## Calculus: Early Transcendentals 8th Edition

$$\frac{dL}{dA}|_{A=12}=1.718(inches/year)$$ $\frac{dL}{dA}|_{A=12}$ denotes the rate of growth of the rockfish at year / age 12. At age 12, the rate of growth of the rockfish is $1.718inches/year$.
$$L=0.0155A^3-0.372A^2+3.95A+1.21$$ 1) Find the derivative of $L$ as $A$ is the variable $$\frac{dL}{dA}=\frac{d}{dA}(0.0155A^3)-\frac{d}{dA}(0.372A^2)+\frac{d}{dA}(3.95A)+\frac{d}{dA}(1.21)$$ $$\frac{dL}{dA}=(0.0155\times3A^2)-(0.372\times2A)+(3.95\times1)+0$$ $$\frac{dL}{dA}=0.0465A^2-0.744A+3.95$$ $\frac{dL}{dA}|_{A=12}$ denotes the derivative when $A=12$. Therefore, $$\frac{dL}{dA}|_{A=12}=0.0465\times12^2-0.744\times12+3.95$$ $$\frac{dL}{dA}|_{A=12}=6.696-8.928+3.95$$ $$\frac{dL}{dA}|_{A=12}=1.718(inches/year)$$ 2) The derivative of a function can be seen as the rate of change of that function. Therefore, the derivative of the length function $L$ would be the rate of change of the length of the rockfish. In other words, the function $\frac{dL}{dA}$ in facts measures the rate of growth of the rockfish over the years (as A is measured in years). That means $\frac{dL}{dA}|_{A=12}$ denotes the rate of growth of the rockfish at year / age 12. At age 12, the rate of growth of the rockfish is $1.718inches/year$.