IB projet maths

To determine the accuracy of these models, I calculated the error between the results predicted by these models and the actual results:

[pic 1]  [pic 2]

Shortened table 6: difference between collected data and data predicted by least squares exponential regression model

As shown in shortened table 6, the average error (the average difference between the data points and the data predicted by my least squares exponential regression model) for all three moulds is 0.1, showing that my three models are exceptionally accurate. However, I should not only rely on mean as the central tendency, as it hides specific data points where the error is especially high. This is important, because my aim is to accurately find how long it takes for the jelly substance to cool down to a certain temperature, so this predicted result cannot be sometimes accurate, sometimes inaccurate. To resolve this problem, I can calculate the standard deviation in the error, which is a measure of the dispersion of the data from the mean (i.e. the average error).

To actually visualise how accurate the model is on the whole, I could also plot a normal distribution curve of the error, which is symmetric about the mean, showing the frequency in occurrence of the data around it.

Although the least squares exponential regression method is an accurate way to model the exponential equation, from considering the accuracy of my model, I experimented with using newly learnt statistical techniques and combine the information they give, considering their advantages and limitations in revealing the full picture of the model’s accuracy. 

By differentiating these equations,  can be found, which will be in the final part of my investigation.[pic 3]

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[pic 6]

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• Move this to part 4
• Write equation for related rates
• Second equation

Resources:

• Modelling the surface area of a ceramic pot
• Cooling curve of coffee
• Cooling curve of tea

• Examiner's comments

• Minimising packaging
• Optimising volume of cuboid
• More exemplar IAs

Advice:

• Define key terms- assume that the examiner does not understand maths and make it understandable
• Personal engagement- relevant to your personal life, show mathematical working
• Reflection- regular, based on progress, show what you learn, results matter less
• Use different methods to prove the same thing

Which part of Maths will you use?

• Functions will be used to model and transform the cooling curve of the jelly substance as the first step to determine the time it takes for it to cool to the appropriate temperature for refrigerating.
• Trigonometry will be used to calculate the surface area of the jelly substance in the different moulds, as this will affect the rate of cooling.
• Calculus will be used to determine the volume of these moulds, find the rate of change of the temperature decrease, and will further be used for optimisation.
• Number and Algebra (systems of linear equations)

*Make a convincing personal engagement study – something that has a more immediate impact on your daily life – use herbal gelatin, healthy to eat every day, refreshing snack

*More complicated cups

According to Stefan-Boltzmann law, the surface area and the surface area both affect the rate of heat transfer by radiation. Thus, I need to calculate the total surface area by multiplying the surface area of the jelly substance by the surface area of the air, and add the surface area of the glass around the jelly substance, multiplied by the surface area of the glass.

Surface area of glass  = 0.88[pic 11]

Surface area of air  = 0.80[pic 12]

Total surface area of mould 1:

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Total surface area of mould 2:

[pic 16]

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Mould 3:

The surface area of the curved surface can be found by revolving the region enclosed by the function , the horizontal lines y = 2.77 and y = 0, the y-axis, through  about the y-axis, using the formula:[pic 19][pic 20]

[pic 21]

Total surface area of mould 3, given that the radius of the surface of 250 cm3 jelly substance (r) and the radius of the base of the mould (r1) were measured to be 5.67cm and 5.00cm respectively:

[pic 22]

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An alternative method is to use Newton’s law of cooling:

[pic 26]

Where:

•  is rate of heat transfer of the body (jelly substance)[pic 27]
•  is the heat transfer coefficient, which is the same for all three moulds as they have the same 1mm thickness and have the same medium[pic 28]
•  is the surface area for heat transfer[pic 29]
•  is the temperature of the body (jelly substance) at time t [pic 30]
•  is the temperature of the environment, which we assume to be 25[pic 31][pic 32]

Given that:

[pic 33]

Where:

•  is the mass of the body (jelly substance)[pic 34]
•  is the specific heat capacity[pic 35]
•  is the change in temperature[pic 36]

Combining the two equations:

[pic 37]

Change to 3 moulds

Steps:

- Real life question: In which jelly mould does the jelly substance cool down the fastest?

- Question: In which jelly mould does 100cm3 of jelly substance take the least time to cool down to 25℃

- Gelatin sets at room temperature

- Same volume for all jelly moulds for fair test

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