topic badge
India
Class XI

Applications of Power Functions

Lesson

Recall that a power function is a function expressible as $f\left(x\right)=ax^r$f(x)=axr where $r$r is a real number. 

We encounter many laws of science expressible as power functions, and one famous one was discovered by Johannes Kepler (1571-1630).

Kepler's third law

Kepler's third law of planetary motion states that the time taken for a planet to orbit the Sun (known as the planets period) is related to its distance away from the Sun.

Specifically, if $p$p is the period of the planet and $d$d is the average distance away from the Sun, then Kepler's 3rd law states that $d^3\propto p^2$d3p2.

By taking cube roots, and introducing a constant of variation $k$k,  the law can be mathematically written as $d=k\times p^{\frac{2}{3}}$d=k×p23 .

Finding k

We can find the constant of variation by considering the data for the planet we live on.

Assuming that the period of Earth is given by $p=365.25$p=365.25 Earth days, and that its distance from the Sun is given by  $d=1.496\times10^8$d=1.496×108 km, then we can write:

$d$d $=$= $k\times p^{\frac{2}{3}}$k×p23
$1.496\times10^8$1.496×108 $=$= $k\times365.25^{\frac{2}{3}}$k×365.2523
$\therefore k$k $=$= $\frac{1.496\times10^8}{365.25^{\frac{2}{3}}}$1.496×108365.2523
  $=$= $2.928\times10^7$2.928×107
     

So, based on the Earth's period and radial distance, we have developed a power model for other planets.

Kepler's 3rd law becomes:

 $d=\left(2.928\times10^7\right)\times p^{\frac{2}{3}}$d=(2.928×107)×p23

where $d$d is the mean distance from the Sun in km and $p$p is the period of the orbit.

 

The Mars-Sun distance

We now can predict the distance Mars' is away from our Sun.  

Mars has a period of $687$687 Earth days and so using this number, our function predicts  $d=\left(2.928\times10^7\right)\times687^{\frac{2}{3}}=2.2797\times10^8$d=(2.928×107)×68723=2.2797×108 km. 

From the internet, Mars has a mean distance from the Sun of $2.279\times10^8$2.279×108 km, so the prediction is quite close.

 

Alternative approach to the 3rd law

Since $d^3\propto p^2$d3p2, we could just as easily made the period $p$p the subject, so that for a different constant of variation, say $k_2$k2,  we have $p=k_2\times d^{\frac{3}{2}}=k_2\times d\sqrt{d}$p=k2×d32=k2×dd

Using the same strategy above, this new constant would have the approximate value of $k_2=1.996\times10^{-10}$k2=1.996×1010, so that the rearranged law would given as:

$p=\left(1.996\times10^{-10}\right)\times d\sqrt{d}$p=(1.996×1010)×dd.

 

Period of Venus

Considering Venus this time, with a average distance from the Sun of $1.082\times10^8$1.082×108 km, we would estimate the orbital period as:

$p=\left(1.996\times10^{-10}\right)\times\left(1.082\times10^8\right)\sqrt{\left(1.082\times10^8\right)}$p=(1.996×1010)×(1.082×108)(1.082×108)

When simplified, we find that $p=224.6$p=224.6 days, which is close to the accepted value of $225$225 days. 

 

 

Worked Examples

QUESTION 1

The surface area of skin on a human body can be approximated by the equation $A=0.007184h^{0.725}w^{0.425}$A=0.007184h0.725w0.425, where $A$A is the skin area in m2 of a person who is $h$h cm tall, and weighs $w$w kg.

In total, $\frac{2}{5}$25 of Britney’s skin is covered in sun spots and sun damage.

  1. Given that Britney weighs $60$60 kg and is $152$152 cm tall, determine the total surface area of her skin that is covered in sun spots and sun damage.

    Give your answer correct to three decimal places.

QUESTION 2

A scientist named Kepler found that the number of days $T$T a planet takes to complete one full revolution around the sun (period of orbit) is related to its distance $D$D (measured in millions of kilometres) from the sun. He found for a given planet, the square of its period of orbit is proportional to the cube of its average distance from the sun.

  1. Using $k$k as the constant of variation, form an equation for $T$T in terms of $D$D.

  2. Mercury is an average $57.9$57.9 million km away from the sun, and takes $88$88 days to orbit the Sun.

    Solve for the value of $k$k to four decimal places.

  3. Saturn is an average of $1427$1427 million km from the Sun. Using the rounded value of $k$k found in the previous part, determine the period of orbit, $T$T, of Saturn.

    Give your answer to the nearest number of days.

Outcomes

11.SF.RF.2

Definition of relation, pictorial diagrams, domain, co-domain and range of a relation. Function as a special kind of relation from one set to another. Pictorial representation of a function, domain, co-domain and range of a function. Real valued function of the real variable, domain and range of these functions, constant, identity, polynomial, rational, modulus, signum and greatest integer functions with their graphs. Sum, difference, product and quotients of functions.

What is Mathspace

About Mathspace