Q4MechanicsStretch

Differentiating Vectors

When a position vector is given as a function of time, you differentiate each component to get velocity, then again for acceleration. This handles variable acceleration in two dimensions.

30 min Kinematics

The big picture

Differentiate each component of the position vector to get velocity, then each velocity component to get acceleration. Speed is $|\mathbf v|$ , and can be nonzero when one velocity component is zero. The rate of change of distance from the origin is a radial rate. It need not equal speed because a particle can move sideways around the origin while maintaining the same distance.

Edexcel A Level Maths: 8.4 Differentiating VectorsCurated by Kepler, not produced by us. Watch it, then work through the notes below.

Open on YouTube

What you'll be able to do

Differentiate a vector component-wise
Find velocity from a position vector
Find acceleration from velocity
Evaluate speed at a given time

Component-wise differentiation

If $\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j}$ , then $\mathbf{v} = \dot{x}\mathbf{i} + \dot{y}\mathbf{j}$ and $\mathbf{a} = \ddot{x}\mathbf{i} + \ddot{y}\mathbf{j}$ — differentiate each component separately.

\mathbf{v} = \frac{d\mathbf{r}}{dt}, \quad \mathbf{a} = \frac{d\mathbf{v}}{dt}

Differentiate each component.

Example 1If

\mathbf{r} = t^2\mathbf{i} + (3t)\mathbf{j}

, find

\mathbf{v}

and

\mathbf{a}

\mathbf{v} = 2t\mathbf{i} + 3\mathbf{j}

\mathbf{a} = 2\mathbf{i}

Answer

\mathbf{v} = 2t\mathbf{i} + 3\mathbf{j}

\mathbf{a} = 2\mathbf{i}

Tip — Speed is the magnitude of the velocity vector: |v| = √(vₓ² + v_y²).

Speed is a magnitude, not a derivative of distance from the origin

Differentiate each component of the position vector to get velocity, then each velocity component to get acceleration. Speed is $|\mathbf v|$ , and can be nonzero when one velocity component is zero.

The rate of change of distance from the origin is a radial rate. It need not equal speed because a particle can move sideways around the origin while maintaining the same distance.

Example 1A particle has

\mathbf r=(t^2+1)\mathbf i+(4t-t^2)\mathbf j

metres. Find its speed and acceleration at t = 1 s.

\mathbf v=2t\mathbf i+(4-2t)\mathbf j

2At t = 1,

\mathbf v=2\mathbf i+2\mathbf j

, so speed

\sqrt{2^2+2^2}=2\sqrt2

m/s.

\mathbf a=2\mathbf i-2\mathbf j

m/s squared, independent of time.

AnswerSpeed

2\sqrt2\approx2.83

m/s; acceleration

2\mathbf i-2\mathbf j

m/s squared.

Think like an examiner

Write the defining equations and show the substitution before evaluating.
Check the domain, units and interpretation of the final result.

Common misconceptions

Differentiating the magnitude instead of each component.

Differentiate the i and j components separately.

Confusing speed with velocity.

Speed is the magnitude of the velocity vector.

Key results to remember

\mathbf{v} = \dot{x}\mathbf{i} + \dot{y}\mathbf{j}

Velocity.

\mathbf{a} = \ddot{x}\mathbf{i} + \ddot{y}\mathbf{j}

Acceleration.

Key takeaways

Differentiate r component-wise to get v, then a.
v = ẋ i + ẏ j; a = ẍ i + ÿ j.
Speed = |v|.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

All 8 lessons