Dimensions (OCR GCSE Maths): Revision Notes

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Dimensions

What Are Dimensions?

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

  1. One Dimension (1D1D):
  • Description: Objects that have just a length.
  • Units of Measurement: cm,mm,m,km,milecm, mm, m, km, mile, etc.
  • Examples: A straight line or a piece of string. image
  1. Two Dimensions (2D2D):
  • Description: Objects that have an area (length and width).
  • Units of Measurement: cm2,mm2,m2cm², mm², m², etc.
  • Examples: A rectangle, square, or circle. image
  1. Three Dimensions (3D3D):
  • Description: Objects that have a volume (length, width, and height).
  • Units of Measurement: cm3,mm3,m3cm³, mm³, m³, etc.
  • Examples: A cube, sphere, or cylinder. image
  1. Four Dimensions (4D4D):
  • Note: 4D4D 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 analyse a formula using dimensions:

  1. Change All Variables to the Letter ll:
  • Variables represent quantities like length, width, and height. For simplicity, change all these variables in the formula to ll.
  1. Ignore Numbers and Constants:
  • Constants such as ππ (pi) or numerical coefficients should be ignored for this step.
  1. Simplify the Formula:
  • After replacing variables with ll, you should simplify the expression.
  1. Interpret the Result:
  • If you end up with ll: The formula is for length (1D1D).
  • If you end up with l2l2: The formula is for area (2D2D).
  • If you end up with l3l3: The formula is for volume (3D3D).
  • 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 analyse the formula 5wh5wh to determine whether it calculates length, area, volume, or nothing at all.

Step-by-Step Breakdown:

  1. Identify the Variables:
  • Here, ww and hh are variables representing lengths. In this analysis, we replace all length variables with DD.
  1. Ignore Constants:
  • The number 55 is a constant, which doesn't affect the dimensionality. We can ignore it for this step.
  1. Simplify the Expression:
  • After replacing ww and hh with D,D, the expression becomes:
5×D×D=5D25×D×D=5D^2
  • We see that the expression simplifies to D2D^2.
  1. Interpret the Result:
  • D2D2 indicates that the formula is for calculating area.
  • Conclusion: The formula 5wh5wh is used to calculate an area.
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Example 2: Analyzing a More Complex Formula

Let's analyse the formula 7h(l−w)+2w27h(l−w)+2w^2 to determine whether it calculates length, area, volume, or nothing.

Step-by-Step Breakdown:

  1. Identify the Variables:
  • The variables l,w,l, w, and hh represent lengths. We replace all these variables with DD to signify their dimensional nature.
  1. Ignore Constants:
  • The numbers 77 and 22 are constants and can be ignored for this step.
  1. Simplify the Expression:
  • After replacing l,w,l, w, and hh with DD, the expression becomes:
7D(D−D)+2D27D(D−D)+2D^2
  • Simplify within the brackets:
7D(0)+2D2=0+2D2=D27D(0)+2D^2=0+2D^2=D^2
  1. Interpret the Result:
  • The final expression simplifies to D2D2, indicating that the formula is for calculating area.

Conclusion: The formula 7h(l−w)+2w27h(l−w)+2w^2 is used to calculate an area.

infoNote

Example 3: Analyzing a Complex Formula

Let's analyse the formula 23h(lh+πw−h2)\frac{2}3h(lh+πw−h^2) to determine whether it calculates length, area, volume, or nothing useful.

Step-by-Step Breakdown:

  1. Identify the Variables:
  • The variables l,wl, w, and hh represent lengths. Replace all these variables with DD to signify their dimensional nature.
  1. Ignore Constants and Fractions:
  • The fraction 23\frac{2}3 and ππ (which is just a number) can be ignored for this analysis as they do not affect the dimensionality.
  1. Simplify the Expression:
  • After replacing the variables with DD, the expression becomes:
23D(D×D+πD−D2)\frac{2}3D(D×D+πD−D^2)
  • Simplify within the brackets:
D(D2+D−D2)=D3+D2−D3D(D^2+D−D^2)=D^3+D^2−D^3
  • Notice that this simplifies to a mixture of D3D^3 and D2D^2.
  1. Interpret the Result:
  • The final expression contains both D2D^2 (which corresponds to area) and D3D^3 (which corresponds to volume). Since a formula should consistently describe either length, area, or volume, the presence of both D2D^2 and D3D^3 indicates that the formula doesn't make sense dimensionally.

Conclusion: The formula 23h(lh+πw−h2)\frac{2}3h(lh+πw−h^2) is not valid as it mixes dimensions incorrectly.

infoNote

Example 4: Analyzing a Complex Volume Formula

Let's analyse the formula:

5h3+2lw2−hlw6\frac{5h^3+2lw^2−hlw}6

​

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

Step-by-Step Breakdown:

  1. Identify the Variables:
  • The variables l,wl, w, and hh represent lengths. We replace all these variables with DD to signify their dimensional nature.
  1. Ignore Constants and Numbers:
  • The number 66 is a constant, so it doesn't affect the dimensionality. Ignore it during the analysis.
  1. Simplify the Expression:
  • Replace l,wl, w, and hh with DD, and simplify the expression:
5D3+2D2D−DDD6\frac{5D^3+2D^2D−DDD}6
  • Simplify the terms within the formula:
5D3+D3−D36=D3+D3−D3=D3\frac{5D^3+D^3−D^3}6=D^3+D^3−D^3=D^3
  1. Interpret the Result:
  • The final expression simplifies to D3D^3, which indicates that the formula is for calculating volume. Conclusion: The formula 5h3+2lw2−hlw6\frac{5h^3+2lw^2−hlw}6 is used to calculate volume.
infoNote

Example 5: Analyzing a Complex Length Formula

Let's analyse the following formula:

kl3+πlw28hl\frac{kl^3+πlw^2}{8hl}

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

Step-by-Step Breakdown:

  1. Identify the Variables:
  • The variables l,wl, w, and hh represent lengths. Replace all these variables with DD to signify their dimensional nature.
  1. Ignore Constants:
  • The constants k,π,k, π, and 88 can be ignored for the purpose of this analysis since they do not affect the dimensionality.
  1. Simplify the Expression:
  • Replace l,w,l, w, and hh with DD, and simplify the expression:
kD3+πDD28DD\frac{kD^3+πDD^2}{8DD}
  • Simplify further by cancelling out the terms:
D3+D3D2=D+D\frac{D^3+D^3}{D^2}=D+D
  • This simplifies to 2D2D, which indicates that the formula is for length.
  1. Interpret the Result:
  • Since the simplified formula results in DD, it indicates that the original formula is used to calculate length.

Conclusion: The formula kl3+πlw28hl\frac{kl^3+πlw^2}{8hl} is used to calculate length

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