Question

A crystal growth furnace is used in research to determine how best to manufacture crystals used in electronic components for the space shuttle. For proper growth of the crystal, the temperature must be controlled accurately by adjusting the input power. Suppose the relationship is given by

\(T\left( w \right) = 0.1{w^2} + 2.155w + 20\)

Where Tis the temperature in degrees Celsius and wis the power input in watts.

(a) How much power is needed to maintain the temperature at\(200^\circ {\rm{C}}\)?

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

(c) In terms of\(\varepsilon \), \(\delta \)the definition of \({\lim _{x \to a}}f\left( x \right) = L\), what is x? What is\(f\left( x \right)\)? What is a? What is L? What value of \(\varepsilon \)is given? What is the corresponding value of\(\delta \)?

Short answer

(a) The \(32.999\;{\rm{watt}}\) of power required to maintain the given temperature.

(b) The range of wattage allowed for the input power is\(32.884 - 33.112\;{\rm{watt}}\).

(c) The values are\(x = w\), \(f\left( x \right) = T\left( w \right)\), \(a = 32.999\), \(L = 200\), \(\varepsilon = 1\), and \(\delta = 0.114\).

Step-by-step solution

Area of the circular disc(a)

The relationship is given by,

\(T\left( w \right) = 0.1{w^2} + 2.155w + 20\)

Find the power needed to maintain the given temperature.

Substitute \(T = 200^\circ {\rm{C}}\) in the given relationship to power.

\(\begin{array}{c}200 = 0.1{w^2} + 2.155w + 20\\0.1{w^2} + 2.155w - 180 = 0\end{array}\)

Use the quadratic formula to find\(w\).

\(\begin{array}{c}w = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ = \frac{{ - 2.155 \pm \sqrt {{{\left( {2.155} \right)}^2} - 4\left( {0.1} \right)\left( { - 180} \right)} }}{{2\left( {0.1} \right)}}\\ = \frac{{ - 2.155 \pm 8.7547}}{{0.2}}\\ = 32.999\;{\rm{or}}\; - 54.549\end{array}\)

Neglect the negative value.

Therefore, the \(32.999\;{\rm{watt}}\) of power required to maintain the given temperature.

(b) Step 3: Find the variation in power

For the upper end of the range, the temperature is,

\(\begin{array}{c}T = 200 + 1\\ = 201^\circ {\rm{C}}\end{array}\)

Substitute \(T = 201^\circ {\rm{C}}\) in the given relationship to power.

\(\begin{array}{c}201 = 0.1{w^2} + 2.155w + 20\\0.1{w^2} + 2.155w - 181 = 0\end{array}\)

Use the quadratic formula to find\(w\).

\(\begin{array}{c}w = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ = \frac{{ - 2.155 \pm \sqrt {{{\left( {2.155} \right)}^2} - 4\left( {0.1} \right)\left( { - 181} \right)} }}{{2\left( {0.1} \right)}}\\ = \frac{{ - 2.155 \pm \sqrt {77.044025} }}{{0.2}}\\ = 33.112\;{\rm{or}}\; - 54.662\end{array}\)

For the lower end of the range, the temperature is,

\(\begin{array}{c}T = 200 - 1\\ = 199^\circ {\rm{C}}\end{array}\)

Substitute \(T = 199^\circ {\rm{C}}\) in the given relationship to power.

\(\begin{array}{c}199 = 0.1{w^2} + 2.155w + 20\\0.1{w^2} + 2.155w - 179 = 0\end{array}\)

Use the quadratic formula to find\(w\).

\(\begin{array}{c}w = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ = \frac{{ - 2.155 \pm \sqrt {{{\left( {2.155} \right)}^2} - 4\left( {0.1} \right)\left( { - 179} \right)} }}{{2\left( {0.1} \right)}}\\ = \frac{{ - 2.155 \pm \sqrt {76.244025} }}{{0.2}}\\ = 32.884\;{\rm{or}}\; - 54.434\end{array}\)

Therefore, the range of wattage allowed for the input power is\(32.884 - 33.112\;{\rm{watt}}\).

(c)Step 3: Find the values of x,\(f\left( x \right)\), a, L, \(\varepsilon \), and the corresponding value of \(\delta \)

From the above result, all the characteristics of\(\mathop {\lim }\limits_{w \to 32.999} T\left( w \right) = 200\). That is to say, for\(\varepsilon = 1\), there exists a \(\delta = 0.114\) such that if \(\left| {w - 32.999} \right| < 0.114\) then\(\left| {T\left( w \right) - 200} \right| < 1\).

All the variables are shown in the following table.

\(x\)	\(w\)
\(f\left( x \right)\)	\(T\left( w \right)\)
\(a\)	\(32.999\)
\(L\)	\(200\)
\(\varepsilon \)	\(1\)
\(\delta \)	\(0.114\)

Therefore, the values are\(x = w\), \(f\left( x \right) = T\left( w \right)\), \(a = 32.999\), \(L = 200\), \(\varepsilon = 1\), and \(\delta = 0.114\).