Question

brain weight b as a function of body weight w in fish has been modeled by the power function b = 0.007w^{2/3}, where b and w are measured in grams. a model for body weight as a function of body length l (measured in centimeters) is w = 0.12l^{2.53}. if, over 10 million years, the average length of a certain species of fish evolved from 15 cm to 24 cm at a constant rate, how fast (in g/yr) was this species brain growing when its average length was 19 cm? (round your answer to four significant figures.)

Explanation:

Step1: Find the derivative of $W$ with respect to $L$

Using the power - rule $\frac{d}{dx}(ax^n)=nax^{n - 1}$, for $W = 0.12L^{2.53}$, we have $\frac{dW}{dL}=0.12\times2.53L^{1.53}=0.3036L^{1.53}$.

Step2: Find the derivative of $B$ with respect to $W$

For $B = 0.007W^{\frac{2}{3}}$, using the power - rule, $\frac{dB}{dW}=0.007\times\frac{2}{3}W^{-\frac{1}{3}}=\frac{0.014}{3}W^{-\frac{1}{3}}$.

Step3: Use the chain - rule $\frac{dB}{dL}=\frac{dB}{dW}\cdot\frac{dW}{dL}$

Substitute $\frac{dB}{dW}$ and $\frac{dW}{dL}$: $\frac{dB}{dL}=\frac{0.014}{3}W^{-\frac{1}{3}}\cdot0.3036L^{1.53}$.
Substitute $W = 0.12L^{2.53}$ into the above formula: $\frac{dB}{dL}=\frac{0.014}{3}(0.12L^{2.53})^{-\frac{1}{3}}\cdot0.3036L^{1.53}$.
Simplify:
\[

$$\begin{align*} \frac{dB}{dL}&=\frac{0.014\times0.3036}{3}\times(0.12)^{-\frac{1}{3}}L^{-\frac{2.53}{3}}L^{1.53}\\ &=\frac{0.014\times0.3036}{3}\times(0.12)^{-\frac{1}{3}}L^{-\frac{2.53}{3}+1.53}\\ &=\frac{0.014\times0.3036}{3}\times(0.12)^{-\frac{1}{3}}L^{\frac{-2.53 + 4.59}{3}}\\ &=\frac{0.014\times0.3036}{3}\times(0.12)^{-\frac{1}{3}}L^{\frac{2.06}{3}} \end{align*}$$

\]

Step4: Calculate the rate of change of length $\frac{dL}{dt}$

The length changes from $L_1 = 15$ cm to $L_2 = 24$ cm over $t = 10\times10^{6}$ years. So $\frac{dL}{dt}=\frac{24 - 15}{10\times10^{6}}=\frac{9}{10\times10^{6}}=9\times10^{-7}$ cm/yr.

Step5: Use the chain - rule $\frac{dB}{dt}=\frac{dB}{dL}\cdot\frac{dL}{dt}$

When $L = 19$ cm:
First, calculate $\frac{dB}{dL}$ at $L = 19$:
\[

$$\begin{align*} \frac{dB}{dL}&=\frac{0.014\times0.3036}{3}\times(0.12)^{-\frac{1}{3}}\times19^{\frac{2.06}{3}}\\ &\approx\frac{0.014\times0.3036}{3}\times(0.12)^{-\frac{1}{3}}\times19^{0.6867}\\ \end{align*}$$

\]
$(0.12)^{-\frac{1}{3}}\approx2.02$, $19^{0.6867}\approx7.37$
\[

$$\begin{align*} \frac{dB}{dL}&=\frac{0.014\times0.3036}{3}\times2.02\times7.37\\ &=\frac{0.014\times0.3036\times2.02\times7.37}{3}\\ &\approx0.0417 \end{align*}$$

\]
Then, since $\frac{dB}{dt}=\frac{dB}{dL}\cdot\frac{dL}{dt}$, and $\frac{dL}{dt}=9\times10^{-7}$ cm/yr
$\frac{dB}{dt}=0.0417\times9\times10^{-7}\approx3.753\times10^{-8}$ g/yr

Answer:

$3.753\times 10^{-8}$ g/yr