# The formula T= 2pi sqrt(L / 980) can be used to find the period (T in seconds, the time it takes...

## Question:

The formula T = 2pi * sqrt(L / 980) can be used to find the period (T in seconds, the time it takes a pendulum to complete one cycle) of a pendulum that is L cm long.

(a) Rewrite this formula using rational exponents. Don't forget to rationalize the denominator.

(b) Solve the formula for L.

## Rational Functions:

A function is said to be rational if it is of the form

{eq}r(x) = \frac{p(x)}{q(x)} {/eq}

where * p(x)* and

*are both polynomials. A rational expression may involve operators such as square roots, usually on terms on the denominator, and therefore for formality, we need to rationalize it. In this problem, the rational function is used to describe the motion of a pendulum of length*

**q(x)****L**, specifically its period

**T**.

## Answer and Explanation:

Given the formula {eq}T = 2\pi \sqrt{\frac{L}{980}} {/eq},

(a) We are to rewrite this formula using rational exponents. The use of rational exponent gets rid of the square root symbol but does not alter the definition of the function. Rewriting the formula gives

{eq}T = 2\pi \sqrt{\frac{L}{980}}\\ T = 2\pi \ \frac{\sqrt{L}}{\sqrt{980}}\\ \text{Rationalizing the denominator gives}\\ T = 2\pi \ \frac{\sqrt{L}}{\sqrt{980}} \cdot \frac{\sqrt{980}}{\sqrt{980}}\\ T = 2\pi \frac{\sqrt{980L}}{980}\\ T = \frac{\pi}{490} \sqrt{980L}\\ T = \frac{\pi}{490} (980L)^{\frac{1}{2}}\\ {/eq}

(b) Here, we are going to solve for L. Using the original function, we have

{eq}T= 2\pi \sqrt{\frac{L}{980}}\\ \text{Squaring both sides gives}\\ T^{2} = [ 2\pi \sqrt{\frac{L}{980}} ]^{2}\\ T^{2} = 4\pi^{2} \frac{L}{980}\\ \frac{980T^{2}}{4\pi^{2}} = L\\ \frac{245}{\pi^{2}} T^{2} = L\\ {/eq}

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