The population (in thousands) of the Tzitzit bird is well described by a function of the form P(t) = ae^kt, where t is the time in years and a and k are constants. If the population was 10 thousand when t-0 and 300 thousand when t=3, determine the constants a and k exactly. Then use the formula for P(t) to find the population when t=(4)
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10000 = ae^0 --> a=10000
300000 = 10000 e^3k
30 = e^3k
3k = ln30
k = 1/3 ln30
P(t) = 10000 * e^(t/3 ln30)
P(4) = 10000 * e^(4/3 ln30) = 932170
Note that since e^ln30 = 30,
P(t) = 10000 * 30^(t/3)
but to evaluate that, you need to fall back to logs anyway.
What it means is that the population grows by a factor of 30 every 3 years. But then, we knew that from the initial conditions given: P(3) = 30*P()
To determine the constants 'a' and 'k' in the function P(t) = ae^kt, we can use the given population values at two different time points.
Given:
When t = 0, P(t) = 10,000 ---- (1)
When t = 3, P(t) = 300,000 ---- (2)
Substituting these values into the function, we have two equations:
Equation (1): 10,000 = a * e^(k*0) = a * e^0 = a
Equation (2): 300,000 = a * e^(k*3)
Simplifying equation (2), we divide both sides by 'a' and substitute the value of 'a' from equation (1):
300,000/10,000 = e^(k*3)
30 = e^(3k)
To solve for 'k', we take the natural logarithm (ln) of both sides:
ln(30) = ln(e^(3k))
ln(30) = 3k * ln(e)
ln(30) = 3k
Now, divide both sides by 3:
k = ln(30) / 3
We can substitute the value of 'k' back into equation (1) to find 'a':
10,000 = a
Therefore, the constants in the function P(t) = ae^kt are:
a = 10,000
k = ln(30) / 3 ≈ 0.693
Now, to find the population when t = 4, we substitute this value into the P(t) function:
P(4) = a * e^(k*4)
P(4) = 10,000 * e^(0.693 * 4)
P(4) ≈ 10,000 * e^2.772
Using a calculator, we can evaluate the expression:
P(4) ≈ 10,000 * 15.956 ≈ 159,560
Therefore, the population when t = 4 is approximately 159,560.