Question

Crystal growth furnaces are used in research to determine how best to manufacture crystals used in electronic components. For proper growth of a crystal,the temperature must be controlled accurately by adjusting the input power. Suppose the relationship is given by

\(T\left( w \right) = {\bf{0}}.{\bf{1}}{w^{\bf{2}}} + {\bf{2}}.{\bf{155}}w + {\bf{20}}\)

where T is the temperature in degrees Celsius and w is the power input in watts.

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

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

(c) In terms of \(\varepsilon \), \(\delta \) definition of \(\mathop {{\bf{lim}}}\limits_{x \to a} f\left( x \right) = L\), 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 power required to maintain the temperature is 33.0 watts.

(b) The range of wattage is \(32.89 < w < 33.11\).

(c) If x represents the input power, then \(f\left( x \right)\) is the temperature. arepresents the target input power, and L represents the target temperature. \(\varepsilon \) is the tolerance in the temperature and \(\delta \) is the tolerance in power.

Step-by-step solution

Step 1:Find the amount of power

Substitute 200 for \(T\) in theequation \(T = 0.1{w^2} + 2.155w + 20\), to find the amount of power required.

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

The procedureto obtain the solution of thequadratic equation by using the online calculator (Symbolab) is shown below:

1. Open the online calculator “Symbolab”.
2. From the “Algebra” tab, select “Equations”.
3. Input the equation \(0.1{w^2} + 2.155w - 180 = 0\), and enter “GO”.

Thus, the obtained result is \(w \approx 33.0\;{\rm{watts}}\) for the temperature \(200^\circ C\).

Find the range of input power

It can be observed from the graph that \(199 \le T \le 201\) and the range of power is \(32.89 < w < 33.11\).

It is given that the variation is \( \pm 1^\circ C\).

So, substitute 201 for \(T\) in the equation\(T = 0.1{w^2} + 2.155w + 20\), to find the amount of power required.

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

The procedure to obtain the solution of the quadratic equation by using the online calculator (Symbolab) is shown below:

4. Open the online calculator “Symbolab”.

5. From the “Algebra” tab, select “Equations”.

6. Input the equation \(0.1{w^2} + 2.155w - 181 = 0\), and enter “GO”.

Thus, the obtained result is \(w \approx 33.11\;{\rm{watts}}\) for the temperature \(201^\circ C\).

Similarly, substitute 199 for \(T\) in the equation \(T = 0.1{w^2} + 2.155w + 20\), to find the amount of power required.

\(\begin{aligned}{c}0.1{w^2} + 2.155w + 20 = 199\\0.1{w^2} + 2.155w - 181 = 179\end{aligned}\)

The procedure to obtain the solution of the quadratic equationby using the online calculator (Symbolab) is shown below:

7. Open the online calculator “Symbolab”.

8. From the “Algebra” tab, select “Equations”.

9. Input the equation \(0.1{w^2} + 2.155w - 179 = 0\), and enter “GO”.

Thus, the obtained result is \(w \approx 32.89\;{\rm{watts}}\) for the temperature \(199^\circ C\).

The figure below represents the graph of temperature and power.

It can be observed from the graph that \(199 \le T \le 201\) and the range of power is \(32.89 < w < 33.11\).

Find the answer for part (c)

If x represents the input power, then \(f\left( x \right)\) is the temperature. arepresents the target input power, and L represents the target temperature. \(\varepsilon \) is the tolerance in the temperature and \(\delta \) is the tolerance in power.