Motion in a plane can be described using Cartesian coordinates, $\vec{A} = A_x\hat{i} + A_y\hat{j}$, where $\hat{i}$ and $\hat{j}$ are unit vectors along the $x$ and $y$ directions. It can equally be described using plane polar coordinates, $\vec{A} = A_r\hat{r} + A_\theta\hat{\theta}$, where $$\hat{r} = \frac{\vec{r}}{r} = \cos\theta\,\hat{i} + \sin\theta\,\hat{j}, \qquad \hat{\theta} = -\sin\theta\,\hat{i} + \cos\theta\,\hat{j}$$ are unit vectors along the directions in which $r$ and $\theta$ increase, respectively. (a) Express $\hat{i}$ and $\hat{j}$ in terms of $\hat{r}$ and $\hat{\theta}$. (b) Show that $\hat{r}$ and $\hat{\theta}$ are both unit vectors and are perpendicular to each other. (c) Show that $\dfrac{d}{dt}(\hat{r}) = \omega\hat{\theta}$ and $\dfrac{d}{dt}(\hat{\theta}) = -\omega\hat{r}$, where $\omega = \dfrac{d\theta}{dt}$. (d) For a particle moving along a spiral given by $\vec{r} = a\theta\,\hat{r}$ with $a = 1$ (unit), find the dimensions of $a$. (e) Find the velocity and acceleration in polar-vector form for the particle moving along the spiral in (d).

Motion in a plane can be described using Cartesian coordinates, vecA = Aₓhati + Ayhatj, where hati and hatj ar

NCERT Exemplar · Q36

\vec{A} = A_x\hat{i} + A_y\hat{j}

(a) Express

\hat{i}

and

\hat{j}

in terms of

\hat{r}

and

\hat{\theta}

(b) Show that

\hat{r}

and

\hat{\theta}

are both unit vectors and are perpendicular to each other.

\dfrac{d}{dt}(\hat{r}) = \omega\hat{\theta}

and

\dfrac{d}{dt}(\hat{\theta}) = -\omega\hat{r}

, where

\omega = \dfrac{d\theta}{dt}

(d) For a particle moving along a spiral given by

\vec{r} = a\theta\,\hat{r}

with

a = 1

(unit), find the dimensions of

a

(e) Find the velocity and acceleration in polar-vector form for the particle moving along the spiral in (d).

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Concept understanding — Kinematics Vector Differentiation

Kinematics Vector Differentiation

Imagine you're tracking a drone flying in the sky. At any instant, it has a position — say, 30 metres east and 40 metres north of you. That's a vector: $\vec{r} = 30\hat{i} + 40\hat{j}$ . A second later, it's moved. The question kinematics asks is: how fast is that position changing? That rate of change is velocity, and to get it, you differentiate the position vector.

But here's the key difference from school calculus: in school, you differentiated a scalar function like $y = x^2$ . Here, you're differentiating a vector function — something that has both magnitude and direction, and both can change with time.

The Intuition First

Think of a vector as an arrow. When time passes, that arrow can do two things:

It can get longer or shorter (magnitude changes).
It can rotate (direction changes).

Velocity is the total rate of change of that arrow. If the drone flies straight away from you, only the length changes. If it flies in a circle around you, only the direction changes. Most real motion does both.

So vector differentiation is just: take the derivative of each component separately, because components are independent scalars.

The Precise Statement

If a position vector is written in Cartesian coordinates as:

\vec{r}(t) = x(t)\,\hat{i} + y(t)\,\hat{j} + z(t)\,\hat{k}

where $\hat{i}, \hat{j}, \hat{k}$ are fixed unit vectors (they don't change direction with time), then:

\frac{d\vec{r}}{dt} = \frac{dx}{dt}\,\hat{i} + \frac{dy}{dt}\,\hat{j} + \frac{dz}{dt}\,\hat{k}

That's it. You differentiate each component function $x(t), y(t), z(t)$ exactly as you would in single-variable calculus, and the unit vectors stay put.

