# IB Mathematics AA SL - Original IA and supplied review Canonical HTML: https://www.prelearning.de/en/ib-mathematics-ia-sample-review/ Note: This machine-readable text version supports search and accessibility. The unaltered original files remain authoritative for layout, formulas, tables and figures. ## Document 1: Originally submitted IA (English, unaltered) An exploration into Cooling Process Internal Assessment: Mathematics Analysis and Approaches Standard Level Formulating a Suitable Equation to Model the Cooling Process of Hibiscus Tea 15 pages February 2, 2026 Personal code: e2u7l1e8r Introduction I was inspired to undertake this investigation because of a habit which has become part of my evening routine. After finishing my homework, I usually brew hibiscus tea and drink it while reading books. The bright red colour and sour taste makes the drink especially enjoyable, but I have a repeated problem: I often begin another task and forget the cup beside me. When I return, it has either cooled into the comfortable range or become too cold. The difference between these outcomes is only several minutes, so I wanted a way of predicting the useful drinking time instead of testing the tea each evening again and again. This situation connected naturally to functions and calculus. Temperature is continuous, it changes at a non-constant rate, and the surrounding room places a lower restriction on the cooling process. I therefore decided to measure the temperature of one preparation of hibiscus tea repeatedly and then formulate an equation which models its cooling. My research question is: How accurately can an exponential model predict the time required for say 300 mL of hibiscus tea to cool below 45°C in a room with an ambient temperature of approximately 22.4°C? The boundary of 45°C is personally relevant because I have checked and below this value the drink tastes flat and I normally reheat it. Research about hot beverages often reports preferred temperatures close to 54-60°C, although preference changes between people and beverages (Brown and Diller, 2008). I used a lower boundary because hibiscus tea is less viscous than coffee and, in my experience, still feels warm at 45°C. This decision is subjective, but it allows the mathematics to answer a clear and useful question. From chemistry and physics, I expected the difference between the tea temperature and the room temperature to somehow influence the cooling rate. A larger difference should cause a faster loss of thermal energy, while a smaller difference should slow the process. The liquid should not cool below the room temperature because at that point the tea and the environment have reached thermal equillibrium. Evaporation, conduction through the mug and convection at the surface all contribute to this behaviour. I anticipated an exponential relationship with a horizontal asymptote at the ambient temperature. Background and Data Collection Methodology Hibiscus beverages may be prepared by hot or cold extraction. Ramirez-Rodrigues et al. compared both methods and showed that hot extraction at 90°C can obtain similar anthocyanin concentrations much faster than cold extraction. For consistency, I prepared the tea using water from the same kettle and the same packet of dried Hibiscus sabdariffa calyces in every trial. This gave the exploration a real context while keeping the mathematical variable of interest as temperature over time. For each trial, 6.0 g of dried hibiscus calyces was placed in a ceramic mug. I heated water to 90°C, measured 300 mL with a graduated cylinder, poured it into the mug, and stirred for 10 seconds. After an extraction time of four minutes the calyces were removed using the same metal strainer. A digital probe thermometer was placed approximately 3 cm below the surface without touching the mug. The first reading was defined as t = 0, and a temperature was recorded every 5 minutes for one hour. The experiment was repeated three times on the same desk. The room temperature was measured before each trial and treated as 22.4°C. A fluctuation of about ±0.3°C was visible during the experiment, but this was ignored in the later model so that a single horizontal asymptote could be used. The same mug, water volume, calyx mass, stirring time, probe depth and desk position were held constant. Windows were closed and no fan was used. The thermometer displayed values to 0.1°C and the measuring cylinder had 5 mL divisions. Using three trials should eliminate random errors and produce data which are more representative than one observation. It was not possible to control the exact initial tea temperature, the humidity, or the small amount of water remaining on the strainer. I also assumed that the thermometer did not significantly cool the liquid. These limitations were considered acceptable because the purpose was to construct a usable mathematical model rather than a full thermodynamic description. Variables • Independent variable: elapsed time, t, measured in minutes. • Dependent variable: temperature of the tea, T, measured in degrees Celsius. • Controlled variables: mug, liquid volume, hibiscus mass, extraction time, probe depth, stirring and room location. Raw Data Table 1 - Temperature over time from Trial 1 Time 0 5 10 15 20 25 30 35 40 45 50 55 60 (min) Temp 78.9 69.9 62.9 55.2 49.9 44.0 38.6 34.9 31.3 28.0 27.1 26.4 25.5 . (°C) Table 1.1 - Temperature over time from Trial 2 Time 0 5 10 15 20 25 30 35 40 45 50 55 60 (min) Temp 78.2 70.7 62.3 55.8 49.5 43.5 39.1 34.5 31.8 28.4 26.7 25.9 26.1 . (°C) Table 1.2 - Temperature over time from Trial 3 Time 0 5 10 15 20 25 30 35 40 45 50 55 60 (min) Temp 78.7 70.0 62.3 55.2 50.0 43.9 39.2 35.0 31.1 28.2 26.9 26.3 25.5 . (°C) Processed Data Table 1.3 - Mean temperature over time Time 0 5 10 15 20 25 30 35 40 45 50 55 60 (min) Temp 78.6 70.2 62.5 55.4 49.8 43.8 39.0 34.8 31.4 28.2 26.9 26.2 25.7 . (°C) Figure 1 - Mean temperature plotted against time for a cup of hibiscus tea A horizontal asymptote at t = 22.4 was added to Figure 1.0 to represent the measured ambient temperature. According to Boyle's gas law, a substance cannot decrease its temperature below that of its environment. This explains why the data approaches 22.4°C rather than continuing downwards in a straight line. The graph shows a negative correlation between temperature and time, together with a curved shape that resembles exponential decay. Temperature falls by 8.4°C in the first five minutes, but only by 0.5°C in the final five minutes. The changing differences imply that a linear equation would not be suitable for the full domain. Also, the y-intercept represents the measured initial temperature, so when t = 0, T = 78.6°C. Producing a Suitable Equation - Attempt A I began by using the appearance of Figure 1.0 rather than applying a known physical law. Since temperature T was plotted against time t, I treated T as y and t as x. An exponential curve with a negative slope can first be represented by raising e to the power of negative time: 𝑇(𝑡) = 𝑒 −𝑡 The entire graph has to be translated upward by the ambient temperature. This creates the translation constant T_a: 𝑇(𝑡) = 𝑒 −𝑡 + 𝑇𝑎 𝑇(𝑡) = 𝑒 −𝑡 + 22.4 This first expression includes the correct horizontal asymptote, but it still has an initial value of only 23.4°C. I graphed it to understand which transformations were missing before calculating parameters from the experimental values. Figure 1.1 - The preliminary model T(t) = e⁻ᵗ + 22.4 Although an exponential relationship is present and the curve approaches the correct asymptote, this graph does not resemble the observed cooling process. Almost all the change occurs close to t = 0. For the positive values of time used in my experiment, the function immediately becomes indistinguishable from 22.4°C. The