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 \(9.8\) ^{m}⁄_{s2}).

# 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

##### 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 m2}⁄_{kg2}.)

##### 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.