends on a parameter $\theta$.
• Asked for $d/d\theta$ at a specific value of $\theta$.
• Or: the parallelogram/parallelepiped involves time-varying vectors; find rate of change of area or volume.

Solution Template

1. Compute the cross product or triple product explicitly (component-by-component).
2. Differentiate the resulting vector or scalar with respect to the parameter.
3. Evaluate at the specified value.

Worked Example

2024 Paper 1, 2024-P1-Q5c (10 marks)

$\bar\alpha=t\hat{i}+(t+1)\hat{j}+(2t+1)\hat{k}$, $\bar\beta=2t\hat{i}+(3t-1)\hat{j}+t\hat{k}$, $\bar\gamma=\hat{i}+3t\hat{j}+\hat{k}$. Find (a) $\dfrac{d}{dt}(\bar\alpha\times\bar\gamma)$ and (b) $\dfrac{dV}{dt}$ at $t=1$.

Part (a). $\bar\alpha\times\bar\gamma = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\t&t+1&2t+1\\1&3t&1\end{vmatrix}$:

• $\hat{i}$: $(t+1)-3t(2t+1) = -6t^2-2t+1$
• $\hat{j}$: $-(t-(2t+1)) = t+1$
• $\hat{k}$: $3t^2-(t+1)$

$\mathbf{A}(t)=(-6t^2-2t+1,\;t+1,\;3t^2-t-1)$.

$\boxed{\tfrac{d\mathbf{A}}{dt}=(-12t-2,\;1,\;6t-1).}$

Part (b). $V = \bar\alpha\cdot(\bar\beta\times\bar\gamma)$; expand the $3\times3$ determinant to get $V(t)=9t^3+2t^2-3t+1$ (computation omitted for brevity). Then $dV/dt=27t^2+4t-3$; at $t=1$: $\boxed{28.}$

Common Traps

• Cross product gives a vector; differentiate component-wise. Don’t try to differentiate $|\bar\alpha\times\bar\gamma|$ unless specifically asked for the scalar area.
• Scalar triple product is a scalar. Differentiating the determinant formula gives $dV/dt=\dot{\bar\alpha}\cdot(\bar\beta\times\bar\gamma)+\bar\alpha\cdot(\dot{\bar\beta}\times\bar\gamma)+\bar\alpha\cdot(\bar\beta\times\dot{\bar\gamma})$ — but computing $V(t)$ explicitly as a polynomial and differentiating is usually faster.

Marks-Aware Writing

For tangent-angle (10 marks): show $\mathbf{r}'$ at both points, display the dot product and magnitudes, state $\cos\theta$ and $\theta$. Three steps suffice.

For motion-components (10 marks): write $\mathbf{v}$ and $\mathbf{a}$ from derivatives, evaluate at $t=0$, then show the two projections separately — including the cross product $\mathbf{r}\times\mathbf{v}$ displayed as a determinant.

For vector-derivative (10 marks): show the cross product or determinant explicitly with each cofactor computed. For the volume part, expanding the determinant fully and then differentiating is clearer than using the product-rule formula.

Practice Set

• 2023-P1-Q5e (10 m) — — Hint: compute $\mathbf{a}\times(\mathbf{b}\times\mathbf{c})$ component-by-component at the two $\theta$ values; track which components depend on $\theta$.