Problem 12 - Calculus: Early Transcendentals

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 $$ T(\omega) = 0.1\omega^2 + 2.155\omega + 20 $$ where \( T \) is the temperature in degrees Celsius and \( \omega \) is the power input in watts. (a) How much power is needed to maintain the temperature at \( 200^{\circ}C \)? (b) If the temperature is allowed to vary from \( 200^{\circ}C \) by up to \( \pm 1^{\circ}C \), what range of wattage is allowed for the input power? (c) In terms of the \( \varepsilon \), \( \delta \) definition of \( \displaystyle \lim_{x \to a} f(x) = L \), 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 \( \delta \)?

Step-by-Step Solution

Answer

(a) The power needed is about 33 watts. (b) The power range is 32.35 to 33.64 watts. (c) \( x = \omega \), \( f(x) = T(\omega) \), \( a = 200 \), \( L = 200 \), \( \varepsilon = 1 \), \( \delta \approx 1.145 \).

Step 1: Understanding the Equation

The equation given is \( T(\omega) = 0.1\omega^2 + 2.155\omega + 20 \), which represents the temperature \( T \) as a function of the power input \( \omega \).

Step 2: Solve for (a) Required Power for 200°C

To find the power \( \omega \) needed to maintain a temperature of 200°C, set \( T(\omega) = 200 \) and solve the equation: \[ 0.1\omega^2 + 2.155\omega + 20 = 200 \]. First, subtract 200 from both sides: \[ 0.1\omega^2 + 2.155\omega - 180 = 0 \]. Solve this quadratic equation using the quadratic formula \( \omega = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 0.1 \), \( b = 2.155 \), \( c = -180 \).

Step 3: Substitute into the Quadratic Formula

Substitute \( a = 0.1 \), \( b = 2.155 \), and \( c = -180 \) into the quadratic formula: \[ \omega = \frac{-2.155 \pm \sqrt{(2.155)^2 - 4 \times 0.1 \times (-180)}}{2 \times 0.1} \] Calculate the discriminant and simplify: \[ \omega = \frac{-2.155 \pm \sqrt{4.641025 + 72}}{0.2} \].

Step 4: Calculate the Roots

The discriminant \( \sqrt{76.641025} \approx 8.754 \). Thus, the solutions are: \[ \omega_1 = \frac{-2.155 + 8.754}{0.2} \approx 33.0 \] \[ \omega_2 = \frac{-2.155 - 8.754}{0.2} \approx -54.5 \]. Since \( \omega \) represents power, it must be positive, so \( \omega \approx 33.0 \) watts.

Step 5: Solve for (b) Power Range with Error Margin

For the temperature to vary from 199°C to 201°C, repeat steps for both: First, set \( T(\omega) = 199 \) and solve:\[ 0.1\omega^2 + 2.155\omega - 179 = 0 \] Calculate using quadratic formula. The roots are approximately \( \omega_1 = 32.35 \), then, set \( T(\omega) = 201 \) and solve:\[ 0.1\omega^2 + 2.155\omega - 181 = 0 \]. The roots are approximately \( \omega_2 = 33.64 \). Thus, the power range is 32.35 to 33.64 watts.

Step 6: Solve for (c) Limits and Definitions

Using the context of \( \varepsilon, \delta \)-definition, identifying terms: - \( x \) is the power input \( \omega \),- \( f(x) \) is \( T(\omega) \),- \( a \) is the target temperature \( 200^{\circ}C \),- \( L \) is \( 200^{\circ}C \),- \( \varepsilon = 1 \)°C (the allowed variance),- \( \delta \) corresponds to the range of \( \omega \) which is about 1.145 watts (from 32.35 to 33.64).

Key Concepts

quadratic equations

Quadratic equations are a central part of algebra and calculus, used to describe various relationships, such as the one given in the exercise where temperature relies on power input. A quadratic equation is generally expressed in the form:
\[ ax^2 + bx + c = 0 \]where:

• \( a \) is the coefficient of the quadratic term \( x^2 \),
• \( b \) is the coefficient of the linear term \( x \), and
• \( c \) is the constant term.

In our context, the equation is related to the power input \( \omega \) of a crystal growth furnace reacting quadratically to control temperature. To solve for the power when the temperature is a specific value, like 200°C, substitute the temperature into the equation to find the corresponding \( \omega \). Use the quadratic formula:
\[ \omega = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula finds the roots, or solutions, of the quadratic equation for \( \omega \). Only positive solutions are physically meaningful here since \( \omega \) – power – must be positive.

epsilon-delta definition

The \( \varepsilon, \delta \)-definition is a fundamental concept in calculus, used to rigorously define what it means for a function to approach a limit. In simple terms, this definition ensures that for any small \( \varepsilon \) margin in the function's output, there's a corresponding \( \delta \) margin in the function's input, making the function's values close to a limit within the set boundaries.
In our problem, the target temperature (the limit \( L \)) is 200°C, but the temperature can vary by \( \varepsilon = \pm 1 \)°C. Here:

• \( x \) represents the power input \( \omega \).
• \( f(x) \) is the temperature function \( T(\omega) \).
• \( a \) is the desired temperature point, 200°C.

Thus, \( \delta \) represents the acceptable range of power inputs to maintain temperatures within the allowed variation, which has been calculated as about 1.145 watts around 33 watts. This tells you the power input must stay between these values to ensure the temperature remains close to 200°C.

temperature control

Temperature control is crucial in processes sensitive to fluctuations, such as crystal growth in a furnace used for electronics. Consistent temperature ensures uniform crystal growth, affecting material properties and performance. In this exercise, the temperature control is managed by adjusting the power input to the furnace, which is modeled using a quadratic equation.
To maintain a certain temperature or allow for minimal deviations, understanding the quadratic relationship between power and temperature becomes essential. By solving the equation with small variations around the target temperature, one determines the necessary adjustments to power to stay within acceptable temperature limits. This is vital for maintaining product quality and reliability.

power input calculation

Calculating the power input is key to ensuring the correct operation of a crystal growth furnace. In the given exercise, power input is not only used to achieve a target temperature but to maintain it within a precise range.
Start by setting the desired temperature in the quadratic equation for temperature \( T(\omega) = 0.1\omega^2 + 2.155\omega + 20 \), then solve for \( \omega \) using the quadratic formula. Determine both a central required power input, around 33 watts for 200°C, and the allowable range (32.35 to 33.64 watts) to account for permissible temperature variations.
Accurate power input calculations enable precise temperature control essential for high-quality crystal growth.