Question

biologists have proposed a cubic polynomial to model the length l of alaskan rockfish at age a: l = 0.0155a^3 - 0.372a^2 + 3.95a + 1.21, where l is measured in inches and a in years. calculate $\frac{dl}{da}\big|_{a = 11}$. (round your answer to three decimal places.) $\frac{dl}{da}\big|_{a = 11}=$ in/yr interpret your answer. a 11 - year old rock fish shrinks at a rate of 1.393 in/yr. a 11 - year old rock fish shrinks at a rate of 40.739 in/yr. a 11 - year old rock fish grows at a rate of 1.393 in/yr. a 11 - year old rock fish grows at a rate of 39.346 in/yr. a 11 - year old rock fish grows at a rate of 40.739 in/yr.

Explanation:

Step1: Differentiate the length - function

Using the power - rule $\frac{d}{dA}(x^n)=nx^{n - 1}$, if $L = 0.0155A^{3}-0.372A^{2}+3.95A + 1.21$, then $\frac{dL}{dA}=0.0155\times3A^{2}-0.372\times2A + 3.95$.
So, $\frac{dL}{dA}=0.0465A^{2}-0.744A + 3.95$.

Step2: Evaluate at $A = 11$

Substitute $A = 11$ into $\frac{dL}{dA}$:
$\frac{dL}{dA}\big|_{A = 11}=0.0465\times(11)^{2}-0.744\times11 + 3.95$.
First, calculate $0.0465\times(11)^{2}=0.0465\times121 = 5.6265$.
Second, calculate $0.744\times11 = 8.184$.
Then, $\frac{dL}{dA}\big|_{A = 11}=5.6265-8.184 + 3.95$.
$5.6265-8.184+3.95=(5.6265 + 3.95)-8.184=9.5765 - 8.184=1.3925\approx1.393$.
Since $\frac{dL}{dA}\big|_{A = 11}=1.393>0$, it means the length of the 11 - year - old rockfish is increasing.

Answer:

$\frac{dL}{dA}\big|_{A = 11}=1.393$ in/yr
A 11 - year old rock fish grows at a rate of 1.393 in/yr.