Question

The length of some fish are modeled by a von Bertalanffy growth function. For Pacific halibut, this function has the form L(t) = 200(1 – 0.956e–0.18t ) where L(t) is the length (in centimeters) of a fish t years old.
(a) Find the rate of change of the length as a function of time
(b) In this part, give you answer to the nearest unit. At what rate is the fish growing at age:______.
t= 0 cm/yr
t= 1 cm/yr
t= 6 cm/yr
(c) When will the fish be growing at a rate of 6 cm/yr? (nearest unit)

Answer 1

Answer:

a) L'(t) = 34.416*e^(-0.18*t)

b) L'(0) = 34 cm/yr , L'(1) =29 cm/yr , L'(6) =12 cm/yr

c) t = 10 year                                          

Step-by-step explanation:

Given:

- The length of fish grows with time. It is modeled by the relation:

                                   L(t) = 200*(1-0.956*e^(-0.18*t))

Where,

L: Is length in centimeter of a fist

t: Is the age of the fish in years.

Find:

(a) Find the rate of change of the length as a function of time

(b) In this part, give you answer to the nearest unit. At what rate is the fish growing at age: t = 0 , t = 1, t = 6

c) When will the fish be growing at a rate of 6 cm/yr? (nearest unit)

Solution:

- The rate of change of length of a fish as it ages each year  can be evaluated by taking a derivative of the Length L(t) function with respect to x. As follows:

                             dL(t)/dt = d(200*(1-0.956*e^(-0.18*t))) / dt

                             dL(t)/dt = 34.416*e^(-0.18*t)

- Then use the above relation to compute:

                            L'(t) = 34.416*e^(-0.18*t)

                            L'(0) = 34.416*e^(-0.18*0) = 34 cm/yr

                            L'(1) = 34.416*e^(-0.18*1) = 29 cm/yr

                            L'(6) = 34.416*e^(-0.18*6) = 12 cm/yr

- Next, again use the derived L'(t) to determine the year when fish is growing at a rate of 6 cm/yr:

                             6 cm/yr = 34.416*e^(-0.18*t)

                             e^(0.18*t) = 34.416 / 6

                             0.18*t = Ln(34.416/6)

                             t = Ln(34.416/6) / 0.18

                             t = 10 year