Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 34

At what rate was this species brain growing when the average length was 18cm?
Given:
$
\begin{array}{c}
B & = & 0.007 W^\frac{2}{3}\\
W & = & 0.12 L^{2.53}
\end{array}
$
where
$
\begin{array}{c}
B & = & \text{brain weight in grams}\\
W & = & \text{body weight in grams}\\
L & = & \text{body length in cm}
\end{array}
$


Required: $\displaystyle \frac{dB}{dt}$, when $L = 18$cm

Getting the equation of brain weigh in terms of its body lengthwe get
$\displaystyle B = 0.007 \left[ 0.12 L^{2.53}\right]^{\frac{2}{3}} \qquad \Longleftarrow \text{ Equation 1}$
Getting the derivative with respect to time

$
\begin{equation}
\begin{aligned}
\frac{dB}{dt} &= \frac{2}{3} (0.007) \left[ 0.12 L^{2.53}\right]^{\frac{-1}{3}} \left( (0.12) (2.53) L^{(2.53-1)} \right) \left(\frac{dL}{dt}\right)\\
\\
\frac{dB}{dt} &= 0.0028724 \left( \frac{dL}{dt} \right) \qquad \Longleftarrow \text{ Equation 2}
\end{aligned}
\end{equation}
$


To get the value of $\displaystyle \frac{dL}{dt}$ we use the values given that the fish evolved from 15cm to 20cm for over 10 million years.
$\displaystyle \frac{d}{dt} = \frac{20-15}{10} = 0.5\frac{\text{cm}}{\text{million years}}$

Now, to get the unknown we use Equation 2
$\displaystyle \frac{dB}{dt} = 0.0028724 (18) 0.68667 (0.5)$
$\boxed{\displaystyle \frac{dB}{dt} = 0.0104513 \frac{\text{grams}}{\text{million years}}}$

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