Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 12

Suppose that the relationships of the temperature $T$ in degrees celcius and the power input $w$ in watts of a certain device is given by $T(w) = 0.1 w^2 + 2.155w + 20$

(a) How much power is needed to maintain the temperature at $200^0 C$

$200 = 0.1 w^2 + 2.155w + 20$

$0.1w^2 + 2.155w - 180 = 0$

Using quadratic formula

$\displaystyle w = \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}$

$\displaystyle w = \frac{-2.155 \pm \sqrt{(2.155)^2 - 4 (0.1)(-180)}}{2(0.1)}$

$w = 33 watts$ and $w = -54.55 \, watts$

We only choose $w = 33 watts$ because there is no such negative power.

(b) If the temperature is allowed to vary from $200^0 C$ by up to $\pm 1^0 C$, what range of wattage is allowed for the input power?

$T(w) = 0.1w^2 + 2.155w + 20$

$0.1w_1^2 + 2.155w_1 + 20 = 200 + 1$

$0.1w_1^2 + 2.155w_1 - 181 = 0$

Using Quadratic Formula

$w_1 = 33.1124 \, watts$

$T(w) = 0.1w^2 + 2.155w + 20$

$0.1w_2^2 + 2.155w_2 + 20 = 200 - 1$

$0.1w_2^2 + 2.155w_2 - 179 = 0$

Using Quadratic Formula

$w_2 = 32.8839$ watts

The allowed wattage should be the closer to the ideal value which is 33.1124 watts and the tolerance can be computed as 33.1124 - 33 = 0.1124 watts.
Therefore, in order to fit in the tolerance of $\pm 1^{\circ} C$ in temperature, the allowed range of input power should be $\pm 0.112$ watts withinn the ideal value 33 watts.
(c) In terms if the $\varepsilon, \delta$ definition of $\lim \limits_{x \to a} f(x) = L$, what is $x$? What is $f(x)$?
What is $a$? What is $L$? What value of $\varepsilon$ is given? What is the corresponding value of $\delta$?


In terms of the definition of the precise limit,


$\begin{equation} \begin{aligned} & x && \text{corresponds to input power in watts}\\ & f(x) && \text{for Temperature}\\ & a && \text{ is the ideal input power of } 33 \, watts\\ & L && \text{ is the target temperature of } 200^0 C\\ & \epsilon && \text{ corresponds to the tolerance } \pm 1^0 C \text{ in temperature}\\ & \delta && \text{ is for the allowed range of input power } \pm 0.112 \, watts \end{aligned} \end{equation}$