NZ Level 7 (NZC) Level 2 (NCEA)

Applications of Power Functions

Lesson

Recall that a power function is a function expressible as $f\left(x\right)=ax^r$`f`(`x`)=`a``x``r` 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 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 $d^3\propto p^2$`d`3∝`p`2.

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`×`p`23 .

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)×`p`23

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

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.

Since $d^3\propto p^2$`d`3∝`p`2, we could just as easily made the period $p$`p` the subject, so that for a different constant of variation, say $k_2$`k`2, we have $p=k_2\times d^{\frac{3}{2}}=k_2\times d\sqrt{d}$`p`=`k`2×`d`32=`k`2×`d`√`d`.

Using the same strategy above, this new constant would have the approximate value of $k_2=1.996\times10^{-10}$`k`2=1.996×10−10, so that the rearranged law would given as:

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

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`=(1.996×10−10)×(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.

The surface area of skin on a human body can be approximated by the equation $A=0.007184h^{0.725}w^{0.425}$`A`=0.007184`h`0.725`w`0.425, where $A$`A` is the skin area in m^{2} 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.

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.

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.

Using $k$

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

Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs

Apply graphical methods in solving problems