The intrinsic rate of increase (r) for northern elephant seals about 0.15 (per annum). We will assume that population regulation is linearly density dependent.
a. If the carrying capacity of elephant seals at a recently recolonized island is 1000 adult females and the initial population size is 100, how many more years (approximately) will it take for the population to grow to K?
b. If the island population overshot carrying capacity to 1900 adult females, how many more years (approximately) will it take for the population to decline to K?
- i know were suppose to use the forumla N(t)= K/1+[(K-N(o)/N(o))]e^-rt, it said how many years it take to grow to K and i set N(t) to 1000 and plugged everything in to solve for T but when i do the math i know the answers wrong
To calculate the approximate number of years required for the northern elephant seal population to reach the carrying capacity (K) in each scenario, we'll use the formula N(t) = K / [1 + ((K - N0) / N0) * e^(-rt)], where:
N(t) represents the population size at time t,
K represents the carrying capacity,
N0 represents the initial population size, and
r represents the intrinsic rate of increase.
a. To find the number of years it takes for the population to grow to K:
Given:
K = 1000 (carrying capacity)
N0 = 100 (initial population size)
r = 0.15 (intrinsic rate of increase per annum)
Plugging these values into the formula, we get:
1000 = 1000 / [1 + ((1000 - 100) / 100) * e^(-0.15t)]
Simplifying, we have:
1 = 1 / [1 + (900 / 100) * e^(-0.15t)]
Taking the reciprocal of both sides, we get:
1 + (900 / 100) * e^(-0.15t) = 1
Simplifying further, we have:
(900 / 100) * e^(-0.15t) = 0
Since e^(-0.15t) cannot be zero, this equation has no solution. It means that the population will not reach the carrying capacity (K) in this scenario.
b. To find the number of years it takes for the population to decline to K:
Given:
K = 1000 (carrying capacity)
N0 = 1900 (initial population size)
r = 0.15 (intrinsic rate of increase per annum)
Plugging these values into the formula, we get:
1000 = 1000 / [1 + ((1000 - 1900) / 1900) * e^(-0.15t)]
Simplifying, we have:
1 = 1 / [1 + (-900 / 1900) * e^(-0.15t)]
Taking the reciprocal of both sides, we get:
1 + (-900 / 1900) * e^(-0.15t) = 1
Simplifying further, we have:
(-900 / 1900) * e^(-0.15t) = 0
Again, since e^(-0.15t) cannot be zero, this equation has no solution. It means that the population will not decline to the carrying capacity (K) in this scenario.
Therefore, according to these calculations, the population will neither grow to the carrying capacity nor decline to the carrying capacity in the given scenarios.
To solve the problems using the formula N(t) = K / (1 + [(K - N(0)) / N(0)])e^(-rt), you need to follow these steps accurately:
a. To find how many more years it will take for the population to grow to carrying capacity (K) of 1000 adult females from an initial population size of 100, you should set N(t) to 1000.
Step 1: Substitute the given values into the formula:
N(t) = 1000 (population size at time t)
N(0) = 100 (initial population size)
K = 1000 (carrying capacity)
r = 0.15 (intrinsic rate of increase)
Step 2: Solve for T (number of years it will take to reach carrying capacity).
1000 = 1000 / (1 + [(1000 - 100) / 100])e^(-0.15T)
Step 3: Simplify the equation and isolate T:
1 = 1 / (1 + [900 / 100])e^(-0.15T)
1 = 1 / (1 + 9)e^(-0.15T)
Step 4: Solve for e^(-0.15T) by multiplying both sides by (1 + 9):
1 + 9 = e^(-0.15T)
Step 5: Simplify:
10 = e^(-0.15T)
Step 6: Take the natural logarithm (ln) of both sides:
ln(10) = ln(e^(-0.15T))
Step 7: Apply the logarithmic property that ln(e^x) = x:
ln(10) = -0.15T
Step 8: Solve for T by dividing both sides by -0.15:
T = ln(10) / -0.15
To approximate the value of T, you can use a calculator to find that T is approximately 8.87 years.
b. To determine how long it will take for the population to decline to carrying capacity (K) of 1000 adult females from an initial population size of 1900, you should set N(t) to 1000.
Step 1: Substitute the given values into the formula:
N(t) = 1000 (population size at time t)
N(0) = 1900 (initial population size)
K = 1000 (carrying capacity)
r = 0.15 (intrinsic rate of increase)
Step 2: Solve for T (number of years it will take to decline to carrying capacity).
1000 = 1000 / (1 + [(1000 - 1900) / 1900])e^(-0.15T)
Step 3: Simplify the equation and isolate T:
1 = 1 / (1 + [-900 / 1900])e^(-0.15T)
1 = 1 / (1 - 0.474)e^(-0.15T)
Step 4: Solve for e^(-0.15T) by multiplying both sides by (1 - 0.474):
1 - 0.474 = e^(-0.15T)
Step 5: Simplify:
0.526 = e^(-0.15T)
Step 6: Take the natural logarithm (ln) of both sides:
ln(0.526) = ln(e^(-0.15T))
Step 7: Apply the logarithmic property that ln(e^x) = x:
ln(0.526) = -0.15T
Step 8: Solve for T by dividing both sides by -0.15:
T = ln(0.526) / -0.15
To approximate the value of T, you can use a calculator to find that T is approximately 3.58 years.