function must therefore be stretched horizontally so that the cooling takes place across sixty minutes. It must also be stretched vertically so that its y-intercept is close to 78.6°C. Let k represent the horizontal stretch factor and m represent the vertical stretch factor. A more useful general expression is: 𝑇(𝑡) = 𝑚𝑒 −𝑘𝑡 + 𝑇𝑎 𝑇(𝑡) = 𝑚𝑒 −𝑘𝑡 + 22.4 The constant m controls the distance between the initial temperature and the asymptote, while k controls the speed of cooling. Both constants should be positive when the negative sign is already included in the exponent. A larger k would make the tea cool more quickly. Values for these constants can be found by using two points from Table 1.3. For Attempt A, I selected the first and last mean observations. These points cover the entire experimental interval and therefore should create the most representative model. This method does not use the intermediate points directly, which is one weakness, but the two endpoints provide two equations for the two unknown constants. Determining the Constants for Attempt A Beginning with the transformed exponential equation: 𝑇 − 𝑇𝑎 = 𝑚𝑒 −𝑘𝑡 Natural logarithms can be used to remove the exponential term: ln(𝑇 − 𝑇𝑎 ) = ln(𝑚) − 𝑘𝑡 Substituting the first point (0, 78.6) and the last point (60, 25.7) gives the following simultaneous equations. All values are kept unsimplified instead of being rounded to three significant figures, in order to ensure the utmost accuracy of the model. ln(78.6 − 22.4) = ln(𝑚) − 0𝑘 ln(25.7 − 22.4) = ln(𝑚) − 60𝑘 Table 1.4 - Simultaneous equations used to determine m and k Equation 1 Equation 2 ln(56.2) = ln(m) ln3.3 = ln(m) − 60k ln(m) = 4.02892 1.19392 = ln(m) − 60k Subtracting Equation 1 from Equation 2 eliminates ln(m): −60𝑘 = ln(3.3) − ln(56.2) 𝑘 = 0.047250 The value of m follows directly from Equation 1: 𝑚 = 𝑒 ln(56.2) = 56.2000 Equation A Substituting m = 56.2000, k = 0.047250 and T_a = 22.4 into the general expression produces Equation A: 𝑇(𝑡) = 56.2000𝑒 −0.047250𝑡 + 22.4 Equation A The equation returns T(0) = 78.6°C and T(60) = 25.7°C, so the two points used in its construction are exact. Because it contains a negative exponential, it also approaches 22.4°C as t becomes large. These features agree with the principal observations from Figure 1.0. Figure 1.2 - Equation A compared with the collected mean temperatures The black curve follows the general trend of the red data points. However, Equation A underpredicts several observations between 10 and 35 minutes. It only fits the endpoints perfectly because those were the points selected for the simultaneous equations. The deviation may be caused by small changes in air change, the thermal mass of the mug or the rate of evaporation. Real cooling data cannot be expected to follow a mathematical function exactly. Nevertheless, the repeated pattern of residuals suggests more than random error. A method which uses every transformed data point may create a more reliable value for k and reduce the influence of one anomalous measurement. Producing a Suitable Equation - Attempt B In Attempt B I linearised the collected data. When a relationship is linear, the gradient and intercept can be estimated using all observations rather than only two. This should account better for natural fluctuations and makes it easier to judge whether an exponential relationship is appropriate. Starting from the same general expression: 𝑇 − 𝑇𝑎 = 𝑚𝑒 −𝑘𝑡 ln(𝑇 − 𝑇𝑎 ) = ln(𝑚) − 𝑘𝑡 This has the linear form y = mx + c: ln(𝑇 − 𝑇𝑎 ) = (−𝑘)𝑡 + ln(𝑚) 𝑦 = ln(𝑇 − 𝑇𝑎 ), 𝑚𝑥 = 𝑘𝑡, 𝑐 = ln(𝑚) Therefore, I plotted ln(T − T_a) on the y-axis against time on the x-axis. If the points are approximately linear, the original temperature data can be treated as exponential. The processed logarithms are shown below. Because logarithms have no units, the transformed values are recorded without °C. Table 1.5 - Values of ln(T − Tₐ) plotted against time Time 0 5 10 15 20 25 30 35 40 45 50 55 60 (s) ln(T − 4.0 3.8 3.6 3.4 3.3 3.0 2.8 2.5 2.1 1.7 1.5 1.3 1.19 Tₐ) 29 67 91 97 11 63 09 18 97 58 04 35 4 The use of a mean ambient temperature is important here. If T_a were changed for every row, the logarithmic transformation would no longer describe one function. I therefore maintained T_a = 22.4°C even though the room thermometer varied slightly. Figure 1.3 - Linearised graph of ln(T − Tₐ) plotted against time The transformed points are distributed closely around the line of best fit, though the later points curve below the line. The coefficient of determination is approximately 0.9871, meaning that 98,71% of the variation in temperature is explained by time. This high value supports the use of a exponential equation. From the regression output: 𝑦 = −0.050997𝑥 + 4.20430 Comparing this with ln(T − T_a) = −kt + ln(m), the gradient equals −k and the intercept equals ln(m). Therefore: 𝑘 = 0.050997, 𝑚 = 𝑒 4.20430 = 66.9735 Substituting both constants gives the second model: 𝑇(𝑡) = 66.9735𝑒 −0.050997𝑡 + 22.4 Equation B Equation B does not pass exactly through the first or last data point because the regression considers all thirteen values. Its y-intercept is 89.37°C, which is higher than the recorded starting temperature. This is an unexpected outcome of the line of best fit and will be considered when the models are compared. Figure 1.4 - Equation B compared with the collected mean temperatures Equation B follows the data during most of the monitored interval, but it overpredicts the temperature at t = 0 and underpredicts some temperatures around 20-35 minutes. The deviation in the beginning is greater than for Equation A. In the final twenty minutes, however, the curve is relatively close to the observations. The linearisaton uses every point, but the logarithmic transformation changes the influence of errors. A difference of 1°C near the ambient temperature produces a much larger proportional change in ln(T − T_a) than the same error at a high temperature. Consequently, fitting the transformed values does not necessarily minimise the errors measured in degrees Celsius. Producing a Suitable Equation - Attempt C Since both equations were created mainly from the shape of the experimental graph, I completed additional research. Newton's Law of Cooling, first investigated in the early eighteenth century, states that the rate of heat loss of a body is directly proportional to the temperature difference between the body and its environment (OpenStax, 2021). In mathematical notation: 𝑑𝑇 ∝ (𝑇 − 𝑇𝑎 ) 𝑑𝑡 Because the hibiscus tea is cooling, the instantaneous change in temperature must be negative. Introducing a positive constant k gives: 𝑑𝑇 = −𝑘(𝑇 − 𝑇𝑎 ) 𝑑𝑡 Solving the Differential Equation: The variables temperature and time can be seperated: 1 𝑑𝑇 = −𝑘 𝑑𝑡 𝑇 − 𝑇𝑎 Integrating both sides gives: 1 ∫ 𝑑𝑇 = ∫ − 𝑘 𝑑𝑡 𝑇 − 𝑇𝑎 ln|𝑇 − 𝑇𝑎 | = −𝑘𝑡 + 𝐶 Taking e to the power of both sides removes the logarithm: 𝑇 − 𝑇𝑎 = 𝑒 −𝑘𝑡+𝐶 = 𝑒 𝐶 𝑒 −𝑘𝑡 The value e^C is another positive arbitrary constant, which I labelled Z. Rewriting the solution with temperature as a function of time gives: 𝑇(𝑡) = 𝑍𝑒 −𝑘𝑡 + 𝑇𝑎 where T is temperature in degrees celsius, t is time in minutes, T_a is the ambient temperature, k is the cooling coefficient and Z is a constant set by an initial condition. For this experiment the room temperature is inserted: 𝑇(𝑡) = 𝑍𝑒 −𝑘𝑡 + 22.4 The derived solution has the same form as my earlier equations, but it is now justified by a differential equation. Newtons law assumes a constant surrounding temperature and a rate coefficient which does not change during the process. These assumptions simplify the combined effects of convection, conduction, radiation and evaporation into one constant. Determining the Constants for Attempt C To determine Z and k, I chose two intermediate observations rather than the endpoints. I used