Rate of change of fish population.

[1irishman]

Homework Statement

The population of a species of fish is given by the equation below. Where t is time in years. What is the rate of change of the fish population after 5 years?

Homework Equations

P(t)= 1.2 x 10^8(1+0.02t)^ -2 where t is time in years

The Attempt at a Solution

Before i go further i just want to know if i am setting it up right from the beginning, here is how i started:

P'(t) = 1.2x10^8(1/sqrt of 1+0.02t)
= 1.2x10^8/sqrt of 1+0.02t

 

[tiny-tim]

hi 1irishman!

(have a square-root; √ and try using the X² icon just above the Reply box )

P'(t) = 1.2x10^8(1/sqrt of 1+0.02t)
= 1.2x10^8/sqrt of 1+0.02t

no, use the https://www.physicsforums.com/library.php?do=view_item&itemid=353" (or a substitution)

 

[1irishman]

that big sci fi constant in front of the brackets messes me up when i try to use the chain rule for some reason, did i do this right to start?
(-2)(1.2x10^8)(1+0.02t)^-3(0.02)

 

[tiny-tim]

(-2)(1.2x10⁸)(1+0.02t)^-3(0.02) ?

yes, that's correct

 

[1irishman]

okay, i think i might be doing the order wrong though from here, here is what i got next:

P'(t) = -2.4x10^8(0.02)(1+0.02t)^-3
= -4.8x10^6(1+0.02t)^-3 this doesn't look right to me somehow?

 

[tiny-tim]

this doesn't look right to me somehow?

why??

 

[1irishman]

okay, i think i might be doing the order wrong though from here, here is what i got next:

P'(t) = -2.4x10^8(0.02)(1+0.02t)^-3
= -4.8x10^6(1+0.02t)^-3 this doesn't look right to me somehow?

Actually, i take that back as it appears when i plug the 5 years into the expression it works out to be the correct answer. so thank you all is well.