Chapter 1.7, Problem 68E

To determine

To find the predicted density of the species and to identify the species that will use most energy using the given information

Answer to Problem 68E

The predicted density of the species with the mass of $1000g$ is $D=56.234perhectare$ .

The species with a mass of $100kg$ will use the most energy.

Explanation of Solution

Given information: $D=c{M}^{\frac{-3}{4}}$

Calculation:

Consider the following function:

$D=c{M}^{\frac{3}{4}}$

Here, $D$ is the population density of a species of mammal, and $M$ is body mass of the species of

mammal.

If an unlikely $1g$ mammal has a density of ${10}^{4}perhectare$ , then we have

$\begin{array}{l}D={10}^{4}perhectare\\ M=1g\end{array}$

Let us substitute ${10}^4$ for $D$ and $1$ for $M$ into equation $\left(1\right)$ , and then solve for $c$ .

  $\begin{array}{l}{10}^4=c{\left(1\right)}^{\frac{3}{4}}\\ {10}^4=c.1\\ {10}^4=c\end{array}$

When finding the predicted density of a species with a mass of $1000g$ , we will substitute $1000$ for $M$ and ${10}^4$ for $c$ into equation $\left(1\right)$

$\begin{array}{l}D=\left({10}^4\right){\left(1000\right)}^{\frac{3}{4}}\\ ={10}^4{\left({10}^3\right)}^{\frac{3}{4}}\\ ={10}^4{.10}^{\left(\frac{9}{4}\right)}\\ ={10}^{\frac{7}{4}}\\ =56.234\end{array}$

Hence, the predicted density of the species with the mass of $1000g$ is

$D=56.234perhectare$ .

The predicted density is less than the initial value of $D={10}^4$ per hectare.

Now, the total mass of the mammals is $100kg$ , so we need to first convert $100kg$ to grams.

We know that

$\begin{array}{l}1kg=1000g\\ 1=1000\frac{g}{kg}\end{array}$

Thus, we get

$\begin{array}{l}100kg=100kg\times 1\\ =100kg\times 1000\text{}\frac{g}{kg}\\ =\left(100000\right)g\\ ={10}^{5}g\end{array}$

Again, to find the predicted density of the species with a mass of ${10}^{5}g$ , we substitute ${10}^{5}g$ for $M$ and ${10}^4$ for $c$ into equation $\left(1\right)$ .

$\begin{array}{l}D={10}^4{\left({10}^5\right)}^{\frac{3}{4}}\\ ={10}^4{.10}^{\left(\frac{15}{4}\right)}\\ ={10}^{\frac{1}{4}}\\ =1.778\end{array}$

Hence, the predicted density of the species with the mass of ${10}^{5}g$ is $D=1.778perhectare$ .

The predicted density is less than the initial value, $D={10}^{4}perhectare$ .

According to the metabolic scaling law,

$E=0.018M^{0.75}\mathrm{.........}\left(2\right)$

$E$ Measured in $Kcal/hr$ and $M$ is measured in grams.

When finding the energy used by a species with a mass of $1000g$ We substitute 1,000 for $M$ into equation $\left(2\right)$

$\begin{array}{l}E=0.018{\left(1000\right)}^{0.75}\\ =0.018\times 177.282\\ =3.201\end{array}$

Hence, the energy used by a species with using down mass of $1000g$ is $3.20Kcal/hr$ .

When finding the energy used by species with mass of $100Kgor{10}^{5}g$ , we substitute ${10}^5$ for $M$ into the equation $\left(2\right)$ .

$\begin{array}{l}E=0.018{\left({10}^5\right)}^{0.75}\\ =0.018\times 5623.413\\ =101.221\end{array}$

Therefore, the energy used by a species with the mass of $1000gis101.221Kcal/hr$ .

So, according to the metabolic scaling law, the species with a mass of $100kg$ will use the most

energy.