T = 55.4°C at t = 15 minutes and T = 28.2°C at t = 45 minutes. The first point is after the steepest early cooling and the second is close to the drinking threshold, so they should better represent the useful part of the curve. I 55.4 = 𝑍𝑒 −15𝑘 + 22.4 II 28.2 = 𝑍𝑒 −45𝑘 + 22.4 Table 1.6 - Simultaneous equations used to determine Z and k Equation 1 Equation 2 33.0 = Ze⁻¹⁵ᵏ 5.8 = Ze⁻⁴⁵ᵏ ln(33.0) = ln(Z) − 15k ln(5.8) = ln(Z) − 45k 3.49651 = ln(Z) − 15k 1.75786 = ln(Z) − 45k Subtracting Equation 1 from Equation 2 removes ln(Z): ln(5.8) − ln(33.0) = −30𝑘 𝑘 = 0.057955 This value is positive, as expected for a decreasing exponential model. I retained six decimal places to prevent the final predicted time from being affected by premature rounding. Equation C Substituting k into Equation 1 determines Z: 𝑍 = (33.0)𝑒 15(0.057955) = 78.7149 The third model is therefore: 𝑇(𝑡) = 78.7149𝑒 −0.57955𝑡 + 22.4 Equation C This equation passes exactly through the observations at 15 and 45 minutes. It also approaches the horizontal asymptote of 22.4°C, satisfying the physical limitation used throughout the exploration. However, its predicted original value is 101.11°C because T(0) = 78.7149 + 22.4. This is impossible for water prepared at 90°C and differs substantially from the recorded 78.6°C. Figure 1.5 - Equation C compared with the collected mean temperatures The largest discrepancy occurs at the beginning. As time increases, the curve moves closer to the data and is especially accurate around the two chosen points. The x-axis is displayed in seconds by the graphing software, although the numerical time values were entered in five-minute intervals. Comparing All Three Developed Models To compare the models numerically, I calculated the temperature predicted by each equation for every recorded time. More decimal places are shown for the predicted values than the thermometer could measure because they were generated directly from the equations. Table 1.7 - Recorded and predicted temperature values Time Recorded T (°C) Equation A (°C) Equation B (°C) Equation C (°C) (min) 0 78.6 78.6000 89.4046 101.1149 5 70.2 66.7746 74.3237 81.3127 10 62.5 57.4375 62.6371 66.4922 15 55.4 50.0650 53.5808 55.4000 20 49.8 44.2438 46.5628 47.0983 25 43.8 39.6475 41.1244 40.8850 30 39.0 36.0184 36.9101 36.2347 35 34.8 33.1528 33.6442 32.7544 40 31.4 30.8903 31.1135 30.1495 45 28.2 29.1038 29.1523 28.2000 50 26.9 27.6932 27.6325 26.7409 55 26.2 26.5794 26.4548 25.6489 60 25.7 25.7000 25.5422 24.8316 Equation A is exact at 0 and 60 minutes, Equation C is exact at 15 and 45 minutes, and Equation B is not exact at any point because it was generated through regression. Visual comparisons alone do not identify the strongest model, so I used the root mean square error. Root Mean Square Error RMSE is the standard deviation of the residuals, which are the differences between the recorded temperatures and the values predicted by a model. Squaring prevents positive and negative residuals from cancelling. A lower result shows that the regression line is a better suit for the data. 𝑛 1 2 RMSE = √ ∑(𝑇𝑖 − 𝑇̂𝑖 ) 𝑛 𝑖=1 Table 1.8 - Root mean square error of each equation Model RMSE (°C) Method used Equation A 3.130 First and last observations Equation B 3.530 Linear regression of logarithms Equation C 7.214 Newton model using t = 15 and 45 Figure 1.6 - All three equations compared with the collected data Equation A has the smallest full-data RMSE, although Equation B is similar. Equation C has the highest error because its extrapolated starting temperature is much too high. This initially suggests that Equation A is the strongest model for the entire sixty-minute interval. Interpretation of the Model Comparison The relative success of Equation A can be linked to its algebraic derivation. It was constructed from the first and last observed values, and its rate constant describes the overall change during the experiment. Its predictions are not exact in the middle, but the errors are spread across the full domain. The model is specific to this mug, volume and room, so the same constants should not be assumed for another cooling process. Equation B also produces a reasonable RMSE. The linearisation uses every point and reduces the influence of high temperature residuals, but the transformed regression has a higher intercept than the actual data. The value R² = 0.9871 seems very high, yet this statistic applies to ln(T − T_a), not to temperature itself. It is possible for a line to fit the transformed graph well while producing larger errors after transforming back. Equation C is the least accurate if all readings are included. Its main flaw is the assumption that the room remained exactly 22.4°C. As the tea cooled, the air immediately surrounding the mug would of increased in temperature. The difference between the tea and its environment therefore became smaller than the model assumed, slowing the real cooling process. The mug also stored heat and returned part of it to the liquid, which is not represented by a single constant k. On closer observation, the underprediction and overprediction from Equation C are concentrated at the start. The research question is not concerned with the initial temperature; it asks when the tea falls below 45°C. This happens after about twenty minutes. Therefore, the early error may be less important for the practical purpose even though it increases the overall RMSE considerably. To test this, I removed the predicted and recorded values from the first ten minutes and recalculated the RMSE of Equation C. This adapted comparison begins at t = 10 and keeps the data near the drinking threshold. Removing these values is justified because I would never drink the tea immediately after brewing. Adapted Comparison of Equation C Figure 1.7 - Equation C compared with the data after disregarding the first ten minutes Table 1.9 - Adapted RMSE for Equation C Comparison RMSE (°C) Equation C, all values 7.214 Equation C, t ≥ 10 minutes 1.991 After disregarding the first ten minutes, the calculated RMSE decreases from 7.214 to 1.991°C. This is smaller than the RMSE of Equations A and B calculated over the whole interval. The improvement shows that Equation C becomes more accurate at higher values of time, especially once the tea is close to the lower end of its preferred temperature range. The comparison is not completely equal because Equation C is now tested with fewer data points than Equations A and B. Nevertheless, it answers the investigation more directly. I therefore selected Equation C as the most suitable equation for predicting when the tea is no longer acceptably hot. It is based on a physical law, fits the later measurements, and contains the required ambient-temperature asymptote. The adapted graph still contains residuals of approximately 2-3°C between 20 and 35 minutes. A prediction near 45°C can consequently be incorrect by several minutes. This limitation should be remembered when a single maximum waiting time is calculated. Answering the Original Question The original question asks for the time required for the hibiscus tea to cool below 45°C. I used Equation C and set T = 45: 45 = 78.7149𝑒 −0.057955𝑡 + 22.4 45 − 22.4 = 𝑒 −0.057955𝑡 78.7149 22.6 ln ( ) = −0.057955𝑡 78.7149 −1.2480 𝑡= = 21.53 −0.057955 𝑡 ≈ 22 minutes I can therefore conclude that approximately 22 minutes is the maximum amount of time which can elapse before my hibiscus tea is no longer considered hot. This agrees with the raw data because the temperature is 49.8°C at 20 minutes and 43.8°C at 25 minutes. Linear interpolation between these readings would place 45°C at around 24 minutes, which is reasonably close to the model. If I begin drinking within twenty minutes, the model predicts the tea will remain above my chosen minimum. The result is