• March 17th 2013, 07:59 PM
munkhuu
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 dont really understand the equation so i dont 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 dont know.
• March 18th 2013, 05:00 AM
majamin
Quote:

Originally Posted by munkhuu
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 dont really understand the equation so i dont 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 dont know.

a) Correct
b) Correct, but you typed in the wrong expression for the derivative (you probably have the right one)
c) The differential of N is $\Delta N = N(t+1) - N(t) = \frac{n(s+1)}{t + 1} - \frac{ns}{t} = \ldots$
d) Just like c) except now the second sample tagged number remains the same, whilst s increases by one.
• March 18th 2013, 07:18 AM
munkhuu
Hey. I just ask one more thing bout different topic. How do u find direction if gradient is given. For example gradient is <12,20,30> and i need to find which directoo. Its going do i find the unit vector or smthn?
• March 18th 2013, 08:08 AM
majamin