Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.3 Exercises: 86

Answer

The rate at which the population of bacteria is growing when $t = 2$ is $31.5501$ bacteria/hour.

Work Step by Step

Rewrite $P(t)$: $P(t) = 500 + \frac{2000t}{50 + t^{2}}$ Find the derivative of $P(t)$ using the Quotient Rule: $P'(t) = \frac{(2000)(50 + t^2) - (2000t)(2t)}{(50 + t^{2})^{2}}$ Find $P'(2)$ to find the rate when $t = 2$: $P'(2) = \frac{(2000)(50 + (2)^2) - (2000(2))(2(2))}{(50 + (2)^{2})^{2}}$ $= 31.5501$ bacteria/hour
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