specific to the 300 mL volume and ceramic mug, so it should not be treated as a universal value for hibiscus tea. Even so, the investigation has transformed an ordinary routine into a practical use of exponential functions, logarithms and differential equations. Reflection The mathematics developed from a visual conjecture into a physically motivated model. Attempt A demonstrated transformations of e⁻ᵗ, Attempt B used linear regression, and Attempt C connected the same form to a differential equation. Comparing the models with RMSE made the final choice less dependent on appearance alone, although the adapted comparison was necessary for the research question. Limitations and Possible Improvements The most significant limitation is the treatment of ambient temperature as constant. The measured room temperature varied by approximately 0.3°C, and the air beside the mug may have been warmer than the air measured farther away. A future experiment could use a second temperature probe next to the mug and model T_a as a function of time instead of a fixed asymptote. The five-minute sampling interval was also too large during the beginning, when temperature changed quickly. Recording every minute would create more data for the regression and make the crossing of 45°C easier to identify. More than three trials would improve the reliability of the mean and permit uncertainty intervals. I did not propagate the ±0.1°C thermometer resolution through the logarithm or the final value of t. A second improvement would be to measure mass rather than volume because evaporation reduced the amount of liquid during the hour. Covering the cup would reduce evaporation but also change the cooling mechanism, so both covered and uncovered mugs could be compared. Different mug materials and surface areas would allow the constant k to be interpreted rather than only fitted. The model assumes one uniform temperature throughout the tea. In reality, the liquid near the surface and the wall may be cooler than liquid in the centre. I stirred only before the first measurement because repeated stirring could increase convection, but this means the probe may not represent the mean temperature of the entire cup. A controlled magnetic stirrer at a low constant speed could make the temperature more uniform. Finally, choosing Equation C after deleting the first ten minutes introduced bias. A better comparison would calculate the RMSE of all three models on exactly the same restricted domain or use a separate validation trial which was not involved in selecting constants. Residual plots could reveal whether the errors are random. A two-term exponential or a model with a variable heat-transfer coefficient may fit the mug and liquid separately, but this would increase complexity and perhaps reduce the clarity of the exploration. Despite these weaknesses, the exploration produced a model that is plausible in the interval relevant to my routine. The final time of about 22 minutes should be interpreted as an estimate, not as an exact rule. Works Cited 1. Brown, Fredericka, and Kenneth R. Diller. “Calculating the Optimum Temperature for Serving Hot Beverages.” Burns, vol. 34, no. 5, 2008, pp. 648-654. doi:10.1016/j.burns.2007.09.012. Accessed 25 Aug. 2026. 2. OpenStax. “4.9 First-order Linear Equations: Newton's Law of Cooling.” Calculus Volume 2, Rice University, 2021, openstax.org/books/calculus-volume-2/pages/4-key-equations. Accessed August 25, 2026. 3. Ramírez-Rodrigues, Milena M., et al. “Physicochemical and Phytochemical Properties of Cold and Hot Water Extraction from Hibiscus sabdariffa.” Journal of Food Science, vol. 76, no. 3, Apr. 2011, pp. C428-C435, doi:10.1111/j.1750-3841.2011.02091.x. 4. Lienhard, John H., and John H. Lienhard. A Heat Transfer Textbook. 5th ed., Phlogiston Press, 2019. ahtt.mit.edu. 5. Khan Academy. “Newton's Law of Cooling.” Differential Equations, 2022, www.khanacademy.org/math/differential-equations/first-order-differential-equations/exponential- models-diff-eq/v/newtons-law-of-cooling. Accessed 25 Aug. 2026. 6. Abraham, John P., et al. “A Review of Hot Beverage Temperatures-Satisfying Consumer Preference and Safety.” Journal of Food Science, vol. 84, no. 8, 2019, pp. 2011-2014. https://doi.org/10.1111/1750-3841.14699. All experimental temperatures and graphs in this exploration were produced for the investigation. Calculations were completed using a spreadsheet and checked with graphing software. ## Document 2: Supplied English review (unaltered) Full Evaluation of the IB Mathematics AA SL Exploration Formulating a Suitable Equation to Model the Cooling Process of Hibiscus Tea Strict teacher review under the current assessment model Review date: 27 August 2026 File reviewed: 22 PDF pages, including a two-page bibliography Course: IB Mathematics: Analysis and Approaches, Standard Level 1. How the final IB Mathematics AA SL grade is calculated For the current Mathematics AA SL course, the final subject grade is assembled from three weighted components. The exploration is important, but it is only one fifth of the subject result. Component | Conditions | Weight in the final subject result Paper 1 | 1 hour 30 minutes; no technology allowed | 40% Paper 2 | 1 hour 30 minutes; technology allowed | 40% Internal assessment: mathematical exploration | Assessed by the teacher and externally moderated | 20% Total | | 100% If , and denote percentage scores for Paper 1, Paper 2 and the exploration, the weighted subject total is For example, Paper 1 at 80%, Paper 2 at 80% and an exploration at 15/20, or 75%, would give This is only an illustration. The final grade from 1 to 7 is awarded from the combined session result using the grade boundaries for that examination session. Therefore, neither 15/20 nor 18/20 on the exploration guarantees a 7 by itself. How the exploration’s 20 raw marks are formed The exploration is marked out of 20. Its five current criteria are: Criterion | Maximum | What it rewards A: Presentation | 4 | A coherent, well-organised and concise exploration B: Mathematical communication | 4 | Relevant, appropriate and consistently correct mathematical language, notation and representation C: Personal engagement | 3 | Authentic, independent engagement that drives the investigation D: Reflection | 3 | Substantial critical reflection throughout the exploration E: Use of mathematics — SL | 6 | Relevant and correct mathematics at an appropriate level, supported by thorough understanding Total | 20 | The raw exploration mark, scaled to 20% of the subject result The current five-criterion structure remains the correct basis for work submitted before the new course’s first assessment in May 2029. The redesigned criteria—Problem specification, Abstraction, Computation and Interpretation—belong to the course first taught from August 2027 and first assessed in May 2029; they should not replace the present criteria for a current exploration. 2. Overall judgement This exploration begins from a worthwhile idea. It has a credible personal context, original experimental data and a sensible connection between exponential functions, logarithms, regression, differential equations and model criticism. That is a stronger foundation than many average explorations. In its present form, however, several important mathematical inconsistencies prevent a secure top mark: Equation C is printed once with instead of , a factor-of-ten error. The parameters of equation B do not agree across the regression graph, prose, displayed equation, prediction table and RMSE calculation. The alternatives are not evaluated on the same observations. The conclusion that C is the best equation for the relevant domain is contradicted by a fair calculation from the displayed data. The adjusted RMSE of cannot be reproduced. Table 1.7 gives approximately for . Boyle’s law is used incorrectly to explain the approach to room temperature. Units, symbols, axes and numerical precision are inconsistent in several places. My strict but realistic estimate is 15/20, with a plausible moderation range of 14–16/20. After a mathematically coherent revision, 18–20/20 is attainable; 19/20 is a realistic high target. The exploration itself does not receive a standalone grade from 1 to 7. 