The Surface Area Of Oddly Shaped Objects Construction Essay

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The purpose of this project is to determine the surface area of oddly shaped objects like hands and leaves. To do this, I tried to cut different sizes of squares (like 3cm, 4cm, etc) and find each of its mass in grams using an electronic scale. Using my collected data, I will make a scatter plot with the area in cm2 on the y-axis and the mass in grams on the x-axis manually and electronically using Graph 4.3. To find the line of best fit, of both scatter plots, I will find and compare the masses of the leaf and the hand. Then, I will estimate the total surface area of the hand and leaf. I will also experiment by identifying the possible errors and solutions. Thus, I will conclude the experiment by find out how scientists determine the surface area accurately and why it is important for scientists to b able to determine surface areas.

The table below illustrates the results of the experiment I performed.

In the graph, the independent variable and the y-axis represent the area of the square in cm2 while the dependent variable and the x-axis represent the mass of the squares in grams. This shows that as the mass of the square increases, the area of the square increases too, making a positive correlation. The shape of my graph is a positive correlation without any strange or unexpected features because the areas of the squares are expected to increase as the mass of the square increases. In other words, the larger the area of the square cutout gets, the larger the mass gets. My model can work for both the values of x and y since my points is very close to the line of best fit.

The equation of my line of best fit I found without using Graph 4.3 is y=35.7x+1.43. This shows that the slope of my line is 35.7 and represents the area of the square in cm2 per mass of the square in grams. However, the y-intercept is 1.43 and illustrates that this is an anomaly since the y-intercept has to be 0.

To find the slope and y-intercept using the slope-intercept form, I tried to find two positions in the line of best fit that I can easily find the coordinates of. For example, I used these two pairs of coordinates to find my equation: (2.2, 80) and (3.6, 130).

Note that the actual equation is y=35.7142857142x+1.428571429 and the equation stated above is an estimate of 3 significant figures. That would mean that the actual slope is 35.7142857142 and the actual y-intercept is 1.428571429.

The graph below shows the comparison of the line of best fit I found manually and the line of best fit using Graph 4.3

As seen from the graph above, the regression model is y=36.210279x-0.099138263 and the estimated regression model in 3 significant figures is y=36.2x+0.10. This means that the slope of the regression model is 36.210279 while the estimated slope in 3 significant figures is 36.2 and the y-intercept of the regression model is -0.099138263 while the estimated y-intercept is -0.10.

Also as seen from the graph above, the line of best fit I found manually and the line of best fit found using Graph 4.3 closely matches while both closely passes through all the points of my graph.

To find the area of the leaf which is the y-value, I first substituted the mass of the leaf which is the x-value to the equation of the line of best fit I found both manually and using Graph 4.3. To find the area of the hand cutout which is also the y-value, I did the same thing except that the substituted amount is the mass of the hand cutout which is also the x-value to the equation of the line of best fit I found both manually and using Graph 4.3 instead of the mass of the leaf.

Let x=mass of leaf in grams

Let y=area of leaf in cm2

Area using the line of best fit found manually Area using the line of best fit found using Graph 4.3

y=35.71428571x + 1.428571429 y=36.210279x - 0.099138263

y=35.71428571(0.84) + 1.428571429 y=36.210279(0.84) - 0.099138263

y=31.42857143 y=29.42525173

y31.4 cm2 (rounded to 3 significant figures) y29.4 cm2 (rounded to 3 significant figures)

Therefore, the area of the leaf using the line of best fit found manually is approximately 31.4 cm2 while the area using the line of best fit found using Graph 4.3 is approximately 29.4 cm2.

Let x=mass of hand cutout in grams

Let y=area of hand cutout in cm2

Area using the line of best fit found manually Area using the line of best fit found using Graph 4.3

y=35.71428571x + 1.428571429 y=36.210279x - 0.099138263

y=35.71428571(4.62) + 1.428571429 y=36.210279(4.62) - 0.099138263

y=166.4285714 y=167.1923507

y166 cm2 (rounded to 3 significant figures) y167 cm2 (rounded to 3 significant figures)

Therefore, the area of the hand cutout using the line of best fit found manually is approximately 166.4 cm2 while the area of the hand cutout using the line of best fit found using Graph 4.3 is approximately 167.2 cm2.

There are some similarities and differences between the area of the leaf and the hand cutout. For example, both areas of the leaf and hand cutout have area differences between their areas found manually and using Graph 4.3 of 1 cm2. The difference of the areas of the leaf and hand cutout is significant by at least 130 cm2.

In order to find the surface area of the leaf, I had to multiply the area by two since the leaf has two sides. That means that the surface area of the leaf is approximately in between 62.8 cm2 and 58.8 cm2 but I also had to identify the measurement of the surface area of the inner fold of the hand. To do this, I added half of the amount of one side of the hand before multiplying it by two. After doing so, the total surface area of the hand is approximately in between 498 cm2 and 501 cm2. I believe that the leaf has a more accurate approximation of the surface area since its width is a lot thinner compared to the hand and the leaf has no inner folds.

Let y= surface area of the leaf

Surface area using line of best fit found manually Surface area using line of best fit using Graph 4.3

Area of leaf31.4 cm2 Area of leaf29.4 cm2

y31.42 y29.42

y62.8 cm2 y58.8 cm2

Therefore, the surface area of the leaf is approximately between 58.8 cm2 and 62.8 cm2

Let y=surface area of the hand

Surface area using line of best fit found manually Surface area using line of best fit using Graph 4.3

Area of hand cutout166 cm2 Area of hand cutout167 cm2

y166 y167

y62 cm2 y63.5 cm2

y166+62 y167+63.5

y228 cm2 y230.5 cm2

y2282 y230.52

y456 cm2 y461 cm2

Therefore, the surface area of the hand is approximately between 456 cm2 and 461 cm2

In this experiment, a lot of errors could have taken place.

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