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McGraw Hill Glencoe Algebra 1, 2012 View details

4. Radical Equations

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Exercise 21 Page 644

a The time for one full swing of a simple pendulum is called the period. It is represented by a function.

t = 2 π \frac{ℓ}{3 2}

In this function,

t

is the period of the pendulum in seconds, and

ℓ

is the length of the pendulum in feet. Given that the period of the Giant Swing is approximately

8

seconds, we will find the length of the pendulum's arm. To do so, let's substitute

t = 8

into the function and solve it for

ℓ .

t = 2 π \frac{ℓ}{3 2}

Substitute

$t = 8$

8 = 2 π \frac{ℓ}{3 2}

▼

Solve for

ℓ

DivEqn

$LHS / 2 = RHS / 2$

4 = π \frac{ℓ}{3 2}

DivEqn

$LHS / π = RHS / π$

\frac{4}{π} = \frac{ℓ}{3 2}

RaiseEqn

$LHS^{2} = RHS^{2}$

(\frac{4}{π})^{2} = \frac{ℓ}{3 2}

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

\frac{4 ^{2}}{π ^{2}} = \frac{ℓ}{3 2}

CalcPow

Calculate power

\frac{1 6}{π ^{2}} = \frac{ℓ}{3 2}

MultEqn

$LHS \cdot 32 = RHS \cdot 32$

\frac{1 6}{π ^{2}} \cdot 32 = ℓ

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

\frac{1 6 \cdot 3 2}{π ^{2}} = ℓ

Multiply

\frac{5 1 2}{π ^{2}} = ℓ

RearrangeEqn

Rearrange equation

ℓ = \frac{5 1 2}{π ^{2}}

UseCalc

Use a calculator

ℓ = 51.876446 \dots

RoundInt

Round to nearest integer

ℓ = 52

The arm of the pendulum measures approximately

52

feet long.

b Consider the given function.

t = 2 π \frac{ℓ}{3 2}

As the length of a pendulum increases, the numerator of the quotient goes up. Therefore, the value of the radical increases, which implies that the period of the pendulum also increases. Let's test our argument by examining the periods of pendulums with different lengths.

$ℓ$	$2 π \frac{ℓ}{3 2}$	$t$
$52$	$2 π \frac{5 2}{3 2}$	$8$
$55$	$2 π \frac{5 5}{3 2}$	$8.24$
$60$	$2 π \frac{6 0}{3 2}$	$8.60$

As we can also see from the table, increasing the length of the pendulum increases the period.

Subchapter links

Check Your Understanding p.644

H.O.T. Problems p.645

Standardized Test Practice p.646

Skills Review p.646