Question

Biologists have proposed a cubic polynomial to model the length L of Alaskan rockfish at ageA:

\(L = 0.0155{A^3} - 0.372{A^2} + 3.95A + 1.21\)

where \(L\) is measured in inches and \(A\) in years. Calculate

\({\left. {\frac{{dL}}{{dA}}} \right|_{A = 12}}\)

and interpret your answer.

Short answer

• The value of\({\left. {\frac{{dL}}{{dA}}} \right|_{A = 12}}\)is \(1.718\).
• The rate of change of length of the fish at the age of 12 is equal to 1.718 inches per year.

Step-by-step solution

Find first derivative and second derivative

The given function of length is \(L = 0.0155{A^3} - 0.372{A^2} + 3.95A + 1.21\). On differentiating this function with respect to \(A\), the following expression is obtained:

\(\begin{aligned}\frac{{dL}}{{dA}} &= \frac{d}{{dA}}\left( {0.0155{A^3} - 0.372{A^2} + 3.95A + 1.21} \right)\\ &= \frac{d}{{dA}}\left( {0.0155{A^3}} \right) - \frac{d}{{dA}}\left( {0.372{A^2}} \right) + \frac{d}{{dA}}\left( {3.95A} \right) + \frac{d}{{dA}}\left( {1.21} \right)\\ &= 0.0465{A^2} - 0.744A + 3.95\end{aligned}\)

Therefore, \(\frac{{dL}}{{dA}} = 0.0465{A^2} - 0.744A + 3.95\).

Find \({\left. {\frac{{dL}}{{dA}}} \right|_{A = 12}}\)

Plug \(A = 12\)into \(\frac{{dL}}{{dA}}\) to find \({\left. {\frac{{dL}}{{dA}}} \right|_{A = 12}}\), and simplify, as shown below:

\(\begin{aligned}frac{{dL}}{{dA}} &= 0.0465{\left( {12} \right)^2} - 0.744\left( {12} \right) + 3.95\\ &= 1.718\end{aligned}\)

Thus, the value of \({\left. {\frac{{dL}}{{dA}}} \right|_{A = 12}}\) is \(1.718\).

Interpret the result

The rate of change of length of the fish at the age of 12 is equal to 1.718 inches per year