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  $29e^{-t}$  cm/year where  $t$  is the age of the fish. What is the average length of the fish over its first five years?

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

Solution:

ExpandStep 1:
Let  $L(t)$  be the length of the fish at age  $t.$
Given the information in the problem, we know
$L'(t)=29e^{-t}$  and  $L(0)=1.$
First, we find  $L(t).$
We have
${\begin{array}{rcl}\displaystyle {L(t)}&=&\displaystyle {\int L'(t)~dt}\\&&\\&=&\displaystyle {\int 29e^{-t}~dt}\\&&\\&=&\displaystyle {-29e^{-t}+C.}\end{array}}$

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

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

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

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