3. Criterion-by-criterion estimate Criterion | Current estimate | Evidence for the estimate | What is needed for the top band A: Presentation | 3/4 | The progression from question to conclusion is generally clear and most figures are placed near the relevant discussion. It is not fully concise: three very similar parameter estimates occupy substantial space, numbering is inconsistent, the page count is wrong and navigation is weak. | Remove repetition, combine raw-data tables, add page numbers, standardise references, and make the model-selection logic shorter and more direct. B: Mathematical communication | 2/4 | There is a good range of representations, but serious inconsistencies remain: instead of , seconds instead of minutes, the factor-of-ten exponent error, missing parameter units, two meanings for , and conflicting regression parameters. | Use one notation table and one reproducible source for every parameter. Check every equation, axis, table, label and unit. C: Personal engagement | 2/3 | The tea routine, chosen threshold and self-collected measurements are authentic. Some development reads as reconstructed for the assessment, particularly the staged sequence from A to B to C and the selection of two favourable points after viewing the data. | Show a genuine mathematical decision: state a prediction, use residuals to reject or refine an estimate, and reserve at least one trial for validation. D: Reflection | 3/3, narrowly | There are recurring reflective points about transformation, an impossible intercept, limitations and model comparison. The strength is reduced because an acknowledged unfair comparison is still used to justify the final choice. | Act on each criticism mathematically and quantify its consequence instead of listing limitations without testing them. E: Use of mathematics — SL | 5/6 | The mathematics is relevant and mostly correct. The differential-equation derivation is above the minimum expected at SL and is largely understood. The wrong exponent, inconsistent regression, incorrect RMSE and unsupported model choice prevent the work from being consistently correct. | Add a direct nonlinear fit, compare methods fairly, show reproducible residual and RMSE calculations, include a sensitivity or uncertainty analysis, and remove all notation errors. Total | 15/20 | A strong investigation with material errors that affect the central conclusion. | Realistic revised target: 19/20. 4. Numbered error and revision table 4.1 Form, structure, data collection and presentation No. | Location | Error or risk, and why it matters | Exact correction or possible replacement wording 1 | Title page | “15 pages” does not match the submitted file. The PDF has 22 pages; excluding the two-page bibliography leaves 20. An incorrect count suggests that the final document was not checked and weakens presentation. | Recount after the final edit. For example: “20 pages excluding the bibliography.” 2 | Whole document | Visible page numbers are missing. This makes a long exploration harder to navigate, moderate and reference. | Add page numbers in the footer. The title page may hide its number, but should normally still be counted. 3 | Title page | “An exploration into Cooling Process” is grammatically incomplete and unnatural. | Use “An exploration of the cooling process of hibiscus tea”, or remove the line. 4 | Whole document | At 20 pages excluding the bibliography, the exploration is at the upper end of the usual guidance and appears close to single spacing in places. Quality, not page count, is decisive, but repetition makes a 4 in criterion A less secure. | Merge the raw-data tables, shorten elementary algebra and replace three near-identical model graphs with one comparison figure. Aim for about 15–17 content-rich pages. 5 | Tables 1–1.9 | Numbering alternates between “Table 1”, “Table 1.1”, “Table 1.2”, and so on. The decimal system implies sub-tables even where the items are independent. | Use Table 1, Table 2, … throughout, or use Table 1a–1d only for deliberately grouped data tables. 6 | Figures 1–1.7 | The prose refers to both “Figure 1.0” and “Figure 1”, while captions use another sequence. | Choose one continuous figure sequence and make every in-text reference match it exactly. 7 | Research question | “for say 300 mL” is filler language and reduces precision. | “How accurately does Newton’s law of cooling model the cooling of 300 mL of hibiscus tea, and what time does the model predict for the temperature to fall below in a room at ?” 8 | Research question | “How accurately” is not operationalised. It is unclear whether accuracy means RMSE, residual behaviour, threshold-time error or performance on a new trial. | Add: “Accuracy will be assessed using RMSE in degrees Celsius, residual plots and error in the predicted crossing time on a validation trial.” 9 | Introduction | “Temperature is continuous” is presented as an experimental fact, although measurements are discrete. Continuity is a modelling assumption. | “I modelled temperature as a continuous function of time, although readings were taken at discrete five-minute intervals.” 10 | Introduction | “The surrounding room places a lower restriction” is physically too absolute. Evaporation can cool a liquid below the dry-bulb ambient temperature. | “Under the simplified Newton cooling model, and neglecting evaporative cooling below ambient temperature, room temperature is the long-term equilibrium value.” 11 | Introduction | Subject–verb agreement is wrong in “The bright red colour and sour taste makes”. | “The bright red colour and sour taste make the drink especially enjoyable.” 12 | Introduction | “thermal equillibrium” is misspelled. | Use “thermal equilibrium.” 13 | Introduction | The claim that hibiscus tea is less viscous than coffee is unsupported and does not explain the chosen drinking temperature. | Remove it, or write: “I chose from repeated personal trials; it is a subjective usability threshold rather than a universal preference.” 14 | Background | The discussion of anthocyanin extraction is only weakly connected to the mathematical question and reduces concision. | Reduce it to: “I used the same hot-infusion procedure in every trial so that preparation differences would not dominate the cooling data.” 15 | Methodology | The three measured room temperatures are not displayed as raw data, so the stated mean of is not reproducible. | Give the initial, final and mean room reading for each trial, or show the ambient time series in a compact table. 16 | Methodology | “A fluctuation of about ” is ambiguous. It could mean measurement uncertainty, observed range or deviation from the mean. | “During the trials, the room reading ranged from … to … ; the thermometer resolution was … .” 17 | Methodology | “Using three trials should eliminate random errors” is false. Repetition does not eliminate random error; it helps estimate variability and can reduce its influence on the mean. | “Three repeated trials do not remove random error, but they allow variability to be estimated and reduce its influence on the mean.” 18 | Methodology | Display resolution and measurement uncertainty are treated as identical. A display reading to does not prove an uncertainty of . | State the manufacturer’s accuracy or calibration result. Otherwise distinguish carefully between resolution and estimated measurement uncertainty. 