The displacement vector of a particle of mass $m$ is given by $\mathbf{r}(t) = \hat{i}\,A\cos\omega t + \hat{j}\,B\sin\omega t$. (a) Show that the trajectory is an ellipse. (b) Show that $\mathbf{F} = -m\omega^2\mathbf{r}$.

The displacement vector of a particle of mass m is given by mathbfr(t) = hatiAcosω t + hatjBsinω t. (a) Show t

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m

(a) Show that the trajectory is an ellipse.

(b) Show that

\mathbf{F} = -m\omega^2\mathbf{r}

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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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The particle moves on an ellipse because its $x$ and $y$ coordinates satisfy the standard ellipse equation. The force is central and obeys Hooke’s law, directly proportional to $\mathbf{r}$ and directed toward the origin, with $\mathbf{F} = -m\omega^2 \mathbf{r}$ .

The displacement vector is given in component form:

x(t) = A\cos\omega t, \qquad y(t) = B\sin\omega t.

This is a parametric description of a curve. To see what curve it is, we eliminate the parameter $t$ .

Eliminate $t$ to find the trajectory. From $x = A\cos\omega t$ , we have $\cos\omega t = x/A$ . From $y = B\sin\omega t$ , we have $\sin\omega t = y/B$ . Using the identity $\cos^2\theta + \sin^2\theta = 1$ :

\left(\frac{x}{A}\right)^2 + \left(\frac{y}{B}\right)^2 = 1.

This is the standard equation of an ellipse centered at the origin, with semi-major axis $A$ along the $x$ -axis and semi-minor axis $B$ along the $y$ -axis (or vice versa, depending on which is larger). So the trajectory is indeed an ellipse.

Find the velocity and acceleration. Differentiate $\mathbf{r}(t)$ with respect to time:

\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = -\hat{i}\,A\omega\sin\omega t + \hat{j}\,B\omega\cos\omega t.

Differentiate again for acceleration:

\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = -\hat{i}\,A\omega^2\cos\omega t - \hat{j}\,B\omega^2\sin\omega t.

Factor out $-\omega^2$ :

\mathbf{a}(t) = -\omega^2 \bigl( \hat{i}\,A\cos\omega t + \hat{j}\,B\sin\omega t \bigr) = -\omega^2 \mathbf{r}(t).

Apply Newton’s second law. The force on the particle is $\mathbf{F} = m\mathbf{a}$ . Substituting the acceleration: $\mathbf{F} = m(-\omega^2 \mathbf{r}) = -m\omega^2 \mathbf{r}.$ …

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

KEAM 2026Set eng-2026-04184 marksMCQ
$x$
›Reveal solutionSolution

With displacement $x=1+t^2$ and velocity $v=2t$ , equality gives $(t-1)^2=0$ , i.e. $t=1\,$ s.

The position/displacement is $x=1+t^2$ (in metres) and the velocity is
$v=\frac{dx}{dt}=2t\ \text{m/s}.$
Set the magnitudes equal: …
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KEAM 2026Set eng-2026-04194 marksMCQ
$y$
›Reveal solutionSolution

The position is quadratic, so acceleration is constant $=2\ \text{m s}^{-2}$ ; the average equals this constant value.

Given $y=t^2+2t+3$ .

Velocity: $v=\dfrac{dy}{dt}=2t+2$ .

Acceleration: $a=\dfrac{dv}{dt}=2\ \text{m s}^{-2}$ , which is constant. …
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KEAM 2026Set eng-2026-04204 marksMCQ
$\vec{x} = (t^3 - 3t^2 + 2)\,\hat{\imath}$
›Reveal solutionSolution

Differentiate the position to get velocity, then set it to zero.

$\vec{x}=(t^3-3t^2+2)\,\hat{\imath}$ , so $v=\dfrac{dx}{dt}=3t^2-6t=3t(t-2)$ . …
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KEAM 2026Set eng-2026-04214 marksMCQ
$\vec{r} = 2t\hat{i} + \sqrt{3}t^{2}\hat{j} + 5\hat{k}$
›Reveal solutionSolution

Differentiate to get $\vec v$ ; at $t=1$ the components give $\tan\theta=\sqrt3$ , so $\theta=60^\circ$ .

$\vec r=2t\hat i+\sqrt3\,t^2\hat j+5\hat k\Rightarrow \vec v=\dfrac{d\vec r}{dt}=2\hat i+2\sqrt3\,t\,\hat j$ .

At $t=1$ s: $\vec v=2\hat i+2\sqrt3\,\hat j$ (the $\hat k$ term is constant, contributes nothing). …
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KEAM 2025Set eng-2025-04234 marksMCQ
$ms^{-1}$
›Reveal solutionSolution
a = dv/dt = 3.0 j, so |a| = 3 m/s^2.

Concept and Intuition

Acceleration is the time derivative of velocity. Only the time-dependent component of v contributes.

Step-by-Step Solution

v = 2.0 i + 3.0t j.

a = dv/dt = 0 i + 3.0 j.

Magnitude |a| = 3.0 m/s^2.

Common Mistakes …
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KEAM 2025Set pha-2025-0429F4 marksMCQ
$t$
›Reveal solutionSolution

Acceleration is the second time-derivative of position; for $\vec r = 8t\hat i + 3t^2\hat j + 3\hat k$ only the $3t^2$ term contributes, giving $\vec a = 6\hat j$ (magnitude 6).

Given $\vec r = 8t\hat i + 3t^2\hat j + 3\hat k$ m, the velocity is
$\vec v = \frac{d\vec r}{dt} = 8\hat i + 6t\hat j,$
and the acceleration is

$\vec a = \frac{d\vec v}{dt} = 6\hat j\;ms^{-2}.$ …
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KEAM 2024Set eng-2024-06054 marksMCQ
$\mathbf{r} = 3\,\mathbf{i} + 2t^{2}\,\mathbf{j}$
›Reveal solutionSolution

Differentiate $\mathbf{r}$ : $\mathbf{v}=4t\,\mathbf{j}$ , so at $t=5$ s speed $=20\ \mathrm{ms^{-1}}$ .

Step 1: Velocity is the time derivative of position:

$\mathbf{v}=\frac{d}{dt}\left(3\mathbf{i}+2t^2\mathbf{j}\right)=0\,\mathbf{i}+4t\,\mathbf{j}.$ …
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Q.The displacement vector of a particle of mass $m$ is given by $\mathbf{r}(t) = \hat{i}\,A\cos\omega t + \hat{j}\,B\sin\omega t$ .

Concept understanding — Kinematics Vector Differentiation

Kinematics Vector Differentiation

The Intuition First

The Precise Statement

Why This Works

A Concrete Example

The One Trap: Non-Constant Unit Vectors

Concept: Parametric Trajectory and Newton's Second Law

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Q.The displacement vector of a particle of mass mmm is given by r(t)=i^ Acos⁡ωt+j^ Bsin⁡ωt\mathbf{r}(t) = \hat{i}\,A\cos\omega t + \hat{j}\,B\sin\omega tr(t)=i^Acosωt+j^​Bsinωt.

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Q.The displacement vector of a particle of mass $m$ is given by $\mathbf{r}(t) = \hat{i}\,A\cos\omega t + \hat{j}\,B\sin\omega t$ .

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