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Radical expressions and functions are commonly found in various fields of mathematics. This lesson will explore the use of radical functions and how they might be applied to a variety of everyday real-life situations, even just watching TV!

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Challenge

How to Calculate the Speed of a Car?

After becoming a subscriber of a streaming platform, Magdalena became interested in a particular crime series. While watching an episode, she found out that after an accident, police can determine the speed of a car before the driver started braking.

The speed of a car can be estimated by using the following formula.

v_{1} (d) = 30 μ d

In this formula,

v_{1} (d)

represents the speed in miles per hour,

μ

is the coefficient of friction between the road surface and the tires, and the braking distance

d

— the length of skid marks on the road — is measured in feet. Assume that

μ = 0.6 .

In another episode, criminals organized illegal drag races.

When crossing the finish line, the speed of a race car with a total weight of

3000

pounds can be modeled by the following function.

v_{2} (p) = 8 3 8 p

In this formula,

p

is the power of the engine of the vehicle in horsepower and

v_{2} (p)

is measured in miles per hour. Because both radical functions measure speed and involve roots, Magdalena does not see any difference between them.

a Determine the domains

D_{1}

and

D_{2}

of the functions.

b Compare the average rate of change of the functions over the interval

[0, 216] .

c Graph the functions. What is their end behavior?

Example

Kleiber's Law

Magdalena shared her streaming account with Kriz so that they can watch a new nature documentary. The documentary is about the work of a Swiss agricultural biologist, Max Kleiber. He researched how the mass of an animal influences the animal's metabolic rate.

Metabolic Rate
The rate at which a person or an animal burns calories to maintain its body weight.

Kleiber's Law states that an animal's metabolic rate

R (w)

can be modeled by the following radical function.

R (w) = 70 4 w^{3}

In this function,

w

is the mass of the animal in kilograms and

R (w)

is measured in kilocalories per day.

a Kriz has a couple of pets. Help Kriz calculate their and their pets' metabolic rates. Each mass is given in the table. Round to

2

decimals if needed and write the metabolic rates in increasing order.

Mammal	Mass
Mouse	$20$ g
Kriz	$81$ kg
Horse	$625$ kg

b Consider two animals with weights

w_{1}

and

w_{2}

such that the first animal metabolizes

64

times more calories per day than the second animal. How many times heavier than the second animal is the first animal?

Hint

a Substitute each mass into the given formula. Use the

n^{th}

Roots of

n^{th}

Powers Property to evaluate the metabolic rate.

b Write the equation for the metabolic rate of a heavier animal. Substitute the formula stated in Kleiber's Law to solve for the weight.

Solution

a The metabolic rate of Kriz and their animals can be calculated using the function stated in Kleiber's Law. Notice that not all masses are given in kilograms. The mass of a mouse can be written in kilograms by using the conversion factor

1 g = 0.001 kg .

w_{mouse} = 20 g = 0.02 kg

Now, substitute

w = 0.02

into

R (w)

to find the metabolic rate of a mouse.

R (w) = 70 4 w^{3}

Substitute

$w = 0.02$

R (0.02) = 70 4 0.02^{3}

▼

Evaluate right-hand side

CalcPow

Calculate power

R (0.02) = 70 4 0.000008

UseCalc

Use a calculator

R (0.02) = 3.722807 \dots

RoundDec

Round to $2$ decimal place(s)

R (0.02) \approx 3.72

Next, Kriz's metabolic rate will be calculated. To do so, substitute

w = 81

into

R (w) .

The

n^{th}

Roots of

n^{th}

Powers Property will be used to evaluate the rate.

R (w) = 70 4 w^{3}

Substitute

$w = 81$

R (81) = 70 4 81^{3}

▼

Evaluate right-hand side

WritePow

Write as a power

R (81) = 70 4 (3^{4})^{3}

$(a^{m})^{n} = (a^{n})^{m}$

R (81) = 70 4 (3^{3})^{4}

$4 a^{4} = ∣ a ∣$

R (81) = 70 ∣ ∣ ∣ 3^{3} ∣ ∣ ∣

CalcPow

Calculate power

R (81) = 70 ∣ 27 ∣

AbsPos

$∣ 27 ∣ = 27$

R (81) = 70 (27)

Multiply

R (81) = 1890

The same process can be used to calculate the metabolic rate of the horse.

R (625) = 70 4 625^{3} = 8750

Finally, the results can be summarized in a table.

Mammal	Mass $(kg)$	Metabolic Rate $(kcal / day)$
Mouse	$0.02$	$\approx 3.72$
Kriz	$81$	$1890$
Horse	$625$	$8750$

b It is given that the metabolic rate

R_{1}

of a heavier animal is

64

times the metabolic rate

R_{2} .

