Dimensions

What Are Dimensions?

Dimensions describe the different aspects of an object that can be measured. There are three primary dimensions:

One Dimension ( $1D$ ):

Description: Objects that have just a length.
Units of Measurement: $cm, mm, m, km, mile$ , etc.
Examples: A straight line or a piece of string.

Two Dimensions ( $2D$ ):

Description: Objects that have an area (length and width).
Units of Measurement: $cm², mm², m²$ , etc.
Examples: A rectangle, square, or circle.

Three Dimensions ( $3D$ ):

Description: Objects that have a volume (length, width, and height).
Units of Measurement: $cm³, mm³, m³$ , etc.
Examples: A cube, sphere, or cylinder.

Four Dimensions ( $4D$ ):

Note: $4D$ involves time, which is not typically considered in GCSE Maths.

Using Dimensions to Understand Formulas

The advantage of understanding dimensions is that it allows you to determine what a formula is calculating: length, area, volume, or something else. Here's how you can analyze a formula using dimensions:

Change All Variables to the Letter $l$ :

Variables represent quantities like length, width, and height. For simplicity, change all these variables in the formula to $l$ .

Ignore Numbers and Constants:

Constants such as $π$ (pi) or numerical coefficients should be ignored for this step.

Simplify the Formula:

After replacing variables with $l$ , you should simplify the expression.

Interpret the Result:

If you end up with $l$ : The formula is for length ( $1D$ ).
If you end up with $l2$ : The formula is for area ( $2D$ ).
If you end up with $l3$ : The formula is for volume ( $3D$ ).
If the result doesn't fit any of these: The formula may be incorrect or not applicable.

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Example 1: Analyzing a Formula

Let's analyze the formula $5wh$ to determine whether it calculates length, area, volume, or nothing at all.

Step-by-Step Breakdown:

Identify the Variables:

Here, $w$ and $h$ are variables representing lengths. In this analysis, we replace all length variables with $D$ .

Ignore Constants:

The number $5$ is a constant, which doesn't affect the dimensionality. We can ignore it for this step.

Simplify the Expression:

After replacing $w$ and $h$ with $D,$ the expression becomes:

5×D×D=5D^2

We see that the expression simplifies to $D^2$ .

Interpret the Result:

$D2$ indicates that the formula is for calculating area.
Conclusion: The formula $5wh$ is used to calculate an area.

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Example 2: Analyzing a More Complex Formula

Let's analyze the formula $7h(l−w)+2w^2$ to determine whether it calculates length, area, volume, or nothing.

Step-by-Step Breakdown:

Identify the Variables:

The variables $l, w,$ and $h$ represent lengths. We replace all these variables with $D$ to signify their dimensional nature.

Ignore Constants:

The numbers $7$ and $2$ are constants and can be ignored for this step.

Simplify the Expression:

After replacing $l, w,$ and $h$ with $D$ , the expression becomes:

7D(D−D)+2D^2

Simplify within the brackets:

7D(0)+2D^2=0+2D^2=D^2

Interpret the Result:

The final expression simplifies to $D2$ , indicating that the formula is for calculating area.

Conclusion: The formula $7h(l−w)+2w^2$ is used to calculate an area.

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Example 3: Analyzing a Complex Formula

Let's analyze the formula $\frac{2}3h(lh+πw−h^2)$ to determine whether it calculates length, area, volume, or nothing useful.

Step-by-Step Breakdown:

Identify the Variables:

The variables $l, w$ , and $h$ represent lengths. Replace all these variables with $D$ to signify their dimensional nature.

Ignore Constants and Fractions:

The fraction $\frac{2}3$ and $π$ (which is just a number) can be ignored for this analysis as they do not affect the dimensionality.

Simplify the Expression:

After replacing the variables with $D$ , the expression becomes:

\frac{2}3D(D×D+πD−D^2)

Simplify within the brackets:

D(D^2+D−D^2)=D^3+D^2−D^3

Notice that this simplifies to a mixture of $D^3$ and $D^2$ .

Interpret the Result:

The final expression contains both $D^2$ (which corresponds to area) and $D^3$ (which corresponds to volume). Since a formula should consistently describe either length, area, or volume, the presence of both $D^2$ and $D^3$ indicates that the formula doesn't make sense dimensionally.

Conclusion: The formula $\frac{2}3h(lh+πw−h^2)$ is not valid as it mixes dimensions incorrectly.

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Example 4: Analyzing a Complex Volume Formula

Let's analyze the formula:

\frac{5h^3+2lw^2−hlw}6

Objective: Determine whether this formula calculates length, area, volume, or something else.

Step-by-Step Breakdown:

Identify the Variables:

The variables $l, w$ , and $h$ represent lengths. We replace all these variables with $D$ to signify their dimensional nature.

Ignore Constants and Numbers:

The number $6$ is a constant, so it doesn't affect the dimensionality. Ignore it during the analysis.

Simplify the Expression:

Replace $l, w$ , and $h$ with $D$ , and simplify the expression:

\frac{5D^3+2D^2D−DDD}6

Simplify the terms within the formula:

\frac{5D^3+D^3−D^3}6=D^3+D^3−D^3=D^3

Interpret the Result:

The final expression simplifies to $D^3$ , which indicates that the formula is for calculating volume. Conclusion: The formula $\frac{5h^3+2lw^2−hlw}6$ is used to calculate volume.

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Example 5: Analyzing a Complex Length Formula

Let's analyze the following formula:

\frac{kl^3+πlw^2}{8hl}

Objective: Determine whether this formula calculates length, area, volume, or nothing.

Step-by-Step Breakdown:

Identify the Variables:

The variables $l, w$ , and $h$ represent lengths. Replace all these variables with $D$ to signify their dimensional nature.

Ignore Constants:

The constants $k, π,$ and $8$ can be ignored for the purpose of this analysis since they do not affect the dimensionality.

Simplify the Expression:

Replace $l, w,$ and $h$ with $D$ , and simplify the expression:

\frac{kD^3+πDD^2}{8DD}

Simplify further by canceling out the terms:

\frac{D^3+D^3}{D^2}=D+D

This simplifies to $2D$ , which indicates that the formula is for length.

Interpret the Result:

Since the simplified formula results in $D$ , it indicates that the original formula is used to calculate length.

Conclusion: The formula $\frac{kl^3+πlw^2}{8hl}$ is used to calculate length

Dimensions Simplified Revision Notes for GCSE OCR Maths

Dimensions

What Are Dimensions?

Using Dimensions to Understand Formulas

Example 1: Analyzing a Formula

Example 2: Analyzing a More Complex Formula

Example 3: Analyzing a Complex Formula

Example 4: Analyzing a Complex Volume Formula

Example 5: Analyzing a Complex Length Formula

500K+ Students Use These Powerful Tools to Master Dimensions For their GCSE Exams.

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