# rates of change

• Nov 24th 2006, 01:24 PM
Jacobpm64
rates of change
With length, $\displaystyle l$, in meters, the period $\displaystyle T$, in seconds, of a pendulum is given by

$\displaystyle T = 2 \pi \sqrt{ \frac{l}{9.8} }.$

(a) How fast does the period increase as $\displaystyle l$ increases?
(b) Does this rate of change increase or decrease as $\displaystyle l$ increases?

I think I am correct for this one, just check please.

(a)
$\displaystyle \frac{dT}{dl} = \pi ( \frac{l}{9.8} )^{- \frac{1}{2} } * ( \frac {1}{9.8} )$
$\displaystyle \frac{dT}{dl} = ( \frac{ \pi \sqrt{9.8} }{9.8 \sqrt{l} } ) \frac{m}{s}$

(b)
$\displaystyle \frac{d^2T}{dl^2} = \frac{- \pi \sqrt{9.8} }{19.6l \sqrt{l}}$ <-- negative.
Decreases
• Nov 26th 2006, 03:12 PM
topsquark
Quote:

Originally Posted by Jacobpm64
With length, $\displaystyle l$, in meters, the period $\displaystyle T$, in seconds, of a pendulum is given by

$\displaystyle T = 2 \pi \sqrt{ \frac{l}{9.8} }.$

(a) How fast does the period increase as $\displaystyle l$ increases?
(b) Does this rate of change increase or decrease as $\displaystyle l$ increases?

I think I am correct for this one, just check please.

(a)
$\displaystyle \frac{dT}{dl} = \pi ( \frac{l}{9.8} )^{- \frac{1}{2} } * ( \frac {1}{9.8} )$
$\displaystyle \frac{dT}{dl} = ( \frac{ \pi \sqrt{9.8} }{9.8 \sqrt{l} } ) \frac{m}{s}$

(b)
$\displaystyle \frac{d^2T}{dl^2} = \frac{- \pi \sqrt{9.8} }{19.6l \sqrt{l}}$ <-- negative.
Decreases

Yup. Ya got this one, too.

-Dan