R_{1} = 64 R_{2}

To determine the number of times the animal is heavier, the given formula for the metabolic rate will be substituted into the equation.

R_{1} = 64 R_{2} ⇓ 70 4 w_{1}^{3} = 64 (70 4 w_{2}^{3})

Now, solve the obtained equation for

w_{1} .

70 4 w_{1}^{3} = 64 (70 4 w_{2}^{3})

▼

Solve for

w_{1}

Multiply

70 4 w_{1}^{3} = 4480 4 w_{2}^{3}

DivEqn

$LHS / 70 = RHS / 70$

4 w_{1}^{3} = 64 4 w_{2}^{3}

RaiseEqn

$LHS^{4} = RHS^{4}$

(4 w_{1}^{3})^{4} = (64 4 w_{2}^{3})^{4}

PowProdII

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

(4 w_{1}^{3})^{4} = 6 4^{4} (4 w_{2}^{3})^{4}

$(n a)^{n} = a$

w_{1}^{3} = 6 4^{4} (w_{2}^{3})

RadicalEqn

$3 LHS = 3 RHS$

3 w_{1}^{3} = 3 6 4^{4} (w_{2}^{3})

RootProd

$3 a \cdot b = 3 a \cdot 3 b$

3 w_{1}^{3} = 3 6 4^{4} \cdot 3 w_{2}^{3}

RootPowToNumber

$3 a^{3} = a$

w_{1} = 3 6 4^{4} \cdot w_{2}

▼

Evaluate right-hand side

RootToPowD

$n a = a^{\frac{1}{n}}$

w_{1} = (6 4^{4})^{\frac{1}{3}} \cdot w_{2}

$(a^{m})^{n} = (a^{n})^{m}$

w_{1} = (6 4^{\frac{1}{3}})^{4} \cdot w_{2}

PowToRootD

$a^{\frac{1}{n}} = n a$

w_{1} = (364)^{4} \cdot w_{2}

CalcRoot

Calculate root

w_{1} = 4^{4} \cdot w_{2}

CalcPowProd

Calculate power and product

w_{1} = 256 w_{2}

This means that an animal that metabolizes

64

times more calories per day is

256

times heavier.

Example

A Ball Thrown Vertically

While watching a basketball game on the streaming platform, Kriz started thinking about the height of a thrown ball and how it relates to how much time passed since the throw.

They once heard that the relationship is given by a quadratic function. However, Kriz wants to check whether this is also the case for a vertical throw. The following radical equation models how much time passed since the vertical throw.

t = \frac{1}{3 2} (v + v^{2} + 64 (s - h))

In this formula,

v

is the vertical velocity in feet per second,

h

is the corresponding height of the ball, and the ball is thrown from a height of

s .

Both

s

and

h

are measured in feet.

a Help Kriz derive the formula for the height of the ball to check if it is modeled by a quadratic function.

b Assume that Kriz throws a ball from a height of

6

feet with an initial vertical velocity of

32

feet per second. What will the height of the ball be after

2

seconds? Calculate the difference between the maximum and the obtained height.

Hint

a Solve the given equation for

h .

Start by isolating the square root. Then, square the equation.

b Use the formula written in Part A. To evaluate the maximum height, find the

x -

coordinate of the vertex of the corresponding parabola.

Solution

a The formula for the height of a ball can be derived by solving the given equation for

h .

To do so, start by isolating the square root.

t = \frac{1}{3 2} (v + v^{2} + 64 (s - h))

▼

Rewrite

MultEqn

$LHS \cdot 32 = RHS \cdot 32$

32 t = 32 (\frac{1}{3 2}) (v + v^{2} + 64 (s - h))

DenomMultFracToNumber

$32 \cdot \frac{a}{3 2} = a$

32 t = 1 (v + v^{2} + 64 (s - h))

IdPropMult

Identity Property of Multiplication

32 t = v + v^{2} + 64 (s - h)

SubEqn

$LHS - v = RHS - v$

32 t - v = v^{2} + 64 (s - h)

RearrangeEqn

Rearrange equation

v^{2} + 64 (s - h) = 32 t - v

Both sides of the equation can now be squared to solve for

h .

v^{2} + 64 (s - h) = 32 t - v

RaiseEqn

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

(v^{2} + 64 (s - h))^{2} = (32 t - v)^{2}

▼

Solve for

h

PowSqrt

$(a)^{2} = a$

v^{2} + 64 (s - h) = (32 t - v)^{2}

Distr

Distribute $64$

v^{2} + 64 s - 64 h = (32 t - v)^{2}

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

v^{2} + 64 s - 64 h = (32 t)^{2} - 2 (32 t) (v) + v^{2}

SubEqn

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

64 s - 64 h = (32 t)^{2} - 2 (32 t) (v)

