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...

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- Apr 2nd 2008, 11:18 AMmathleteRates!!
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 PMTheEmptySet
With respect to what?

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

If you double the length 2l then

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

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

So doubling the length increases the period by $\displaystyle \sqrt{2} \approx 1.414$ - Apr 2nd 2008, 12:46 PMmathlete
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 PMTheEmptySet
$\displaystyle T=2\pi\sqrt{\frac{l}{9.8}}=\frac{2\pi}{\sqrt{9.8}} \cdot \sqrt{l}= \frac{2 \pi}{\sqrt{9.8}} \cdot l^{1/2}$

Now taking the derivative we get

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