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Average rate of change of an exponential population model

A population of bacteria is modeled by P(t)=500(1.03)tP(t) = 500(1.03)^t, where tt is measured in hours and P(t)P(t) is the number of bacteria.

Find the average rate of change of P(t)=500(1.03)tP(t) = 500(1.03)^t on the interval [0,10][0, 10]. This corresponds to the slope of the secant line through the points (0,P(0))(0, P(0)) and (10,P(10))(10, P(10)) on the graph of PP. What is this average rate of change, in bacteria per hour?

Hint

The average rate of change on [a,b][a,b] is P(b)−P(a)b−a\dfrac{P(b) - P(a)}{b - a}. Compute P(0)P(0) and P(10)P(10) first, then divide by the length of the interval.

Step-by-step solution

The average rate of change on [0,10][0,10] is given by the slope formula P(10)−P(0)10−0\dfrac{P(10) - P(0)}{10 - 0}, which is exactly the slope of the secant line connecting (0,P(0))(0, P(0)) and (10,P(10))(10, P(10)) on the graph of PP.

Evaluate P(0)=500(1.03)0=500P(0) = 500(1.03)^0 = 500 and P(10)=500(1.03)10≈671.958P(10) = 500(1.03)^{10} \approx 671.958. Then the average rate of change is 671.958−50010≈17.196\dfrac{671.958 - 500}{10} \approx 17.196 bacteria per hour.

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