# 007B Sample Midterm 3, Problem 3 Detailed Solution

For a fish that starts life with a length of 1cm and has a maximum length of 30cm, the von Bertalanffy growth model predicts that the growth rate is  ${\displaystyle 29e^{-t}}$  cm/year where  ${\displaystyle t}$  is the age of the fish. What is the average length of the fish over its first five years?

Background Information:
The average value of a function  ${\displaystyle f(x)}$  on an interval  ${\displaystyle [a,b]}$  is given by
${\displaystyle f_{\text{avg}}={\frac {1}{b-a}}\int _{a}^{b}f(x)~dx.}$

Solution:

Step 1:
Let  ${\displaystyle L(t)}$  be the length of the fish at age  ${\displaystyle t.}$
Given the information in the problem, we know
${\displaystyle L'(t)=29e^{-t}}$  and  ${\displaystyle L(0)=1.}$
First, we find  ${\displaystyle L(t).}$
We have

${\displaystyle {\begin{array}{rcl}\displaystyle {L(t)}&=&\displaystyle {\int L'(t)~dt}\\&&\\&=&\displaystyle {\int 29e^{-t}~dt}\\&&\\&=&\displaystyle {-29e^{-t}+C.}\end{array}}}$

Step 2:
Next, we need to solve for  ${\displaystyle C.}$
We have
${\displaystyle {\begin{array}{rcl}\displaystyle {1}&=&\displaystyle {L(0)}\\&&\\&=&\displaystyle {-29e^{0}+C}\\&&\\&=&\displaystyle {-29+C.}\end{array}}}$
So, we get  ${\displaystyle C=30.}$
Therefore,

${\displaystyle L(t)=-29e^{-t}+30.}$

Step 3:
Finally, we need to find the average value of  ${\displaystyle L(t)}$  over the interval  ${\displaystyle [0,5].}$
We have
${\displaystyle {\begin{array}{rcl}\displaystyle {L_{\text{avg}}}&=&\displaystyle {{\frac {1}{5-0}}\int _{0}^{5}L(t)~dt}\\&&\\&=&\displaystyle {{\frac {1}{5}}\int _{0}^{5}-29e^{-t}+30~dt}\\&&\\&=&\displaystyle {{\frac {1}{5}}(29e^{-t}+30t){\bigg |}_{0}^{5}}\\&&\\&=&\displaystyle {{\frac {1}{5}}(29e^{-5}+30(5))-{\frac {1}{5}}(29e^{0}+0)}\\&&\\&=&\displaystyle {{\frac {1}{5}}(29e^{-5}+121){\text{ cm}}.}\end{array}}}$

${\displaystyle {\frac {1}{5}}(29e^{-5}+121){\text{ cm}}}$