# If the instantaneous rate of change of a population is #50t^2 - 100t^(3/2)# (measured in individuals per year) and the initial population is 25000 then what is the population after t years?

In order to respond to this question, you must identify a specific antiderivative.

The population is changing at a rate, which indicates that we will soon learn about the derivative, or the rate of change. Stop right there.

(Multiplying by the reciprocal is what divides.)

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The population after ( t ) years is given by the antiderivative of the rate of change function plus the initial population. Thus, the population function is:

[ P(t) = \int (50t^2 - 100t^{3/2}) dt + 25000 ]

[ P(t) = \frac{50}{3}t^3 - \frac{200}{5/2}t^{5/2} + 25000 + C ]

[ P(t) = \frac{50}{3}t^3 - \frac{400}{\sqrt{t}} + 25000 + C ]

Since the initial population is 25000, we can solve for ( C ):

[ P(0) = \frac{50}{3}(0)^3 - \frac{400}{\sqrt{0}} + 25000 + C = 25000 ]

[ 25000 = 0 - \frac{400}{0} + 25000 + C ]

[ 25000 = 25000 + C ]

[ C = 0 ]

Thus, the population function is:

[ P(t) = \frac{50}{3}t^3 - \frac{400}{\sqrt{t}} + 25000 ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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