If there are 6438 fish left in the lake, and the pollution is increasing at the rate of 5.6 ppm/year. what is the rate at which the fish population of this lake is decreasing? F= 78000/(8+√ x )

Question

F= fish of a certain species in the lake

x=  parts per million (ppm)

Answer 1

I'll take a stab at this.....

First take the equation, & inputting 6438 in for F, and solving for x yields

x=16.937865718660782452279061849393 ppm

Next, find the derivative of F with respect to x

F= 78000/(8+√ x )

dF/dx=78000 * derivative of [ (8+√ x )^-1] = -39000 / (16x + √ x (x+64) )

Substituting x=16.937865718660782452279061849393 into this derivative yields

dF/dx = -39000 / [ 16 * 16.937865718660782452279061849393 + √16.937865718660782452279061849393 (16.937865718660782452279061849393+64) ]

= -39000 / [ 271.00585149857251923646498959029 + √16.937865718660782452279061849393 (80.937865718660782452279061849393} ]

= -39000 / [ 271.00585149857251923646498959029 + 333.10495341435788938421653040719 ]

= -39000 / 604.11080491293040862068151999748

= 64.55769319607023411371237458194 fish / year . This is the rate of decrease at the time when the fish is at 6438.