\frac{d}{dt}\big(f(t)\,\hat{u}\big) = \frac{df}{dt}\,\hat{u} \quad\text{(if $\hat{u}$ is constant)}

Why This Works

The derivative of a vector is defined the same way as for a scalar — as a limit:

\frac{d\vec{r}}{dt} = \lim_{\Delta t \to 0} \frac{\vec{r}(t+\Delta t) - \vec{r}(t)}{\Delta t}

The numerator is a vector difference. When you write $\vec{r}$ in components, the difference splits into component differences. The limit then acts on each component separately because the unit vectors are constant. So the definition forces component-wise differentiation.

A Concrete Example

A particle moves such that:

\vec{r}(t) = (3t^2)\,\hat{i} + (5\sin t)\,\hat{j} + (2e^{-t})\,\hat{k}

Its velocity is:

\vec{v}(t) = \frac{d\vec{r}}{dt} = (6t)\,\hat{i} + (5\cos t)\,\hat{j} + (-2e^{-t})\,\hat{k}

Notice: the $x$ -component grows linearly, the $y$ -component oscillates, the $z$ -component decays. Each derivative is just the ordinary derivative of that component's function.

The One Trap: Non-Constant Unit Vectors

The rule above assumes $\hat{i}, \hat{j}, \hat{k}$ are fixed. That's true in Cartesian coordinates. But in polar coordinates, the unit vectors $\hat{r}$ and $\hat{\theta}$ rotate as the particle moves. Differentiating a vector in polar coordinates requires the product rule because the unit vectors themselves depend on time. …

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Parts (a)-(c) are algebra with the given definitions of $\hat{r}$ and $\hat{\theta}$ : invert them, check magnitudes and the dot product, and differentiate. In (d) the relation $r=a\theta$ with dimensionless $\theta$ forces $a$ to carry the dimension of length. In (e) differentiate $\vec{r}=a\theta\hat{r}$ using the results of (c).

(a) $\hat{i}$ and $\hat{j}$ in terms of $\hat{r},\hat{\theta}$

Given $\hat{r}=\cos\theta\,\hat{i}+\sin\theta\,\hat{j}$ and $\hat{\theta}=-\sin\theta\,\hat{i}+\cos\theta\,\hat{j}$ . Form the combinations:

\cos\theta\,\hat{r}-\sin\theta\,\hat{\theta} = (\cos^2\theta+\sin^2\theta)\hat{i} = \hat{i},

\sin\theta\,\hat{r}+\cos\theta\,\hat{\theta} = (\sin^2\theta+\cos^2\theta)\hat{j} = \hat{j}.

So $\hat{i}=\cos\theta\,\hat{r}-\sin\theta\,\hat{\theta}$ and $\hat{j}=\sin\theta\,\hat{r}+\cos\theta\,\hat{\theta}$ .

(b) Unit vectors and perpendicularity

|\hat{r}|^2=\cos^2\theta+\sin^2\theta=1,\qquad |\hat{\theta}|^2=\sin^2\theta+\cos^2\theta=1,

so both are unit vectors. Their dot product

\hat{r}\cdot\hat{\theta} = (\cos\theta)(-\sin\theta)+(\sin\theta)(\cos\theta)=0,

so they are perpendicular.

(c) Time derivatives

With $\omega=d\theta/dt$ ,

\frac{d\hat{r}}{dt} = (-\sin\theta\,\hat{i}+\cos\theta\,\hat{j})\frac{d\theta}{dt} = \omega\,\hat{\theta},

\frac{d\hat{\theta}}{dt} = (-\cos\theta\,\hat{i}-\sin\theta\,\hat{j})\frac{d\theta}{dt} = -\omega(\cos\theta\,\hat{i}+\sin\theta\,\hat{j}) = -\omega\,\hat{r}.

(d) Dimensions of $a$

For the spiral, the position vector is $\vec{r}=a\theta\,\hat{r}$ , so its magnitude is $r=a\theta$ . Here $r$ is a length, dimension $[\text{L}]$ , while $\theta$ (an angle in radians) is dimensionless. Hence

[a] = \frac{[r]}{[\theta]} = [\text{L}] = \text{M}^0\text{L}^1\text{T}^0,

so $a$ has the dimension of length (it is measured in metres).

(e) Velocity and acceleration on the spiral

Start from $\vec{r}=a\theta\,\hat{r}$ . Differentiate, using $\dot{\hat{r}}=\dot\theta\,\hat{\theta}$ :

\vec{v}=\frac{d}{dt}(a\theta\,\hat{r}) = a\dot\theta\,\hat{r} + a\theta\,\dot{\hat{r}} = a\dot\theta\,\hat{r} + a\theta\dot\theta\,\hat{\theta} = a\dot\theta\,(\hat{r}+\theta\,\hat{\theta}).

