# For a population of female African Elephants, the weight w(t) (in kilograms) at age t (in years)...

## Question:

For a population of female African Elephants, the weight w(t) (in kilograms) at age t (in years) may be approximated by a Von Bathanlanffy growth function w such that {eq}W(t) = 2600(1 - 0.51e^{-0.0075t})^3 {/eq}.

a. Approximate the weight and the rate of growth of a newborn. (You may use a calculator to assist you in the solution.)

b. Assuming that an adult female weighs 1800 kg, estimate her age and rate of growth at present.

## Functions:

Regardless of whether function is defined by an exponential expression or polynomial expression, the output of a function can be determined for an input value by replacing the appropriate input variable with an input value before simplifying the resulting expression.

## Answer and Explanation:

Given: {eq}W(t) = 2600(1-0.51e^{-0.0075t})^3 {/eq}

a. In this situation, the strategy is assume the newborn has {eq}t {/eq} value of {eq}0 {/eq} for the approximation. To approximate the weight of a newborn, the strategy is to substitute {eq}0 {/eq} for {eq}t {/eq} in the expression defining {eq}W(t) {/eq} before simplifying the resulting expression.

{eq}\begin{align*} \text{ Weight of a Newborn } &= W(0) \\ &= 2600(1-0.51e^{-0.0075(0)})^3 \\ &= 2600(1-0.51e^{0})^3 \\ &= 2600(1-0.51(1))^3 \\ &= 2600(1-0.51)^3 \\ &= 2600(0.49)^3 \\ &= 2600(0.117649) \\ &= 305.8874 \text{ kilograms } \\ \end{align*} {/eq}

To approximate the rate of growth of a newborn, the strategy is to take the derivative of {eq}W(t) {/eq} before evaluating the resulting expression at {eq}t = 0 {/eq}.

{eq}\begin{align*} W'(t) &= \frac{d}{dt}(W(t)) \\ &= \frac{d}{dt}(2600(1-0.51e^{-0.0075t})^3) \\ &= (\frac{d}{dt}(2600(1-0.51e^{-0.0075t})^3))(\frac{d}{dt}(1-0.51e^{-0.0075t})) &\text{ [Chain Rule]} \\ &= (3\cdot 2600(1-0.51e^{-0.0075t})^{3-1})(0-0.51e^{-0.0075t}\cdot -0.0075) &\text{ [Chain Rule applied to Last Term in Second Group]} \\ &= (7800(1-0.51e^{-0.0075t})^2)(0.003825e^{-0.0075t}) \\ &= 29.835e^{-0.0075t}(1-0.51e^{-0.0075t})^2 \\ \end{align*} {/eq}

At {eq}t = 0 {/eq}:

{eq}\begin{align*} \text{ Rate of Growth of a Newborn } &= W'(0) \\ &= 29.835e^{-0.0075(0)}(1-0.51e^{-0.0075(0)})^2 \\ &= 29.835e^{0}(1-0.51e^{0})^2 \\ &= 29.835(1)(1-0.51(1))^2 \\ &= 29.835(1-0.51)^2 \\ &= 29.835(0.49)^2 \\ &= 29.835(0.0.2401) \\ &= 7.1633835 \text{ kilograms per year } \\ \end{align*} {/eq}

b. In order to estimate the age of the the female elephant when she weighs {eq}1800 {/eq} kilograms is to replace {eq}W(t) {/eq} with {eq}1800 {/eq} and solve for the unknown {eq}t {/eq} value.

{eq}\begin{align*} W(t) = 1800 &\Rightarrow 2600(1-0.51e^{-0.0075t})^3 = 1800 \\ &\Rightarrow 2600(1-0.51e^{-0.0075t})^3\div 2600 = 1800\div 2600 \\ &\Rightarrow (1-0.51e^{-0.0075t})^3 = 0.6923076923 \\ &\Rightarrow \sqrt[3]{(1-0.51e^{-0.0075t})^3} = \sqrt[3]{0.6923076923} \\ &\Rightarrow 1-0.51e^{-0.0075t} = 0.8846396193 \\ &\Rightarrow 1-0.51e^{-0.0075t}-1 = 0.8846396193-1 \\ &\Rightarrow -0.51e^{-0.0075t} = -0.1153603807 \\ &\Rightarrow -0.51e^{-0.0075t}\div -0.51 = -0.1153603807\div -0.51 \\ &\Rightarrow e^{-0.0075t} = 0.2261968249 \\ &\Rightarrow \ln(e^{-0.0075t}) = \ln(0.2261968249) \\ &\Rightarrow -0.0075t = -1.486349752 \\ &\Rightarrow -0.0075t\div -0.0075 = -1.486349752\div -0.0075 \\ &\Rightarrow t = 198.179967 \end{align*} {/eq}

Therefore, the age of the adult female elephant is approximately {eq}198 {/eq} years.

From part a's derivative expression found and the adult female's age, the strategy is to evaluate the rate of growth equation at the adult's female age found above.

At {eq}t = 0 {/eq}:

{eq}\begin{align*} \text{ Rate of Growth of Adult Female Elephant } &= W'(198) \\ &= 29.835e^{-0.0075(198)}(1-0.51e^{-0.0075(198)})^2 \\ &= 29.835e^{-1.485}(1-0.51e^{-1.485})^2 \\ &= 29.835(0.2265023407)(1-0.51(0.2265023407))^2 \\ &= 6.757697334(1-0.1155161938)^2 \\ &= 6.757697334(0.8844838062)^2 \\ &= 6.757697334(0.7823116034) \\ &= 5.286625037 \text{ kilograms per year } \\ \end{align*} {/eq}