   SEARCH HOME Math Central Quandaries & Queries  Question from Ashley, a student: The period T of a pendulum is given in terms of its length, l, by T=2pi sqrt(l/g) where g is the acceleration due to gravity(a constant) a. find dT/dl b. what is the sign of dT/dl c. what does the sign of dT/dl tell you about the period of the pendulums? Hi Ashley,

The equation

$T = 2 \pi \sqrt{\frac{l}{g}}$

describes the motion of a pendulum in a vacuum where there is no air resistance or friction. $l$ is the length of the pendulum, $g$ is a constant which is the acceleration due to gravity and $T,$ the period is the time it takes the pendulum to swing from one side to the other and then back to where is started. This period depends on the length $l$ of the pendulum and that is what this problem is about. If you increase the length of the pendulum does the period increase or decrease?

Use you knowledge of differentiation find $\large \frac{dT}{dl}$ remembering that $g$ is a constant. $l, g$ and $\pi$ are all positive so is $\large \frac{dT}{dl}$ positive or negative? What does the sign of the derivative $\large \frac{dT}{dl}$ tell you about $T$ as a function of $l?$

Penny     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.