Differentiate again, using $\dot{\hat{r}}=\dot\theta\,\hat{\theta}$ and $\dot{\hat{\theta}}=-\dot\theta\,\hat{r}$ : …

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Concept: Treat $(\hat r,\hat\theta)$ as a Rotated Copy of $(\hat\imath,\hat\jmath)$ , Using Rotation-Matrix and Rotating-Frame Facts

Method: Rotation Matrix for (a)-(b), the $\vec\omega\times\hat e$ Rule for (c), and the General Polar Velocity/Acceleration Formulas for (e)

Instead of inverting and differentiating the given sine/cosine expressions component by component, this method recognizes $(\hat r,\hat\theta)$ as the ordinary axes $(\hat\imath,\hat\jmath)$ rotated by angle $\theta$ , and applies three standard facts about rotations that make parts (a), (b), and (c) nearly automatic.

Steps

(a) Recognize the given definitions as a rotation matrix: $\begin{pmatrix}\hat r\\\hat\theta\end{pmatrix} = \begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}\begin{pmatrix}\hat\imath\\\hat\jmath\end{pmatrix}$ . A rotation matrix's inverse is simply the rotation by the opposite angle (equivalently, its transpose, since rotation matrices are orthogonal):

\hat\imath = \cos\theta\,\hat r - \sin\theta\,\hat\theta, \qquad \hat\jmath = \sin\theta\,\hat r + \cos\theta\,\hat\theta.

(b) Unit length and perpendicularity follow automatically from "this is a rotation." A rotation matrix is orthogonal by construction (it preserves lengths and angles) -- so $\hat r,\hat\theta$ being unit vectors and mutually perpendicular doesn't need a separate dot-product calculation; it's guaranteed by rotating the orthonormal pair $(\hat\imath,\hat\jmath)$ .
(c) Use the standard rotating-unit-vector rule $\dot{\hat e} = \vec\omega\times\hat e$ (with $\vec\omega=\dot\theta\,\hat z$ for planar rotation), instead of differentiating $\cos\theta,\sin\theta$ directly:

\dot{\hat r} = \dot\theta\,\hat z\times\hat r = \dot\theta\,\hat\theta \;(=\omega\hat\theta), \qquad \dot{\hat\theta} = \dot\theta\,\hat z\times\hat\theta = -\dot\theta\,\hat r\;(=-\omega\hat r),

using the standard planar identities $\hat z\times\hat r=\hat\theta$ and $\hat z\times\hat\theta=-\hat r$ (true for any angle $\theta$ , since $\hat r,\hat\theta,\hat z$ always form a right-handed triad).

(d) Dimensional consistency of $r=a\theta$ : $\theta$ (radians) is dimensionless, $r$ has dimension $[\text L]$ , so $[a]=[\text L]$ -- unchanged from the direct method.
(e) Apply the general polar velocity/acceleration formulas (which follow from the product rule combined with the rotating-vector rule in step 3, and are worth memorizing once for any polar-coordinate problem):

\vec v = \dot r\,\hat r + r\dot\theta\,\hat\theta, \qquad \vec a = (\ddot r - r\dot\theta^2)\,\hat r + (r\ddot\theta+2\dot r\dot\theta)\,\hat\theta.

Substitute $r=a\theta$ , $\dot r=a\dot\theta$ , $\ddot r=a\ddot\theta$ : …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-A1 markMCQ
$y=9 x^2$
›Reveal solutionSolution
The key idea is that constant horizontal velocity means zero horizontal acceleration, so the given acceleration is purely vertical. Using the path constraint $y=9x^2$ and differentiating twice with respect to time gives a relation between $v_x$ and the vertical acceleration, yielding $v_x = \frac{1}{3}\,\text{m/s}$ .

We are told the particle moves on the parabola $y = 9x^2$ , and its $x$ -component of velocity is constant. That means $v_x = \dot{x}$ is constant, so $\ddot{x} = 0$ . The acceleration vector is given as $2\,\hat{j}\,\text{m/s}^2$ , meaning the only acceleration is vertical. This is a classic constrained-motion problem: the path geometry links $x$ and $y$ , so we can differentiate that relation to connect velocities and accelerations.

