Lisaly Per1 Precalproject Nuts

Lisa Ly Period 1 11/11/09 Pre Calculus Project-Hawks Dropping Nuts INTRODUCTION For this project, hawks are breaking large nuts. I was to decide which function best fit the set of data given to me. From the set of data, it shows how high the hawks are breaking the large nuts and how many times they had to break the large nuts at that height. NUMBER OF DROPS PER HEIGHT TO BREAK LARGE NUTS Height of drop (feet) 5.5 6.5 9.4 13.3 18.2 20.5 22.7 26.0 32.5 45.0

Number of drops 42.0 21.0 10.3 6.8 5.1 4.8 4.4 4.1 3.7 3.2

The x-variables of the graph is the height of drop (feet) and the y variables of the graph is the number of drops. The graph shows that at certain heights, it takes a certain number of drops for a nut to break its shell. By increasing the height, the number of drops it takes for the nut to break decreases. EXPONENTIAL FUCTION MODELING LARGE NUTS DATA SET Height of drop (feet) 5.5 6.5 9.4 13.3 18.2 20.5 22.7 26.0 32.5 45.0

Number of drops 42.0 21.0 10.3 6.8 5.1 4.8 4.4 4.1 3.7 3.2

Exponential function 20.3 16.6 10.7 7.0 4.8 4.2 3.7 3.1 2.4 1.6

The type of function that seems to model the behavior of the graph is an exponential function which is f(x) = a^-x. I thought that this function would be the closest fit for the graph because the height of drop and the number of drops will never equal zero and the number of drops only decreased but it's possible for the number of drops to continuously decrease if the height continued to increase. For the graph, I have came up with the equation: f(x)=156.9*x^(-1.2) by figuring out how close the f(x) could get to the y-variable which is height of drop (feet) all together as the whole graph and not how close it could get to one point. I chose the power trend line because it traced the graph the best but also, the equation the trend line was similar to the function I have came up with. The equation of the trend line was y=179.2x^-1.153 and r^2=0.8974. The only difference was that a=156.9 in the equation I came up with and a=179.2 in the equation of the trend line. EXPONENTIAL FUCTION MODELING LARGE NUTS DATA SET VS. EXPONENTIAL DECAY MODEL MODELING LARGE NUTS DATA SET Height of drop (feet) 5.5 6.5 9.4 13.3 18.2 20.5 22.7 26.0 32.5 45.0

Number of drops 42.0 21.0 10.3 6.8 5.1 4.8 4.4 4.1 3.7 3.2

Exponential function 20.3 16.6 10.7 7.0 4.8 4.2 3.7 3.1 2.4 1.6

Exponentia l decay model 24.3 22.2 17.0 11.9 7.6 6.2 5.1 3.8 2.1 0.7

Another function that closely models the data is an exponential decay model and the equation was f(x)=40*e^(-1/11*x) but it's still doesn't closely fit the large nut data like the exponential function I have came up with. More points of the exponential function were closer to the original points than the points of the exponential decay model.

NUMBER OF DROPS PER HEIGHT TO BREAK MEDIUM NUTS Height of drop (feet) 4.8 6.5 9.8 13.0 16.3 19.5 22.8 26.0 32.5 48.8

Number of drops

Using original exponential function 23.9 16.6

Changes in exponentia l function 38.2 29.1

27.1 18.3 12.2 11.1 7.4 7.6 5.8 3.6

Two y-variables were missing in which I had to use my original exponential to find them. My original exponential equation gave me 23.9 drops for 4.8 ft high and 16.6 drops for 6.5 ft high. I had to adjust my x from -1.2 to -0.9 so that 23.9 drops changed to be 38.2 drops and 16.6 drops changed to be 29.1 drops. The only thing I made sure of was that the drops wouldn't be significantly different than the drops it would take at the height of 9.8 ft. It wouldn't make sense if it took 27.1 drops to break a nut at 9.8 ft and it took 16.6 drops to break a nut at 6.5 ft, unless, there were force being used while dropping the nut 16.6 times to break a nut at 6.5 ft.

NUMBER OF DROPS PER HIEGHT TO BREAK SMALL NUTS Height of drop (feet) 4.8 6.5 9.4 13.0 16.3 19.5 22.8 26.0 32.5 48.8

Number of drops

Using original exponential function 23.9 16.6 10.7

Changes in exponentia l function 71.6 61.5 51.2

57 19 14.7 12.3 9.7 13.3 9.5

Three y-variables were missing in which I had to use my original exponential to find them. My original exponential equation gave me 23.9 drops for 4.8 ft high; 16.6 drops for 6.5 ft high; and 10.7 drops for 9.4 ft high. I had to adjust my x from -1.2 to -0.5 so that 23.9 drops changed to be 71.6 drops;16.6 drops changed to be 61.5 drops; and 10.7 drops changed to 51.2 drops. The only thing I made sure of was that the drops would be significantly different than the drops it would take at the height of 13.0 ft because it wouldn't make sense if at 9.4 ft, it'll take 10.7 drops for the nut to break while at 13.0 ft, it'll take 57 drops to break, unless, there were force being used when it took 10.7 drops to break a nut at 9.4 ft.

CONCLUSION The higher the hawks dropped the nuts, the less they had to drop it. The function I have chosen was to model the data was a exponential function for the large, medium, and small nut data set but the only thing I had to change to fit each data set was the x which influenced how many drops it took for the hawk to. It's possible to use a exponential decay model but it didn't closely model the graph like the exponential function.