Differentiating Vectors Revision Notes for Leaving Cert Applied Maths

Differentiating Vectors

What is vector differentiation?

Vector differentiation allows us to find how vectors change with respect to time. This is particularly useful in physics when dealing with motion, where we need to find velocity from displacement and acceleration from velocity.

The key principle is that we differentiate vectors by treating each component separately. If a vector has i and j components, we differentiate each component independently and then combine the results.

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This component-wise approach makes vector differentiation straightforward - instead of dealing with a complex vector operation, we simply apply standard differentiation rules to each component individually.

The method

To differentiate a vector, we follow a systematic four-step process:

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Vector Differentiation Steps:

Identify the i and j components
Differentiate the i component with respect to time
Differentiate the j component with respect to time
Combine the results to form the new vector

Key relationships in motion

When dealing with motion problems involving vectors, understanding the relationship between displacement, velocity, and acceleration is crucial:

Displacement vector (s̄) → differentiate → Velocity vector (v̄)
Velocity vector (v̄) → differentiate → Acceleration vector (ā)

Remember that speed is the magnitude of the velocity vector, whilst velocity includes both magnitude and direction.

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This sequence forms the foundation of kinematic analysis. Each differentiation step reveals how the motion quantity changes instantaneously with time.

Worked example

Let's work through a complete example to see how this works in practice.

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Worked Example: Vector Differentiation in Motion

Given: A displacement vector $\bar{s} = (3t^3 - 1)\bar{i} + (100 - 4t^3)\bar{j}$

Finding displacement at a specific time

To find the displacement at $t = 3$ s, we substitute:

$\bar{s} = (3(3)^3 - 1)\bar{i} + (100 - 4(3)^3)\bar{j}$
$\bar{s} = (81 - 1)\bar{i} + (100 - 108)\bar{j}$
$\bar{s} = 80\bar{i} - 8\bar{j}$

Finding velocity

Differentiate the displacement vector:

$\bar{s} = (3t^3 - 1)\bar{i} + (100 - 4t^3)\bar{j}$
$\bar{v} = 9t^2\bar{i} - 12t^2\bar{j}$

Finding speed

Speed is the magnitude of velocity. At $t = 3$ s:

$\bar{v} = 9(3)^2\bar{i} - 12(3)^2\bar{j} = 81\bar{i} - 108\bar{j}$
Speed = $|\bar{v}| = \sqrt{81^2 + (-108)^2} = \sqrt{6561 + 11664} = \sqrt{18225} = 135$ m/s

Finding acceleration

Differentiate the velocity vector:

$\bar{v} = 9t^2\bar{i} - 12t^2\bar{j}$
$\bar{a} = 18t\bar{i} - 24t\bar{j}$

Finding magnitude of acceleration

At $t = 10$ s:

$\bar{a} = 18(10)\bar{i} - 24(10)\bar{j} = 180\bar{i} - 240\bar{j}$
$|\bar{a}| = \sqrt{180^2 + (-240)^2} = \sqrt{32400 + 57600} = \sqrt{90000} = 300$ m/s²

Calculating magnitude

To find the magnitude (or length) of any vector with components $a\bar{i} + b\bar{j}$ :

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Magnitude Formula:

$\text{Magnitude} = \sqrt{a^2 + b^2}$

This formula comes from Pythagoras' theorem and gives us the "size" of the vector without considering direction.

Exam tips

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Essential Exam Strategies:

Always differentiate each component separately
Remember to substitute values after differentiating, not before
Check units - displacement (m), velocity (m/s), acceleration (m/s²)
For magnitude calculations, square both components before taking the square root
Speed is always positive (it's a magnitude), whilst velocity can be negative

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Key Points to Remember:

Vector differentiation: Treat i and j components separately when differentiating
Motion sequence: Displacement → Velocity → Acceleration (each step involves differentiation)
Speed vs velocity: Speed is the magnitude of the velocity vector
Magnitude formula: For vector $a\bar{i} + b\bar{j}$ , magnitude = $\sqrt{a^2 + b^2}$
Always substitute after differentiating to avoid calculation errors

Differentiating Vectors (Leaving Cert Applied Maths): Revision Notes