Respuesta :
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
Based on the given function, the growth rate of the fish reduces as the fish
gets older.
(a) The rate of change of the length as a function of time, is presented as
follows;
- [tex]\underline{\dfrac{L(t) }{dt} =34.416\cdot e^{-0.18 \cdot t }}[/tex]
(b) The rate the fish is growing at age;
- t = 0 is approximately 34 cm/yr.
- t = 1 is approximately 29 cm/yr.
- t = 6 is approximately 12 cm/yr.
(c) The fish will be growing at 6 cm/yr at approximately 10 years of age
Reasons:
The given function for the length of the fish is [tex]L(t) = 200 \cdot \left (1 - 0.956 \cdot e^{-0.18 \cdot t } \right)[/tex]
L(t) = The length of the fish
t = The age of the fish in years
(a) Required:
The rate of change of the length as a function of time
The rate of change of the length as a function of time is given by the
derivative of the given function.
[tex]\dfrac{d}{dt} \left( 200 \cdot \left (1 - 0.956 \cdot e^{-0.18 \cdot t } \right) \right) = \dfrac{34.416\cdot e^{0.18 \cdot t }}{\left(e^{0.18 \cdot t }\right)^2} = 34.416\cdot e^{-0.18 \cdot t }[/tex]
The rate of change of the length as a function of time, [tex]L'(t)[/tex], is therefore;
[tex]L'(t) = 34.416\cdot e^{-0.18 \cdot t }[/tex]
(b) At t = 0, [tex]L'(0) = 34.416\cdot e^{-0.18 \times 0}[/tex] = 34.416
- The rate at which the fish is growing at age t = 0 is approximately 34 cm/yr.
At t = 1, [tex]L'(1) = 34.416\cdot e^{-0.18 \times 1}[/tex] ≈ 28.75 cm/yr
- The rate at which the fish is growing at age t = 1 is approximately 29 cm/yr.
At t = 6, [tex]L'(6) = 34.416\cdot e^{-0.18 \times 6}[/tex] ≈ 11.69 cm/yr
- The rate at which the fish is growing at age t = 6 is approximately 12 cm/yr.
(c) The age at which the fish will be growing at 6 cm/yr. is given as follows;
[tex]L'(t) = 34.416\cdot e^{-0.18 \cdot t } = 6[/tex]
Therefore;
- [tex]t = \dfrac{ln \left(\dfrac{6}{34.416} \right)}{-0.18} \approx 9.7[/tex]
Which gives
The age at which the fish will be growing at 6 cm/yr is when it is approximately 10 years old.
Learn more here:
https://brainly.com/question/14810533
https://brainly.com/question/17364771
https://brainly.com/question/1866154
