# Math Help - rates of change

1. ## rates of change

With length, $l$, in meters, the period $T$, in seconds, of a pendulum is given by

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

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

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

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

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

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

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

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

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

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

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

Yup. Ya got this one, too.

-Dan