# Rates!!

• Apr 2nd 2008, 11:18 AM
mathlete
Rates!!
The period T, in seconds, of a pendulum is given by the equation below with length, l, in meters.

T = 2pi*sqrt(l/9.8)

(a) How fast does the period increase?

need some help please, these rate problems give me trouble...
• Apr 2nd 2008, 12:40 PM
TheEmptySet
Quote:

Originally Posted by mathlete
The period T, in seconds, of a pendulum is given by the equation below with length, l, in meters.

T = 2pi*sqrt(l/9.8)

(a) How fast does the period increase?

need some help please, these rate problems give me trouble...

With respect to what?

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

If you double the length 2l then

$T(2l)=2 \pi \sqrt{\frac{2l}{9.8}} =2 \pi \sqrt{2 \cdot \frac{l}{9.8}}=2 \pi \sqrt{2} \cdot \sqrt{\frac{l}{9.8}}$

$T(2l)=\sqrt{2} \cdot \left( 2 \pi \sqrt{\frac{l}{9.8}} \right)=\sqrt{2} \cdot T$

So doubling the length increases the period by $\sqrt{2} \approx 1.414$
• Apr 2nd 2008, 12:46 PM
mathlete
no sorry, it wants us to find dT/dl, so we're supposed to differentiate with respect to l, sorry...
• Apr 2nd 2008, 04:52 PM
TheEmptySet
Quote:

Originally Posted by mathlete
The period T, in seconds, of a pendulum is given by the equation below with length, l, in meters.

T = 2pi*sqrt(l/9.8)

(a) How fast does the period increase?

need some help please, these rate problems give me trouble...

$T=2\pi\sqrt{\frac{l}{9.8}}=\frac{2\pi}{\sqrt{9.8}} \cdot \sqrt{l}= \frac{2 \pi}{\sqrt{9.8}} \cdot l^{1/2}$

Now taking the derivative we get

$\frac{dT}{dl}=\frac{2\pi}{\sqrt{9.8}}\cdot \left( \frac{1}{2}l^{-1/2}\right)=\frac{\pi}{\sqrt{9.8l}}$