Chapter 4 - Calculating the Derivative - 4.2 Derivatives of Products and Quotients - 4.2 Exercises - Page 217: 50

the total number (in millions) of bacteria present in a culture at a certain time $t$ (in hours) is given by $$ \begin{aligned} N(t)&=3t(t-10)^{2}+40 \\ &=3t(t^{2}-20t +100)+40 \\ &=3t^{3}-60t^{2} +300t+40 \end{aligned} $$ Then, we can find that, (a) $$ \begin{aligned} N^{\prime}(t) &=3 (3)t^{2}-60 (2)t +300+0,\\ &=9 t^{2} -120t +300 \end{aligned} $$ the rate at which the population of bacteria is changing at (b) 8 hours $$ \begin{aligned} N^{\prime}(8) &=9 (8)^{2} -120(8)+300\\ &=9 (64) -960+300\\ &=-84 \quad \text { million per hour} \end{aligned} $$ (c) 11 hours $$ \begin{aligned} N^{\prime}(11) &=9 (11)^{2} -120(11)+300\\ &=9 (121) -1320+300\\ &=69 \quad \text { million per hour} \end{aligned} $$

the total number (in millions) of bacteria present in a culture at a certain time $t$ (in hours) is given by $$ \begin{aligned} N(t)&=3t(t-10)^{2}+40 \\ &=3t(t^{2}-20t +100)+40 \\ &=3t^{3}-60t^{2} +300t+40 \end{aligned} $$ Then, we can find that, (a) $$ \begin{aligned} N^{\prime}(t) &=3 (3)t^{2}-60 (2)t +300+0,\\ &=9 t^{2} -120t +300 \end{aligned} $$ the rate at which the population of bacteria is changing at (b) 8 hours $$ \begin{aligned} N^{\prime}(8) &=9 (8)^{2} -120(8)+300\\ &=9 (64) -960+300\\ &=-84 \quad \text { million per hour} \end{aligned} $$ (c) 11 hours $$ \begin{aligned} N^{\prime}(11) &=9 (11)^{2} -120(11)+300\\ &=9 (121) -1320+300\\ &=69 \quad \text { million per hour} \end{aligned} $$