Question
Asked Dec 17, 2019
131 views

The rate of growth of the fish population was modeled by the equation G(t) = (60,000e^-0.6t)/(1+5e^-0.6t)^2, where t is measured in years since 2000 and G in kilograms per year. If the biomass was 25,000 kg in the year 2000, what is the predicted biomass for the year 2020?

check_circle

Expert Answer

Step 1

Given:

help_outline

Image Transcriptionclose

-0.6f G(t)= (1+ Se*«) -0.61 where t is measured in years since 2000. And the biomass was 25,000kg in the year 2000.

fullscreen
Step 2

The net change from 2000 to 2020 is,

help_outline

Image Transcriptionclose

60,000e-0.61 dt (1+5eo) 20 20 0.6f (1+5e061 Let u =1+5e -0.61 0.61 dt → du = -3e Then t =0 implies, u =1+ 5(1) = 6 t = 20 implies, u=1+5(e1?) =1+3.0721061×10¯³ z 1

fullscreen
Step 3

Therefore...

help_outline

Image Transcriptionclose

20 (60,000e -0s") -0.61 60,000 dt du 3 = + Se-0.61 60,000 iu²du 3 6. = -20,000|u du = 20,000[ 1" ), J6 = 20,000 6. 1 z 16,666 kg

fullscreen

Want to see the full answer?

See Solution

Check out a sample Q&A here.

Want to see this answer and more?

Solutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*

See Solution
*Response times may vary by subject and question.
Tagged in

Math

Calculus