19 | Methodology | Timer resolution, reading delay and probe response time are omitted. These effects matter near the threshold. | State the timer resolution, typical reading delay and probe response time; discuss or test their likely effect on crossing-time estimates. 20 | Raw data | Three almost identical tables occupy substantial space and make between-trial comparison awkward. | Use one table with columns , , , mean and sample standard deviation. 21 | Processed data | Only the mean is reported. Without a standard deviation or error bars, the reproducibility of the trials is invisible. | Calculate at every time and show it in the table or as graph error bars. 22 | Processed data | Averaging first and fitting only the mean curve hides variation in fitted parameters and threshold times. | Fit each trial separately, or at minimum calculate the crossing time for each trial. 23 | Figure 1 | Colour alone distinguishes points and the asymptote, which is poor for greyscale printing and colour-vision accessibility. | Use different markers and line styles; do not describe series only as “red” and “black”. 24 | Data/method | Humidity is described as uncontrollable but was not measured. Because evaporation is later used as an explanation, the claim remains untested. | Measure humidity, or identify evaporation only as one possible and unconfirmed explanation. 4.2 Mathematics, modelling and statistics No. | Location | Error or risk, and why it matters | Exact correction or possible replacement wording 25 | After Figure 1 | “A horizontal asymptote at ” confuses time with temperature. | “A horizontal asymptote at was added.” 26 | After Figure 1 | Boyle’s law does not explain the asymptote. Boyle’s law relates gas pressure and volume at constant temperature. | “In the simplified Newton cooling model, net heat transfer tends to zero as approaches the ambient temperature .” 27 | After Figure 1 | “A substance cannot decrease its temperature below that of its environment” is too absolute and ignores evaporative cooling. | “Without an additional cooling mechanism such as evaporation, the Newton model approaches asymptotically and does not cross it.” 28 | After Figure 1 | “Negative correlation” is imprecise for a curved time series, and no correlation coefficient is calculated. | “The data show a monotonic, nonlinear decrease in temperature with time.” 29 | Attempt A | Beginning with is dimensionally incomplete: the right side is dimensionless while has temperature units. | Write the shape first as ; after scaling, use with in and in . 30 | Attempt A | is called a “horizontal stretch factor”. In , is a rate and is the characteristic time scale. | “The cooling-rate constant has units ; the characteristic time is .” 31 | Attempt A | is treated as a vertical scale factor but no unit is given. | Use and define it: “ is the initial excess temperature, measured in .” 32 | Attempt A | The assertion that the first and last observations “should create the most representative model” is unsupported. Endpoints may contain error and all intermediate readings are ignored. | “This endpoint estimate captures the total observed change, but it is sensitive to both endpoint readings and is used only as a preliminary benchmark.” 33 | Determining constants A | “Utmost accuracy” and six reported decimal places are false precision when temperatures are rounded to . | Retain unrounded values internally, but report, for example, and explain the calculation precision. 34 | Logarithm steps | formally takes the logarithm of a dimensional quantity. Saying that logarithms have no units does not resolve this. | Define and use . At SL, it is also acceptable to explain clearly that numerical values of the temperature difference measured in degrees Celsius are being transformed. 35 | Attempt B | The symbol already denotes exponential amplitude and is then reused as the slope in . | Use , where and . 36 | Attempt B | The line “” has the wrong sign and is not a correct variable mapping. | Write , , and , so . 37 | Table 1.5 | The heading is “Time (s)” even though every observation and calculation uses minutes. | Change it to “Time (min)”. 38 | Attempt B | The claim that changing for each row would mean the equation no longer describes a function is mathematically false. A time-dependent ambient temperature still defines a function, but not this simple linearisation. | “A time-dependent ambient temperature would produce a non-autonomous differential equation and invalidate this simple linearisation.” 39 | Figure 1.3 | “98.71% of the variation in temperature is explained by time” misinterprets . The regression response is , not . | “ means that 98.70% of the variation in the transformed response is accounted for by its linear relation with time in this data set.” 40 | Figure 1.3 | “a exponential equation” is grammatically incorrect. | Use “an exponential equation.” 41 | Regression B | The graph shows approximately , while the prose uses and . The rounded visible values give . | Use one unrounded regression output. From the visible means: . 42 | Equation B/Table 1.7 | The Table 1.7 predictions match the graph version of B with , not the printed equation B with . The equation, table and RMSE therefore cannot all be reproduced from one another. | Generate the table and RMSE automatically from the same final spreadsheet formula or equation; do not retype parameters in several places. 43 | Tables 1.7/parameters | Four to six decimal places in predicted temperatures imply accuracy unsupported by the instrument or model. | Keep more digits internally, but report approximately or in the text and state the uncertainty. 44 | After Figure 1.2 | A “repeated pattern of residuals” is asserted without a residual plot. Criteria D and E require evidence for this claim. | Plot against for each serious candidate, add a zero line and interpret the pattern. 45 | After Figure 1.2 | Air movement, thermal mass and evaporation are presented as causes without being measured or separated. | “Possible explanations include …, but the present data cannot identify which mechanism caused the residual pattern.” 46 | Newton’s law | The unit of is omitted, so the differential equation is not dimensionally documented. | Define in , and in , and in minutes. 47 | Differential equation | The derivation moves from to without explaining why the absolute value may be removed. | “All measured temperatures satisfy , so is positive and the absolute value may be removed on the observed domain.” 48 | Differential-equation solution | A new parameter is introduced even though its meaning follows immediately from the initial condition; it is also left without a unit. | Use . Then . 49 | Attempt C | The justification for selecting and is weak. is not close to the drinking threshold of . | If a threshold-local two-point estimate is required, use the 20- and 25-minute observations. Prefer fitting all selected data by nonlinear regression. 50 | Attempt C | The two points were selected after looking at the data. This creates selection bias and can make C look artificially good in a chosen interval. | State the selection rule before analysis, or use separate training and validation data. 51 | Equation C | The printed equation is , but the calculations use . The printed decay rate is ten times too large. | Use consistently, then regenerate every table and graph. 52 | Figure 1.5 | The horizontal axis reads “Time (seconds)” although the numerical values are minutes. A sentence below the graph does not repair the incorrect axis. | Correct the graph itself to “Time (minutes)”. 53 | Models A–C | A, B and C are not three different physical models. All have the form ; only the parameter-estimation method changes. | Retitle the section “Comparing three parameter-estimation methods for the same Newton cooling model.” 