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Kruka [31]
3 years ago
15

Brain weight B as a function of body weight W in fish has been modeled by the power function B = 0.007W2/3, where B and W are me

asured in grams. A model for body weight as a function of body length L (measured in centimeters) is W = 0.12L2.53. If, over 10 million years, the average length of a certain species of fish evolved from 14 cm to 23 cm at a constant rate, how fast was this species' brain growing when the average length was 19 cm? (Round your answer to four significant figures.)
Mathematics
1 answer:
Norma-Jean [14]3 years ago
6 0

Answer:

\frac{\delta B}{\delta t} = 1.952\ x\ 10^{-8}\ g/y

Step-by-step explanation:

We need to know how fast the brain of the species grows at the point where its average length was 19 cm.

In other words we need to find:

\frac{\delta B}{\delta t}

B = 0.007W^{\frac{2}{3}}

\frac{\delta B}{\delta t} = 0.007({\frac{2}{3}})W^{-\frac{1}{3}}(\frac{\delta W}{\delta t})

Now we need to find \frac{\delta W}{\delta t}

In the statement of the problem it is said that \frac{\delta L}{\delta t}  is constant.

It is also said that the length changed from 14 to 23 cm in 10 ^ 7 years.

So:

\frac{\delta L}{\delta t} = \frac{23-14}{10^7}

\frac{\delta L}{\delta t} = \frac{9}{10^7}

Now we find \frac{\delta W}{\delta t}

\frac{\delta W}{\delta t} = 0.12(2.53)L^{1.53}(\frac{\delta L}{\delta t})\\\\\frac{\delta W}{\delta t} = 0.12(2.53)L^{1.53}(\frac{9}{10^7})\\

Now we find W and \frac{\delta W}{\delta t} for L = 19

W = 0.12(19)^{2.53}\\\\W = 206.27

\frac{\delta W}{\delta t} = 0.12(2.53)(19)^{1.53}(\frac{9}{10^7})\\\\\frac{\delta W}{\delta t} = 2.4719\ x\ 10^{-5}

Now replace  \frac{\delta W}{\delta t} and W in the main equation of \frac{\delta B}{\delta t}

\frac{\delta B}{\delta t} = 0.007({\frac{2}{3}})(206.27)^{-\frac{1}{3}}(2.4719\ x\ 10^{-5})\\\\\frac{\delta B}{\delta t} = 1.952\ x\ 10^{-8}\ g/y

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