- Level: International Baccalaureate
- Subject: Maths
- Document length: 4414 words
Math IA - Logan's Logo
Extracts from this essay...
Introduction
IBO INTERNAL ASSESMENT LOGAN'S LOGO MATHEMATICS SL TYPE II INTRODUCTION Logan has designed the logo at the right. The diagram shows a square which is divided into three regions by two curves. The logo is the shaded region between the two curves. He wishes to find mathematical functions that model these curves. In order to find these functions, we will need to overlay the logo on graph paper, so we can interpret data points to be able to plot them. Take note that in the "modeling the data" in the next section, the logo was not resized, but set as transparent so that data points could be read. Also, take into consideration the uncertainty of the measurements (± 0.25 units). For modeling purposes, the uncertainties are not included in the data calculations; however this should not be overlooked. MODELING THE DATA NOTE: Data tables and their graphs are included on the next page. TOP CURVE : In order to find a function to model the top curve, there are various methods that we can use. One is to overlay the logo onto a set of axes and estimate points for the function. Once we obtain these points, we can then plot them onto a new set of axes. Judging from the logo itself, at first glance it appears that a sine function would fit the data. The sine function would have to undergo a series of transformations to eventually fit the curve. Using the axes and logo depicted above, I estimated 13 points for and recorded them in the data table below: NOTE: Due to the limited precision of the graph, I was only able to estimate to the nearest tenth. Maximum and minimum points have been shaded. X Y -2.5 -1.0 -2.0 -2.5 -1.6 -2.8 -1.0 -2.4 -0.5 -1.3 0.0 0.0 0.5 1.5 1.0 2.6 1.5 3.4 1.8 3.5 2.0 3.4 2.5 2.5 2.6 1.9 After determining these data points, I then plotted them onto a separate set of axes: From here, it is obvious that a sine function would fit the data.
Middle
Now that we've found the highest and lowest points of the sine curve, we must divide the difference between the two y-values (subtracting the highest point on the y-axis from the lowest will give us the actual height of the curve) by 2: The value of a, 2.2 units, is the vertical dilation of the curve because it reflects the stretch factor compared to the original sine curve (defined as ), where the amplitude is 1 unit. However, 2.2 is not the final value of a. It can be seen from the above graph that the curve begins with a negative slope (it goes downwards first and then upwards). This indicates to us that we must place a negative sign before the a value, so the curve will begin with a negative slope. Thus, the final value of a=-2.2. TO FIND b: As stated above, changing the variable b will affect the horizontal dilation of the sine curve. This dilation occurs parallel to the x-axis, which means that the period of the graph is altered. A period is defined as the length it takes for the curve to start repeating itself. Thus in order to determine the variable b's value, we first need to look at the period of the graph. For the original sine curve, the period is, or 360º . Thus the formula that relates the value of b to the period ? of the dilated function is given by: , Since the graph of will show b cycles in 2? radians. From this equation we see that as the value of b is increased (the horizontal dilation of the curve is greater), the period becomes smaller. We know this because b is the denominator, meaning that increasing its value would decrease the overall fraction's value. To find ?, it was easier to find half of the period first, and then double it to ensure accuracy.
Conclusion
- b 3. Horizontal shift- c 4. Vertical shift- d Numbers 1 and 4 both stretch or shrink the curve parallel to the x-axis, which means that the amplitude a and vertical shift d will only be affected by the manipulated range (factor of or). On the other hand, numbers 2 and 3, the horizontal dilation and shift, will stretch or shrink the curve parallel to the y-axis. Again, this means that they will only be affected by the manipulated domain (factor ofor). Now we are able to apply the corresponding shrink factors to their appropriate variables in order to obtain a new curve for Logan's business cards. TOP CURVE The final equation for the top curve is given by the following equation: with a period of 6.3. Thus we must make the following changes to each of the variables. For a, the amplitude or vertical dilation factor: For b, the horizontal dilation factor: For c, the horizontal shift factor: For d, the vertical shift factor: Thus the final equation for the logo with dimensions compatible with that of a standard business card is: with parameters of: and. The graph of the manipulated curve is shown below, on a 9cm × 5cm business card. BOTTOM CURVE The final equation for the top curve is given by the following equation: with a period of 6.0. Thus we must make the following changes to each of the variables. For a, the amplitude or vertical dilation factor: For b, the horizontal dilation factor: For c, the horizontal shift factor: For d, the vertical shift factor: Thus the final equation for the logo with dimensions compatible with that of a standard business card is: with parameters of: and. The graph of the manipulated curve is shown on the next page, on a 9cm × 5cm business card. The final logo on the business card is shown below. Note that the dimensions are in units which, shown through the calculations above, have been converted to ensure they match the ratio and proportion of a standard business card. ?? ?? ?? ?? 26 | Page
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