54 | Modelling method | Because RMSE in degrees Celsius is the chosen criterion, the most direct method is missing: nonlinear least squares in the original temperature scale. | Minimise and compare this result with the log-linear estimate. 55 | RMSE section | RMSE is not generally “the standard deviation of the residuals”. That interpretation requires additional assumptions and a different denominator or centring convention. | “RMSE is the square root of the mean squared residual and measures typical prediction error in the original temperature units.” 56 | RMSE section | “A lower result shows that the regression line is a better suit” is grammatically and mathematically imprecise; not all candidates are regression lines. | “When models are evaluated on the same observations, a lower RMSE indicates a closer fit in degrees Celsius.” 57 | RMSE calculation | No worked RMSE example or unambiguous spreadsheet formula is supplied. This lack of reproducibility allowed the incorrect adjusted value to pass unnoticed. | Include a residual table or one fully substituted calculation, and name the software and formula used. 58 | Table 1.9 | The stated for C on cannot be reproduced from the displayed values. Table 1.7 gives about . | Recalculate: . 59 | Adjusted comparison | C is evaluated on , while A and B are quoted on . RMSE values from different observation sets cannot be compared directly. | Evaluate every equation on the same domain. From the visible data, for : A , B and C . 60 | Selection of C | Under a fair comparison, B—not C—is best among A, B and C for . The central choice in the exploration is therefore not supported by its own data. | Select B, or preferably use a direct nonlinear fit and test it on a separate trial. 61 | Interpretation of C | C’s high initial-temperature estimate is mainly caused by fitting two observations and extrapolating backwards, not by holding constant. | “The unrealistic intercept results from fitting only the 15- and 45-minute observations and extrapolating that curve back to .” 62 | Interpretation | Warming of the local air and heat returned by the cup are described as established explanations, but the data do not separate these mechanisms. | Label them as hypotheses. A coupled cup–liquid model may be mentioned as an extension, not as a demonstrated cause. 63 | Removing 0–10 min | “I would never drink the tea immediately” does not automatically justify deleting early data because the requested time is still measured from . | Define the relevant fitting domain before analysis, retain the original time origin and fit and test every method on the same domain. 64 | Validation | The same data are used both to choose or fit the model and to assess it. This measures in-sample fit, not independent predictive accuracy. | Fit with Trials 1 and 2 and validate on Trial 3, then rotate the held-out trial, or collect a fourth independent validation trial. 4.3 Conclusion, reflection, sources and language No. | Location | Error or risk, and why it matters | Exact correction or possible replacement wording 65 | Answering the question | The model value of 21.53 min and the linearly interpolated value of 24.00 min differ by 2.47 min. “Reasonably close” is too vague without quantification. | “The model predicts 21.53 min, whereas interpolation between the adjacent observations gives 24.00 min; the model predicts the crossing 2.47 min earlier.” 66 | Conclusion | “Maximum amount of time” sounds more exact than the data and model justify. | “The selected model gives an estimated crossing time of about 22 minutes; the five-minute observations place the crossing between 20 and 25 minutes.” 67 | Conclusion | “If I begin drinking within twenty minutes, the model predicts the tea will remain above …” is ambiguous: above the threshold at the beginning, or throughout drinking? | “The model predicts that the tea is still above at min.” 68 | Result | No uncertainty interval is given for the threshold time. One decimal value is not adequate with five-minute sampling and visible model bias. | Report the trial-by-trial interpolations: approximately 24.15, 23.75 and 24.10 min; mean min and sample SD about min. Discuss model uncertainty separately. 69 | Limitations | “More than three trials … permit uncertainty intervals” is false. Three trials already permit a preliminary spread estimate; more trials improve its reliability. | “Three trials allow a preliminary estimate of variability; additional trials would make an uncertainty interval more reliable.” 70 | Limitations | “Recording every minute” is reasonable but not targeted to the research question. | Measure every 30–60 seconds between about 18 and 28 minutes; longer intervals may be sufficient elsewhere. 71 | Limitations | Measuring mass instead of volume is suggested because of evaporation, but one initial mass would not show evaporative loss. | Weigh cup and tea during the experiment, or compare initial and final mass without materially altering heat transfer. 72 | Limitations | Making a function of time is mathematically promising, but the resulting equation is not stated, so the suggestion remains decorative. | Either test in a sensitivity analysis or discuss and an appropriate solution method. 73 | Limitations | A magnetic stirrer would improve temperature homogeneity but also alter convection and therefore . | Treat stirring as a new controlled design, not as a neutral correction to the existing trials. 74 | Reflection | The short “Reflection” section largely summarises methods. The strongest reflection is scattered elsewhere. | Use the section to answer a mathematical question: why did a high linearised not produce the lowest temperature RMSE, and how did that change the model choice? 75 | Assessment-style tone | “Since both equations were created mainly from the shape … I completed additional research” sounds like a retrospective rubric narrative rather than a genuine mathematical decision. | “The residual pattern suggested that a fitted exponential alone did not explain why this form should arise, so I derived the model from a rate equation.” 76 | Assessment-style tone | “The mathematics developed from a visual conjecture into a physically motivated model” echoes assessment language rather than a personal thought process. | “At first I chose the exponential form from the graph. Deriving Newton’s law later showed which assumptions were hidden in that choice.” 77 | Assessment-style tone | Stock transitions such as “This initially suggests …”, “On closer observation …” and “Nevertheless, it answers the investigation more directly” sound formulaic, especially when the numerical decision does not follow. | State the numbers directly: “On , model B has RMSE and model C ; therefore B fits this domain more closely.” 78 | Assessment-style tone | “The investigation has transformed an ordinary routine into a practical use of …” is a generic closing sentence. | “For my preparation method, the data place the crossing between 20 and 25 minutes. The remaining uncertainty is large enough that I would use 20 minutes as a cautious reminder.” 79 | Sources | Lienhard, Khan Academy and Abraham et al. appear in the bibliography but are not clearly cited in the body. This obstructs traceability and may look like bibliography padding. | Cite every source at the exact claim it supports, and remove unused sources. 80 | Introduction image | The hibiscus image has no visible credit. The statement about experimental temperatures and graphs does not cover a third-party image. | Remove the image, or give creator, URL and licence in the caption and bibliography. 81 | Works Cited | Citation style, DOI/URL format and access dates are inconsistent. | Apply one style throughout, such as MLA 9 or APA 7; give DOIs as https://doi.org/... where available. 