PowProdII

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

64 s - 64 h = 3 2^{2} t^{2} - 2 (32 t) (v)

CalcPowProd

Calculate power and product

64 s - 64 h = 1024 t^{2} - 64 v t

SubEqn

$LHS - 64 s = RHS - 64 s$

- 64 h = 1024 t^{2} - 64 v t - 64 s

DivEqn

$LHS / (- 64) = RHS / (- 64)$

h = \frac{1}{- 6 4} (1024 t^{2} - 64 v t - 64 s)

Distr

Distribute $\frac{1}{- 6 4}$

h = - 16 t^{2} + v t + s

It has been calculated that

h = - 16 t^{2} + v t + s .

This means that the height of the ball is indeed modeled by a quadratic function.

b To calculate the ball's height

2

seconds after the throw, the formula derived in Part A will be used. It is given that the ball is thrown from a height of

6

feet with an initial velocity of

32

feet per second. Therefore,

s = 6

and

v = 32

will be substituted into the formula.

h = - 16 t^{2} + v t + s ⇓ h = - 16 t^{2} + 32 t + 6

Now, substitute

t = 2

into the obtained quadratic equation to determine the desired height.

h = - 16 t^{2} + 32 t + 6

Substitute

$t = 2$

h = - 16 (2)^{2} + 32 (2) + 6

▼

Evaluate right-hand side

CalcPow

Calculate power

h = - 16 (4) + 32 (2) + 6

Multiply

h = - 64 + 64 + 6

AddTerms

Add terms

h = 6

After

2

seconds, the ball will be at a height of

6

feet. This is the height at which the ball was initially thrown. Now, the maximum height of the ball will be found. Notice that the function is already written in standard form.

h = - 16 t^{2} + 32 t + 6

To calculate the maximum value of the function, start by determining the

x -

coordinate of the vertex of the corresponding of parabola.

x = - \frac{b}{2 a}

SubstituteII

$a = - 16$ , $b = 32$

x = - \frac{3 2}{2 ( - 1 6 )}

▼

MultPosNeg

$a (- b) = - a \cdot b$

x = - \frac{3 2}{- 3 2}

RemoveNegFracAndDenom

$- \frac{a}{- b} = \frac{a}{b}$

x = \frac{3 2}{3 2}

QuotOne

$\frac{a}{a} = 1$

x = 1

The

x -

coordinate of the vertex is

1 .

The maximum value

h_{max}

can be calculated by substituting

t = 1

into the function rule.

h_{max} = - 16 t^{2} + 32 t + 6

Substitute

$t = 1$

h_{max} = - 16 (1)^{2} + 32 (1) + 6

▼

Evaluate right-hand side

BaseOne

$1^{a} = 1$

h_{max} = - 16 (1) + 32 (1) + 6

IdPropMult

Identity Property of Multiplication

h_{max} = - 16 + 32 + 6

AddTerms

Add terms

h_{max} = 22

Finally, the difference between the maximum height and the height after

2

seconds will be calculated.

h_{max} - h = 22 - 6 = 16

Example

A Quarter Marathon

Magdalena and Kriz see a news story about the quarter marathon they ran last weekend. The distance of the marathon was

10

kilometers. Participants were divided into groups starting at equal intervals.

Magdalena and Kriz had different strategies for how to adjust their running pace to their physical condition. Magdalena's distance covered after

t

minutes can be modeled by the following linear function.

m (t) = \frac{1}{5} t

In this function,

m (t)

is given in kilometers. On the other hand, Kriz's progress after

t

minutes is modeled by a square root function.

k (t) = 2 t - 16

Similarly,

k (t)

is measured in kilometers. For the following questions, assume that the marathon starts at

t = 0 .

a What was the delay between Magdalena's and Kriz's start times?

b At what time(s) did Magdalena and Kriz run side by side? Solve the corresponding system of equations by graphing.

Hint

a Determine at what values of

t

Magdalena and Kriz start the marathon.

b The desired time is given by the values of

t

such that the distances covered by Magdalena and Kriz are equal. Solve the equation by graphing.

Solution

a It is given that the marathon started at

t = 0 .

Notice that when

t

is greater than

0,

the distance covered by Magdalena was also greater than

0 .

t > 0 \Rightarrow \frac{1}{5} t > 0

This means that Magdalena started the race in the first group. In Kriz's case, the distance covered is given by a square root function. Since the distance

k (t) = 2 t - 16

cannot be an imaginary number, this function is defined for values of

t

such that the radicand is non-negative.

