Chapter D, Problem 12E

(a).

To determine

To calculate:

Power is required to maintain the temperature. $200°C$ .

Answer to Problem 12E

Power supply required to maintain the temperature $200°C$ is $w_o=32.9984watt$ .

Explanation of Solution

Given information:

  $T\left(w\right)=0.1w^2+2.155w+20$

Calculation:

Suppose in case $T$ be the temperature in degree Celsius and $w$ be the power input in watts.

Relationship of $T$ , $T\left(w\right)=0.1w^2+2.155w+20$ …(1)

For solving $T\left(w\right)=200$ ,

By using graphic utility.

Now,

  $\begin{array}{l}T\left(w\right)=200\\ 0.1w^2+2.155w+20=200\end{array}$

So here plot the functions $T\left(w\right)=0.1w^2+2.155w+20$ and $T=200$ is given as:

  

From the graph it is observed that both $T\left(w\right)=0.1w^2+2.155w+20$ and $T=200$ intersect at the point.

Since the $x-$ coordinate of the point of intersection gives the solution of $T\left(w\right)=200$ .

Thus, the power supply required to maintain the temperature $200°C$ is $w_o=32.9984watt$ .

(b)

To determine

To calculate:

Range of wattage is allowed for input power when temperature vary up to $\pm 1°C$ .

Answer to Problem 12E

Variation of wattage is allowed up to $\pm 0.1116watt$ .

Explanation of Solution

Given information:

  $T\left(w\right)=0.1w^2+2.155w+20$

Calculation:

Let’s assume temperature allowed to vary from $200°C$ up to $\pm 1°C$ .

Then one required to find the tolerance $\delta >0$ in power supply such that,

  $\left|w-w_0\right|<\delta$ implies $\left|T\left(w\right)-200\right|<1$

We have,

  $\begin{array}{l}\left|T\left(w\right)-200\right|<1\\ -1<T\left(w\right)-200<200+1\\ 199<T\left(w\right)<201\end{array}$

So here we have drawn horizontal lines $T=199,T=201$ along with the curve $T\left(w\right)$ as given,

  

From the graph, we can see that in case one restrict the values of $w$ between $32.885$ and $33.11$ , then graph of (1) lies fully between $T=199,T=201$ .

The distance of exact power $w_0=32.9984$ from $32.885$ and $33.11$ are

  $\begin{array}{l}33.1100-32.9984=0.1116\\ 32.9984-32.8850=0.1134\end{array}$

In case we select $\delta =0.1116$ then,

  $\left|w-w_0\right|<\delta$ implies $\left|T\left(w\right)-200\right|<1$

Thus, variation of wattage is allowed up to $\pm 0.1116watt$ .

(c)

To determine

To calculate:

Value of $\epsilon ,\delta$ and $a,L$ .

Answer to Problem 12E

Given value of $\epsilon =1$ and the corresponding $\delta =0.1116$ .

Explanation of Solution

Given information:

  $T\left(w\right)=0.1w^2+2.155w+20$

Calculation:

In terms of $\epsilon ,\delta$ definition of limit $\underset{x\to a}{\lim }f\left(x\right)=L$

Here $\begin{array}{l}T=200\\ x=w\\ f\left(x\right)=T\left(w\right)\\ \delta =0.1116\end{array}$

The $x$ represents the power supply, the function $f\left(x\right)$ represents the relationship between temperature and wattage $w$ , the $a$ represents the ideal power which produce $200°C$ .

The $L$ represents the temperature $200°C$ .

The given value of $\epsilon =1$ and the corresponding $\delta =0.1116$ .