82 | OpenStax source | The listed OpenStax address “4-key-equations” does not clearly identify the Newton-cooling section being discussed. | Verify the exact section title and cite the direct address. 83 | Reproducibility | “Spreadsheet and graphing software” does not name the program, version, regression option or formula range. | State the software and method, for example: “I regressed on in … and calculated RMSE using …” Explain the central calculation in the text as well. 84 | Language | Repeated errors reduce the professional impression: “seperated”, “linearisaton”, “Newtons law”, “degrees celsius”, missing articles and inconsistent capitalisation. | Use separated, linearisation, Newton’s law, degrees Celsius, then complete a full spelling and grammar check. 85 | Academic integrity | If AI-generated wording or calculations are included in the submitted exploration, current IB guidance requires them to be credited in the body and appropriately referenced. Uncredited AI-generated material is not the student’s own work. | Understand and verify every mathematical idea, then rewrite it in the student’s own voice. Disclose any retained AI-generated material according to the school’s and IB’s requirements; do not submit this review as personal work. 5. Essential mathematical rebuild 5.1 One model, several parameter-estimation methods A, B and C should not be presented as different physical models. Their common assumption is Newton’s law of cooling: with solution Only the method used to estimate and changes: an endpoint estimate; linear regression after logarithmic transformation; a two-point estimate using observations chosen after inspection; direct nonlinear regression in the original temperature scale. The fourth method is aligned with an RMSE criterion in degrees Celsius because it directly minimises 5.2 Independent calculation from the displayed mean data With fixed, direct nonlinear least squares on the visible mean values gives approximately where is measured in minutes. This yields and These values are an independent check based on the rounded numbers visible in the PDF. They should be recalculated from the original unrounded observations and understood by the student before use. 5.3 Fair comparison of the displayed equations Method | RMSE, all observations | RMSE for min | Predicted A: endpoint estimate | | | min B: corrected log regression | | | min C: 15/45-minute points | | | min Direct nonlinear regression | | | min C is neither the best global estimate nor the best estimate for . The observed crossing time obtained by linear interpolation between 20 and 25 minutes is approximately for the three trials, with mean about min and sample standard deviation about min. The difference between min and the direct nonlinear prediction of min is about min. This is valuable reflection: between-trial variability is small, yet the simple one-exponential model has a systematic predictive discrepancy. 6. Revision plan for a top mark Priority 1 — essential before submission Correct the factor-of-ten error in equation C and every seconds/minutes error. Recreate equation B from one parameter set; generate its table, graph and RMSE from the same source. Remove Boyle’s law and explain the asymptote through Newton’s cooling model. Compare all candidates on the same observations. Correct the adjusted RMSE and rewrite the model choice. Priority 2 — largest likely mark gain Add direct nonlinear regression in the original temperature scale. Add and interpret a residual plot for the preferred model. Use Trial 3 as independent validation, rotate the held-out trial, or collect a fourth trial. Report the threshold time for each trial and quantify its spread. Add a notation-and-units table for , , , , , and RMSE. Priority 3 — from strong to excellent Reduce the exploration by about 3–5 pages without removing substantive mathematics. Show personal engagement through consequential decisions, not repeated use of “I”. Attach a quantitative consequence to each important limitation. Remove unused sources and the uncredited image. Standardise every figure, table, page number, unit and decimal place. 7. Natural replacement passages These passages are deliberately direct and less rubric-like. They should be understood, checked against the student’s actual work and rewritten in the student’s own voice rather than copied mechanically. Research question and measure of accuracy Research question. How well does Newton’s law of cooling describe the temperature of 300 mL of hibiscus tea in a room at approximately , and what time does the model predict for the tea to fall below my chosen threshold of ? I will assess the model using residuals, RMSE in degrees Celsius and its error when predicting a trial that was not used to estimate the parameters. Physical modelling assumption I treated room temperature as constant and assumed that the net cooling rate was proportional to the tea’s temperature difference from the room. With and , this gives . Because every recorded tea temperature was above , solving the equation gives , where is the initial excess temperature and is measured in . Model fitting Linearising the data is useful for checking whether an exponential pattern is plausible, but it changes the weighting of errors. Since my final accuracy measure is RMSE in degrees Celsius, I also fitted directly by minimising squared residuals in the original temperature scale. This avoids choosing two convenient observations and uses every value in the selected training set. Fair comparison I evaluated every parameter-estimation method on the same observations. On the restricted domain min, the RMSE values for A, B and C were approximately , and , respectively. C was therefore not the most accurate of these three estimates on the domain relevant to the threshold. The direct nonlinear fit gave the lowest full-data RMSE, but its residuals still showed a systematic pattern, so the one-exponential model should not be treated as exact. Conclusion The direct nonlinear model predicts that the tea reaches after about 22.3 minutes. Linear interpolation of the measurements surrounding the threshold gives about 24.0 minutes. The difference of roughly 1.7 minutes is larger than the variation between the three interpolated trial values, suggesting systematic model error rather than only random measurement variation. I would therefore report a practical crossing window of 22–24 minutes, while the present five-minute sampling establishes only that the crossing occurred between 20 and 25 minutes. Critical reflection The high value of the linearised graph initially made the exponential model appear stronger than it was. However, described the transformed variable , whereas my practical question concerns errors in degrees Celsius and minutes. Refitting the model in the original scale changed the preferred parameter estimates. A visually straight transformed graph is therefore evidence for a model form, but it is not proof that its predictions are most accurate for the quantity I care about. 8. Assessment sources International Baccalaureate, Mathematics: analysis and approaches subject brief — first assessments 2021. International Baccalaureate, Mathematics: analysis and approaches updates, including the new course timeline and May 2029 first assessment. International Baccalaureate, Artificial intelligence in learning, teaching, and assessment. Final assessment The investigation has strong potential. The decisive improvements are consistency, fair validation and honest quantitative criticism of the model—not additional pages or more complicated mathematics. If the five central mathematical errors are corrected and direct regression, residual analysis and validation are integrated coherently, the exploration can become a convincing top-band submission.