2 t - 16 \geq 0 \Leftrightarrow t \geq 8

Because the function is not defined for values of

t

less than

8,

Kriz must not have started the race yet. Moreover, the square root of a positive number is also positive.

t > 8 \Rightarrow 2 t - 16 > 0

This means that Kriz started running

8

minutes after Magdalena, at

t = 8 .

Consequently, the delay was

8

minutes.

b Recall that Magdalena and Kriz ran in the same marathon. Since the delay between their starts was found in Part A, it can be assumed that both given functions share the same timeline

t .

Consider the functions modeling the distance each student ran one more time.

m (t) k (t) = \frac{1}{5} t = 2 t - 16

Any times at which Magdalena and Kriz ran side by side will be given by the values of

t

such that their covered distances are equal.

m (t) = k (t) ⇓ \frac{1}{5} t = 2 t - 16

The combined equation can be solved by graphing. Start with the linear function. To do so, the function rule must be written in slope-intercept form. Then, the slope

m

and

y -

intercept

b

can be easily identified.

y = m t + b ⇓ y = \frac{1}{5} t + 0

Now, the obtained equation can be graphed by plotting its

y -

intercept. Then, the slope will be used to determine another point that satisfies the equation, and the points will be connected with a line.

Next, the second equation $y = 2 t - 16$ will be graphed in the same coordinate plane by using a table of values. Remember to substitute $t \geq 8$ so that the square root can be calculated right away.

$t$	$2 t - 16$	$y = 2 t - 16$
$8$	$2 (8) - 16$	$0 = 0$
$10$	$2 (10) - 16$	$4 = 2$
$16$	$2 (16) - 16$	$16 = 4$
$26$	$2 (26) - 16$	$36 = 6$
$40$	$2 (40) - 16$	$64 = 8$
$58$	$2 (58) - 16$	$100 = 10$

All these points will be plotted and connected with a smooth curve.

According to the graph, Magdalena ran the first

2

kilometers in

10

minutes, while Kriz ran the first

2

kilometers in

2

minutes. They met

10

minutes from the start of the marathon. After this time, Kriz was in front of Magdalena for a while.

t = 10 \Rightarrow m (t) = k (t)

However, Kriz's rhythm was declining while Magdalena's remained constant. Then,

40

minutes after the start of the race, Magdalena caught up with Kriz and from then on, she took the lead until the end of the marathon. Nevertheless, notice that both ran the marathon in

50

minutes.

t = 40 \Rightarrow m (t) = k (t)

It can be concluded that the distances covered by Magdalena and Kriz were equal at

t = 10

and

t = 40 .

This means that they ran side by side

10

minutes and

40

minutes after the start of the marathon.

Example

Period of a Pendulum

A mathematical pendulum is an idealized model of a simple pendulum. It consists of an object suspended from a fixed point with a string. The string cannot be stretched. Consider a pendulum with a period of

0.5

seconds.

The period

T

of a mathematical pendulum, the time it takes to complete one full cycle in seconds, can be approximated for small swings by the following square root function.

T = 2 π \frac{L}{g}

Here,

L

is the length in of the pendulum in feet and

g

is gravity acceleration in feet per second squared. The pendulum that Kriz built has a length of

3

feet.

a Assume that

g

32

feet per second squared and

π \approx 3.14 .

Write the function for the period of Kriz's pendulum depending on the number of feet

x

by which the pendulum is lengthened. Round the constant to

2

decimals if needed.

b Graph the obtained function. Determine the domain and range of the function. Choose the correct answer.

c How does the value of

x

influence the pendulum's period?

Hint

a What is the length of Kriz's pendulum after it is lengthened by

x ?

b Make a table of values to graph the function. Keep in mind that a real square root is defined for a non-negative radicand.

c What is the end behavior of the obtained function?

Solution

a It is given that Kriz's pendulum has a length of

3

feet. Because the pendulum is about to be lengthened by

x,

the total length is given by the following sum.

L = 3 + x

To write the desired function, substitute

3 + x

for

L

into the given function for the period of a pendulum.

T = 2 π \frac{L}{g} ⇓ T = 2 π \frac{3 + x}{g}

Next, substitute

g = 32

and

π \approx 3.14

to simplify the function rule.

T = 2 π \frac{3 + x}{g}

Substitute

$g = 32$

T = 2 π \frac{3 + x}{3 2}

$π \approx 3.14$

T = 2 (3.14) \frac{3 + x}{3 2}

▼

Simplify right-hand side

Multiply

T = 6.28 \cdot \frac{3 + x}{3 2}

SqrtQuot

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

T = 6.28 \cdot \frac{3 + x}{3 2}

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

T

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Catch-Up and Review

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution