Calculus problem about population count

[munkhuu1]

Homework Statement

Calculus question about tag and recapture?
Here is the problem, i tried to do most of it but don't understand what its asking on c and d.
Tag and recapture is used to estimate populations of animals in the wild. First, sample of animals are captured, number=n. They are tagged and released back into the wild. Sometime later, another sample of animals are captured, number=s. Of the s animals in the second sample t are found to be tagged. The estimate of total animal N is ofund from N/n=s/t.
Supposed n=100, s=60, t=15. In the second sample 1/4 of the animals were tagged.
a) what is the total animal population based on these results.
i got 400
b) what is dN/dt of n=100 s=60 t=15
i got -26.67 idk what the negative is saying. i said N=ns/t and dN/dt= -st/t^2 i hope its right.
c) what is the differential change in N if one more animal had been captured in the second sample and it was found to be tagged. express your answer in whole animals. I don't really understand the equation so i don't know what to do.
i just substitued the 100 for n and 61 for s and 16 for t for the dN/dt= -st/t^2 equation and got -23.83
d) what is the the differential change in N if one more animal had been captured in the second sample and it was not found to be tagged. This is almost the same as c so i don't know.
Please help me atleast understand what c and d is asking. and how to start it.

 

[HallsofIvy]

Homework Statement

Calculus question about tag and recapture?
Here is the problem, i tried to do most of it but don't understand what its asking on c and d.
Tag and recapture is used to estimate populations of animals in the wild. First, sample of animals are captured, number=n. They are tagged and released back into the wild. Sometime later, another sample of animals are captured, number=s. Of the s animals in the second sample t are found to be tagged. The estimate of total animal N is ofund from N/n=s/t.
Supposed n=100, s=60, t=15. In the second sample 1/4 of the animals were tagged.
a) what is the total animal population based on these results.
i got 400

Yes, that is correct.

b) what is dN/dt of n=100 s=60 t=15
i got -26.67 idk what the negative is saying. i said N=ns/t and dN/dt= -st/t^2 i hope its right.

You mean, I think, dN/dt= -sn/t^2. Now, evaluate that at n=100, s= 60, t= 15.

c) what is the differential change in N if one more animal had been captured in the second sample and it was found to be tagged. express your answer in whole animals. I don't really understand the equation so i don't know what to do.
i just substitued the 100 for n and 61 for s and 16 for t for the dN/dt= -st/t^2 equation and got -23.83

"s" was the number of animals in the second saple and t was the number tagged.
"if one more animal had been captured in the second sample and it was found to be tagged"
So replace s with s+ 1 and t with t+ 1. That is, s= 61 and t= 16, just as you say.

d) what is the the differential change in N if one more animal had been captured in the second sample and it was not found to be tagged. This is almost the same as c so i don't know.

Yes, it is the same. Now s= s+1= 61 but t remains 15.

Please help me atleast understand what c and d is asking. and how to start it.

 

[munkhuu1]

thank you very much, but just few questions. Why is the answer for c and d are negative? and when it says express your answer in whole animals, i subtract 23.83 from 26.67 and get 2.84. Is the answer 2 or 3?

 

[HallsofIvy]

(c) and (d) ask for derivatives- the rate at which the populations are changing. If those derivatives are negative, it means the populations are declining. "2.84" is closer to 2 than to three so would be rounded up to 3.