# Math-gradient function

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Introduction

GRADIENT FUNCTION

The aim of this assignment is to find the relationship between the gradient function. For this assignment I will use the height of drops and the number of drops needed to crack open nuts of different sizes; large, medium and small. The models are then compared in terms of the usefulness of the models.

The table below shows the average number of drops it takes to break open large nuts from varying heights.

Large nuts

Height of drop (m) | Number of drops |

1.7 | 42.0 |

2.0 | 21.0 |

2.9 | 10.3 |

4.1 | 6.8 |

5.6 | 5.1 |

6.3 | 4.8 |

7.0 | 4.4 |

8.0 | 4.1 |

10.0 | 3.7 |

13.9 | 3.2 |

Let (h) metres be the height of drop and (n) be the number of drops. The graph below is of n against h

Middle

(ax+b)

By use of GDC, geogebra i found the equation by moving the line of the curve to fit accurately. 1 ÷ (0.06x+0.012)-0.08) +1.48

Another function that models the data is the logistic function. This is because it has a horizontal asymptote of 0.

3.6

1-1.55e^ (8-0.31x)

The difference between the reciprocal function and the logistic function is that the logistic function fits more accurately. However, it is still hard to predict other values from the equations

The table below shows the

Conclusion

The logistic equation f (n) for medium nuts= -0.26÷(1-0.99e^0.01x)

Small nuts

Height of drop (m) | Number of drops |

1.5 | Undefined |

2.0 | Undefined |

3.0 | undefined |

4.0 | 57.0 |

5.0 | 19.0 |

6.0 | 14.7 |

7.0 | 12.3 |

8.0 | 9.7 |

10.0 | 13.3 |

15.0 | 9.5 |

1/(0.02x-0.08)

The equation for logistic function f(n) for small nuts is

10.62÷ (1.897e^ (-0.6x)

The first model does not fit the data accurately for nuts of different sizes and it is therefore hard to predict a trend from this. The models were unable to predict undefined values and this were therefore the limitations.

This student written piece of work is one of many that can be found in our GCSE Gradient Function section.

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