Write the path equation and differentiate once The path is $y = 9x^2$ . Differentiate with respect to time $t$ :

$\frac{dy}{dt} = 9 \cdot 2x \frac{dx}{dt} = 18x \, v_x.$
So the vertical velocity is $v_y = 18x\,v_x$ .

Differentiate again to get acceleration Differentiate $v_y = 18x\,v_x$ with respect to time. Since $v_x$ is constant, we treat it as a constant factor:

$a_y = \frac{d}{dt}(18x\,v_x) = 18 v_x \frac{dx}{dt} = 18 v_x \cdot v_x = 18 v_x^2.$
Notice: we used $\frac{dx}{dt} = v_x$ and $v_x$ constant, so no product rule term from $v_x$ itself.

Equate to given acceleration The problem states the acceleration is $2\,\hat{j}\,\text{m/s}^2$ , so $a_y = 2$ . Thus:
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KCET 2023Set A-31 markMCQ
$v_0$
›Reveal solutionSolution

Average velocity is displacement over time; use $v^2-v_0^2=2as$ to express the displacement.

Step 1 — Definition.
$\bar v = \frac{\text{displacement}}{\text{time}} = \frac{s}{t}.$
Step 2 — Get $s$ from a kinematic equation.

For uniform acceleration the time-free equation is
$v^{2} = v_0^{2} + 2as \;\Longrightarrow\; s = \frac{v^{2}-v_0^{2}}{2a}.$
We use this particular equation because the options are written in terms of $v^2$ and $v_0^2$ .

Step 3 — Divide by $t$ .
$\bar v = \frac{s}{t} = \frac{v^{2}-v_0^{2}}{2at}.$
Step 4 — Cross-check with the standard result.

For constant acceleration the velocity–time graph is a straight line, so

$\bar v = \frac{v+v_0}{2}.$ …
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KCET 2019Set A-11 markMCQ
$\vec{F} = (-3\hat{i} + 4\hat{j})$
›Reveal solutionSolution
The key idea is that the velocity component along the x-axis must become zero for the velocity to be purely along the y-axis. Using Newton’s second law and kinematic equations, the time comes out to be 10 s.

The problem asks for the time when the velocity is only along the y-axis. That means the x-component of velocity must be zero at that instant. The force is constant, so the acceleration is constant — this is a straightforward application of the equations of motion under uniform acceleration.

Let’s break it down.

Find the acceleration vector. Newton’s second law: $\vec{F} = m \vec{a}$ . Given $m = 5$ kg and $\vec{F} = (-3\hat{i} + 4\hat{j})$ N,

$\vec{a} = \frac{\vec{F}}{m} = \left( \frac{-3}{5} \hat{i} + \frac{4}{5} \hat{j} \right) \text{ m/s}^2.$

Write the velocity as a function of time. For constant acceleration,

$\vec{v}(t) = \vec{u} + \vec{a} t,$
where initial velocity $\vec{u} = (6\hat{i} - 12\hat{j})$ m/s.

So
$\vec{v}(t) = \left( 6 - \frac{3}{5}t \right) \hat{i} + \left( -12 + \frac{4}{5}t \right) \hat{j}.$

Set the x-component to zero. For velocity to be purely along the y-axis, the $\hat{i}$ component must vanish:

$6 - \frac{3}{5}t = 0.$
Solve:
$\frac{3}{5}t = 6 \quad \Rightarrow \quad t = 6 \times \frac{5}{3} = 10 \text{ s}.$

Check the y-component at that time. At $t = 10$ s, $v_y = -12 + \frac{4}{5}(10) = -12 + 8 = -4 \text{ m/s}.$ …
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(a) i^\hat{i}i^ and j^\hat{j}j^​ in terms of r^,θ^\hat{r},\hat{\theta}r^,θ^

(b) Unit vectors and perpendicularity

(c) Time derivatives

(d) Dimensions of aaa

(e) Velocity and acceleration on the spiral

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(a) $\hat{i}$ and $\hat{j}$ in terms of $\hat{r},\hat{\theta}$

(d) Dimensions of $a$

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