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
