The instantaneous growth rate r of a colony of bacteria t hours after the start of an experiment is given by the function #r(t)=0.01t^3-0.07t^2+ 0.07t +0.15#. What are the times for which the instantaneous growth rate is zero?

1 Answer
Steve M
Nov 5, 2016

#t=0.569# or #4.097# (hours) to 3dp

Explanation:

We have, # r(t) = 0.01t^3 - 0.07t^2 + 0.07t + 0.15 #

The rate of growth is the derivative of #r# wrt #t#, so we have

# dotr(t) = (3)0.01t^2 - (2)0.07t^1 + (1)0.07 #
# :. dotr(t) = 0.03t^2 - 0.14t + 0.07 #

The instantaneous growth rate is zero when # dotr(t) = 0 #
# => 0.03t^2 - 0.14t^1 + 0.07 = 0 #
# :. 3t^2 - 14t + 7 = 0 #

Which we will solve using the quadratic equation, to give:
# t = (-(-14) +- sqrt((-14)^2-4(3)(7))) / ((2)(3)) #
# :. t = (14 +- sqrt(196 - 84)) / 6 #
# :. t = (14 +- sqrt(112)) / 6 #
# :. t = (14 +- sqrt(7*16)) / 6 #
# :. t = (14 +- 4sqrt(7)) / 6 #

Hence, # t=7/3+2/3sqrt7 #, #t=7/3-2/3sqrt7 #
Or, #t=0.569#, #4.097# (3dp)

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