Here are my solutions to the three sets of data:
GRAPH 1:
This graph resembles an inverse relation between the x values (radius) and the y values (resistance). As the radius (mm) increases, the resistance (Ohms) decreases. Since this is a direct square relation, we cannot find a general slope for the graph. We can, however, use derivatives to find the slope at instantaneous points. To find the general slope of the curve (not accurate) we must first linearize the graph. Here is the same graph with a linear trend line:
To linearize this function I plotted 'y' vs '1/x' since this has an inverse relationship. As seen in the equation given, the slope of the line is m=(0.9667)Ohms/mm --> this means that the resistance increases by 0.9667Ohms for every 1 mm that the radius increases. The R² of the exponential trend line is 0.9254 while the R² of the linear trend line is 0.9984. This means the the linear trend line is much more accurate and better displays the expected results.
GRAPH 2:
This graph resembles an direct relation between the x values (Force) and the y values (Acceleration). As the Force (N) increases, the Acceleration (m/s²) increases in a linear manner. We use a trend line to find the equation of the graph. Using the equation we can find the slope, which is m=0.2902m/s²N = 0.2902 (m/s²)/(kgm/s²) = 0.2902kg-1--> the R² of the trend line is 0.9939, meaning that it is extremely accurate. The reason for this direct relation is because of the famous equation F=ma, where F is Force (N), m is mass (kg) and a is acceleration (m/s²). Given that the mass stays the same, as the force increases so must the acceleration to keep the equation balanced.
GRAPH 3:
This graph resembles an direct square relation between the x values (radius) and the y values (resistance). As the time (s) increases, the position (m) increases exponentially. Since this is a direct square relation, we cannot find a general slope for the graph. To find the general slope of the curve (not accurate) we must first linearize the graph. Here is the same graph with a linear trend line:
To linearize this function I plotted 'y' vs 'x^2'. As seen in the equation given, the slope of the line is m=(20.733)m/s² --> this means that the position increases by 20.733m for every 1 second that passes. The R² of the linear trend line is 0.8274 while the R² of the exponential trend line is 0.9372. This means the the exponential trend line is much more accurate and better displays the expected results.
GRAPH 1:
This graph resembles an inverse relation between the x values (radius) and the y values (resistance). As the radius (mm) increases, the resistance (Ohms) decreases. Since this is a direct square relation, we cannot find a general slope for the graph. We can, however, use derivatives to find the slope at instantaneous points. To find the general slope of the curve (not accurate) we must first linearize the graph. Here is the same graph with a linear trend line:
To linearize this function I plotted 'y' vs '1/x' since this has an inverse relationship. As seen in the equation given, the slope of the line is m=(0.9667)Ohms/mm --> this means that the resistance increases by 0.9667Ohms for every 1 mm that the radius increases. The R² of the exponential trend line is 0.9254 while the R² of the linear trend line is 0.9984. This means the the linear trend line is much more accurate and better displays the expected results.
GRAPH 2:
This graph resembles an direct relation between the x values (Force) and the y values (Acceleration). As the Force (N) increases, the Acceleration (m/s²) increases in a linear manner. We use a trend line to find the equation of the graph. Using the equation we can find the slope, which is m=0.2902m/s²N = 0.2902 (m/s²)/(kgm/s²) = 0.2902kg-1--> the R² of the trend line is 0.9939, meaning that it is extremely accurate. The reason for this direct relation is because of the famous equation F=ma, where F is Force (N), m is mass (kg) and a is acceleration (m/s²). Given that the mass stays the same, as the force increases so must the acceleration to keep the equation balanced.
GRAPH 3:
This graph resembles an direct square relation between the x values (radius) and the y values (resistance). As the time (s) increases, the position (m) increases exponentially. Since this is a direct square relation, we cannot find a general slope for the graph. To find the general slope of the curve (not accurate) we must first linearize the graph. Here is the same graph with a linear trend line:
To linearize this function I plotted 'y' vs 'x^2'. As seen in the equation given, the slope of the line is m=(20.733)m/s² --> this means that the position increases by 20.733m for every 1 second that passes. The R² of the linear trend line is 0.8274 while the R² of the exponential trend line is 0.9372. This means the the exponential trend line is much more accurate and better displays the expected results.
Good Graph!
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