The growth of stray cat population, n(t) in a small town can be modeled as n(t) = no e 0.15 t where t is time in years.
Given that in year 1995, the town had 20 stray cats.
a) Find the projected population after 10 years
b) Find the number of years required for the stray-cat population to reach 400.
Assuming the given expression is n(t) = n(0) * e^(0.15t),
Given:
In the year 1995, t = 0
n(0) = 20
a) t = 10
n(t) = n(0) * e^(0.15t)
= 20 * e^(0.15*10)
= 20 * e^(1.5)
b)
n(t) = 400
n(t) = n(0) * e^(0.15t)
= 20 * e^(0.15*t)
400 = 20 * e^(0.15*t)
Solve for t.
To find the projected population after 10 years, we can substitute the value of t = 10 into the equation n(t) = no e^(0.15t).
a) projected population after 10 years:
n(10) = 20e^(0.15 * 10)
= 20e^(1.5)
≈ 20 * 4.48168907
≈ 89.63
Therefore, the projected population after 10 years is approximately 89.63 stray cats.
To find the number of years required for the stray-cat population to reach 400, we can substitute the value of n(t) = 400 into the equation n(t) = no e^(0.15t) and solve for t.
b) number of years required for the stray-cat population to reach 400:
400 = 20e^(0.15t)
To solve for t, we need to isolate the variable t. Divide both sides of the equation by 20:
400/20 = e^(0.15t)
Simplify the left side:
20 = e^(0.15t)
To eliminate the exponent, take the natural logarithm (ln) of both sides:
ln(20) = ln(e^(0.15t))
By the property of logarithms, the exponent can be brought down:
ln(20) = 0.15t * ln(e)
Since ln(e) = 1, we have:
ln(20) = 0.15t
Now, divide both sides of the equation by 0.15 to solve for t:
t = ln(20) / 0.15
≈ 9.038
Therefore, it would take approximately 9.038 years for the stray-cat population to reach 400.