Chapter 1.8, Problem 62E

a .

To determine

To find: and interpret $\left(N\circ T\right)\left(t\right)$

Answer to Problem 62E

  $\left(N\circ T\right)\left(t\right)=90t^2+60t+600$

It gives the number of bacteria in a refrigerated food, t hours after it is removed from refrigeration.

Explanation of Solution

Given: The number N of bacteria in a refrigerated food is given by $N\left(T\right)=10T^2-20T+600$ , $2\le T\le 20$ where T is the temperature of the food in degrees Celsius.

Temperature of food after it is removed from refrigeration is given by $T\left(t\right)=3t+2$ , $0\le t\le 6$ where t is the time in hours.

Calculation:

Now consider,

  $\begin{array}{c}\left(N\circ T\right)\left(t\right)=N\left(T\left(t\right)\right)\\ =N\left(3t+2\right)\\ =10{\left(3t+2\right)}^2-20\left(3t+2\right)+600\end{array}$

  $\begin{array}{c}=10\left(9t^2+12t+4\right)-60t-40+600\\ =90t^2+120t+40-60t-40+600\\ =90t^2+60t+600\end{array}$

Here $\left(N\circ T\right)\left(t\right)=N\left(t\right)$ represents the number of bacteria in a refrigerated food, t hours after it is removed from refrigeration.

b.

To determine

To find: bacteria count after 0.5 hours

Answer to Problem 62E

  $653$

Explanation of Solution

Calculation:

From part (a), the number of bacteria in a refrigerated food, t hours after it is removed from refrigeration is given by:

  $N\left(t\right)=90t^2+60t+600$ , $0\le t\le 6$

Now number of bacteria after 0.5 hours is:

  $\begin{array}{c}N\left(0.5\right)=90{\left(0.5\right)}^2+60\left(0.5\right)+600\\ =90\left(0.25\right)+60\left(0.5\right)+600\\ =22.5+30+600\\ =652.5\end{array}$

Conclusion:

So, after 0.5 hours there will be approximately 653 bacteria.

c.

To determine

To find: the time when the bacteria count reaches 1500.

Answer to Problem 62E

  $2.85\text{ hours}$

Explanation of Solution

Concept Used:

Quadratic formula: The roots of the quadratic equation $a{x}^2+bx+c=0$ are $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Calculation:

From part (a), the number of bacteria in a refrigerated food, t hours after it is removed from refrigeration is given by:

  $N\left(t\right)=90t^2+60t+600$ , $0\le t\le 6$

Now to find t for which $N\left(t\right)=1500$ ,

  $\begin{array}{c}N\left(t\right)=1500\\ 90t^2+60t+600=1500\\ 90t^2+60t-900=0\\ 3t^2+2t-30=0\end{array}$

By quadratic formula the solutions of the above equation are:

  $\begin{array}{c}t=\frac{-2\pm \sqrt{2^2-4\left(3\right)\left(-30\right)}}{2\left(3\right)}\\ =\frac{-2\pm \sqrt{4+360}}{6}\\ =\frac{-2\pm \sqrt{364}}{6}\end{array}$

  $\begin{array}{c}=\frac{-2\pm 2\sqrt{91}}{6}\\ =\frac{-1\pm \sqrt{91}}{3}\end{array}$

This gives that,

  $\begin{array}{c}{t}_1=\frac{-1-\sqrt{91}}{3}\\ \approx -3.51\end{array}$

  $\begin{array}{c}{t}_2=\frac{-1+\sqrt{91}}{3}\\ \approx 2.85\end{array}$

Since $t\in \left[0,6\right]$ , so $t=2.85$

Conclusion:

Thus, the bacteria count reaches 1500 in about 2.85 hours.