# 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?

Background 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:

Step 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

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

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