Chapter 1.8, Problem 60E

(a)

To determine

To find: The composition of $\left(N\circ T\right)\left(t\right)$ and interpret its meaning in the context.

Answer to Problem 60E

The number of bacteria in the food as the function of $t$ is $N\left(T\left(t\right)\right)=30\left(3t^2+2t+20\right)$ .

Explanation of Solution

Given:

The expression for the number of bacteria in the Food is $N\left(T\right)=10T^2-20T+600,2\le T\le 20$ .

The temperature is $T$ of the food in degree Celsius when the food is removed from the refrigeration. The temperature of the food is $T\left(t\right)=3t+2,0\le t\le 6$ .

Calculation:

Consider the function for the number of bacteria is,

  $N\left(T\right)=10T^2-20T+600,2\le T\le 20$

Consider the expression for the temperature of food is,

  $T\left(t\right)=3t+2,0\le t\le 6$

Then the number of bacteria is,

  $\begin{array}{c}N\left(T\left(t\right)\right)=10{\left(3t+2\right)}^2-20\left(3t+2\right)+600\\ =30\left(3t^2+2t+20\right)\end{array}$

Thus, the number of bacteria in the food as the function of $t$ is,

  $N\left(T\left(t\right)\right)=30\left(3t^2+2t+20\right)$

(b)

To determine

To find: The count of bacteria after $0.5$ hours.

Answer to Problem 60E

The number of bacteria is $635$ .

Explanation of Solution

Given:

The expression for the number of bacteria in the Food is $N\left(T\right)=10T^2-20T+600,2\le T\le 20$ .

The temperature is $T$ of the food in degree Celsius when the food is removed from the refrigeration. The temperature of the food is $T\left(t\right)=3t+2,0\le t\le 6$ .

Calculation:

Consider the number of bacteria in the food as the function of $t$ is,

  $N\left(T\left(t\right)\right)=30\left(3t^2+2t+20\right)$

Then, the number of bacteria after 0.5 hours is,

  $\begin{array}{c}N\left(0.5\right)=30\left(3{\left(0.5\right)}^2+2\left(0.\right)+20\right)\\ =635\end{array}$

Thus, the number of bacteria is $635$ .

(c)

To determine

To find: The timeat which the bacteria count is 1500.

Answer to Problem 60E

The time after which the count of bacteria is 1500 is after

  $2.846$ hours.

Explanation of Solution

Given:

The expression for the number of bacteria in the Food is $N\left(T\right)=10T^2-20T+600,2\le T\le 20$ .

The temperature is $T$ of the food in degree Celsius when the food is removed from the refrigeration. The temperature of the food is $T\left(t\right)=3t+2,0\le t\le 6$ .

Calculation:

Consider the number of bacteria in the food as the function of $t$ is,

  $N\left(T\left(t\right)\right)=30\left(3t^2+2t+20\right)$

Then,

  $\begin{array}{l}1500=30\left(3t^2+2t+20\right)\\ 3t^2+2t-30=0\\ t=\frac{-2\pm \sqrt{2^2+4\left(3\right)\left(30\right)}}{2\left(3\right)}\\ t=2.846\end{array}$

The time after which the count of bacteria is 1500 is after

  $2.846$ hours.