a slow growing population of animals increases with time (years). the size of the population is given by P(t)=100+0.6t^2-0.004t^3. how fast is the population changing after 10 years?
clue: "how fast is ... changing" means you want the derivative.
dP/dt = 1.2t - 0.012t^2
So, plug in t=10
If you haven't yet covered derivatives, you must have studied an approximation method, such as
[f(10+h)-f(10)]/h
for small values of h.
can someone please help
To find out how fast the population is changing after 10 years, we need to calculate the derivative of the population function with respect to time (t). The derivative will give us the rate at which the population is changing at any given time.
The population function is given by:
P(t) = 100 + 0.6t^2 - 0.004t^3
To calculate the derivative, we can differentiate each term of the function separately:
dP/dt = d(100)/dt + d(0.6t^2)/dt - d(0.004t^3)/dt
The derivative of a constant term (in this case, 100) is zero, so the first term becomes zero.
dP/dt = 0 + d(0.6t^2)/dt - d(0.004t^3)/dt
Now, let's differentiate the second and third terms using the power rule:
For the term 0.6t^2, the derivative is:
d(0.6t^2)/dt = 2 * 0.6 * t^(2-1) = 1.2t
For the term -0.004t^3, the derivative is:
d(-0.004t^3)/dt = 3 * -0.004 * t^(3-1) = -0.012t^2
Substituting these derivatives back into the equation:
dP/dt = 0 + 1.2t - (-0.012t^2)
Simplifying, we have:
dP/dt = 1.2t + 0.012t^2
Now, we can substitute t = 10 years into the derivative equation to find the rate at which the population is changing after 10 years:
dP/dt = 1.2(10) + 0.012(10)^2
= 12 + 1.2(100)
= 12 + 120
= 132
Therefore, the population is changing at a rate of 132 animals per year after 10 years.