# Activity5.11Derivative Formulas and Function Behavior¶ permalink

Derivative formulas can give us much information about the behavior of a function. For example, the derivative formula for $\fe{f}{x}=x^2$ is $\fe{\fd{f}}{x}=2x\text{.}$ Clearly $\fd{f}$ is negative when $x$ is negative and $\fd{f}$ is positive when $x$ is positive. This tells us that $f$ is decreasing when $x$ is negative and that $f$ is increasing when $x$ is positive. This matches the behavior of the parabola $y=x^2\text{.}$

# Subsection5.11.1Exercises

The amount of time (seconds), $T\text{,}$ required for a pendulum to complete one period is a function of the pendulum's length (meters), $L\text{.}$ Specifically, $T=2\pi\sqrt{\frac{L}{g}}$ where $g$ is the gravitational acceleration constant for Earth (roughly $9.8$ ms2).

##### 1

Find $\lz{T}{L}$ after first rewriting the formula for $T$ as a constant times $\sqrt{L}\text{.}$

##### 2

The sign on $\lz{T}{L}$ is the same regardless of the value of $L\text{.}$ What is this sign and what does it tell you about the relative periods of two pendulums with different lengths?

The gravitational force (Newtons) between two objects of masses $m_1$ and $m_2$ (kg) is a function of the distance (meters) between the objects' centers of mass, $r\text{.}$ Specifically, $\fe{F}{r}=\frac{Gm_1m_2}{r^2}$ where $G$ is the universal gravitational constant (which is approximately $6.7\times10^{-11}$N m2kg2.)

##### 3

Leaving $G\text{,}$ $m_1\text{,}$ and $m_2$ as constants, find $\fe{\fd{F}}{r}$ after first rewriting the formula for $F$ as a constant times a power of $r\text{.}$

##### 4

The sign on $\fe{\fd{F}}{r}$ is the same regardless of the value of $r\text{.}$ What is this sign and what does it tell you about the effect on the gravitational force between two objects when the distance between the objects is changed?

##### 5

Find each of $\fe{\fd{F}}{1.00\times10^{12}}\text{,}$ $\fe{\fd{F}}{1.01\times10^{12}}\text{,}$ and $\fe{\fd{F}}{1.02\times10^{12}}\text{,}$ leaving $G\text{,}$ $m_1\text{,}$ and $m_2$ as constants.

##### 6

Calculate $\frac{\fe{F}{1.02\times10^{12}}-\fe{F}{1.00\times10^{12}}}{1.02\times10^{12}-1.00\times10^{12}}\text{.}$ Which of the quantities found in Exercise 5.11.1.5 comes closest to this value? Draw a sketch of $F$ and discuss why this result makes sense.