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Math
CalculusSingle Variable Calculus: Early Transcendentals, Volume I8th Edition

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Chapter 2.4, Problem 12E
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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

Solutions

Chapter
Section
Problem 1E
Problem 2E
Problem 3E
Problem 4E
Problem 5E
Problem 6E
Problem 7E
Problem 8E
Problem 9E
Problem 10E
Problem 11E
Problem 12E
Problem 13E
Problem 14E
Problem 15E
Problem 16E
Problem 17E
Problem 18E
Problem 19E
Problem 20E
Problem 21E
Problem 22E
Problem 23E
Problem 24E
Problem 25E
Problem 26E
Problem 27E
Problem 28E
Problem 29E
Problem 30E
Problem 31E
Problem 32E
Problem 33E
Problem 34E
Problem 36E
Problem 37E
Problem 38E
Problem 39E
Problem 40E
Problem 41E
Problem 42E
Problem 43E
Problem 44E
BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

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(w) = 0.lw2 + 2.155w + 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 al 200°C?

(b) If the temperature is allowed to vary from 200°C by up to ± 1°C, what range of wan age is allowed for the input power?

(c) In terms of thee, ε, δ definition of lim x → a f ( x ) = L , what is x? What is f(x)? What is a? What is L.? What value of ε. is given? What is the corresponding va1ue of δ?

(a)

To determine

To find: The Power need to maintain the temperature at 200°C.

Explanation

Given:

Suppose the relationship is T(ω)=0.1ω2+2.155ω+20 and the temperature is

T=200°C. Since, 200=0.1ω2+2.155ω+20 (1)

Result used:

The solution for quadratic form of the equation is ax2+bx+c=0 as follows:

x=−b±b2−4ac2a

Calculation:

Multiply both sides to (1) with 1000.

1000(0.1ω2+2.155ω+20)=1000×200100ω2+2155ω+20000=200000100ω2+2155ω+20000−200000=0100ω2+2155ω+180000=0

By Quadratic equation formula,

ω=−2155±21552−4⋅ 100(−180000

(b)

To determine

To find: The range of wattage is allowed for the input power.

(c)

To determine

To explain: The terms in ε,δ definition of limx→af(x)=L.

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