# A pendulum of length L feet has a period of: #T= 3.14(L^.5)/2(2^.5)# seconds. If you increase the length of a 1-foot pendulum by .01 feet, how much does its period increase?

After deriving we have

so

or also

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see below

you seem to have

but by approximation

We can plough the same furrow as Newton (and Taylor expansions), but in a way that is simpler, using a generalised binomial expansion, ie this idea

It looks just like a Taylor Expansion, ie based in Calculus. In fact I think it was Newton that generalised the binomial expansion in this way.

So

which is simply

That is a c.1/2% increase

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App. increase in period

Enjoy Mats.!

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To find the increase in period when the length of a 1-foot pendulum is increased by 0.01 feet, you can use the formula for the period of a pendulum:

T = 3.14 * (L^0.5) / (2 * (2^0.5))

First, find the period of the original pendulum with a length of 1 foot:

T1 = 3.14 * (1^0.5) / (2 * (2^0.5)) = 3.14 * (1) / (2 * 1.414) = 3.14 / 2.828 ≈ 1.113 seconds

Now, find the period of the pendulum with a length increased by 0.01 feet:

New length = 1 + 0.01 = 1.01 feet

T2 = 3.14 * (1.01^0.5) / (2 * (2^0.5)) ≈ 3.14 * 1.00498756211 / 2.828 ≈ 1.114 seconds

The increase in period is the difference between T2 and T1:

Increase in period ≈ T2 - T1 ≈ 1.114 - 1.113 ≈ 0.001 seconds

So, the period increases by approximately 0.001 seconds.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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