Calculus question I need help with:
You have 2 adults. Each year, 1 adult can build 1 house. Each year, every 2 adults make 6 children. Children mature in 6 years. What is the rate of change of house building?
So far, I've got:
The number of children born in a year is the number of adults times three. C(n) = 3 * A(n)
The number of children that mature in a year is the number of children born six years agao. G(n) = C(n-6)
The number of adults in a given year is the number of adults in the previous year plus how many children mature. A(n) = A(n-1) + G(n)
The number of houses built in a year equals the number of adults. H(n) = A(n)
I need the first derivative of the number of houses being built. H'(n)
So:
C(n) = 3 * A(n)
G(n) = C(n-6)
H(n) = A(n)
/ 0 where n < 0.
A(n) = | 2 where 0 <= n < 6.
\ A(n-1) + (3 * A(n-6)) where n >= 6.
That's all well and good, but I have no clue how to take the derivative of a recursive function, so I need a way to make that explicit. I do not know how to do this. Any help?
So far, I've got:
The number of children born in a year is the number of adults times three. C(n) = 3 * A(n)
The number of children that mature in a year is the number of children born six years agao. G(n) = C(n-6)
The number of adults in a given year is the number of adults in the previous year plus how many children mature. A(n) = A(n-1) + G(n)
The number of houses built in a year equals the number of adults. H(n) = A(n)
I need the first derivative of the number of houses being built. H'(n)
So:
C(n) = 3 * A(n)
G(n) = C(n-6)
H(n) = A(n)
/ 0 where n < 0.
A(n) = | 2 where 0 <= n < 6.
\ A(n-1) + (3 * A(n-6)) where n >= 6.
That's all well and good, but I have no clue how to take the derivative of a recursive function, so I need a way to make that explicit. I do not know how to do this. Any help?