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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 $?

(a) $w=32.998$(b) from 32.884 to 33.112 watts(c) see explanation

05:00

Madi S.

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 4

The Precise Definition of a Limit

Limits

Derivatives

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So in this problem were given this relationship for a crystal growth furnace where the temperature T and T is in degrees centigrade is 0.1 W squared plus 2.155 W was 20 where w is the heat input in was So the first question we were asked is If we want to maintain a temperature 200° centigrade, so that's T. So we just put 200 in there for TFW. So 200 equals 0.1 W squared plus 2.155, W plus 20. So now if we solve the if we put this in standard form then we have zero equals 0.1, W squared Plus 2.155, W minus 180. Because we would subtract 200 to the other side. Now, if you remember the quadratic formula quadratic equation, then this is a, this is B, this is cr coefficients and rick quadratic formula is X equals minus B plus or minus the square root, B squared minus four A C. All over to a. So Let's substitute these ends, that's -2.155. What's the minus the square root? 2.155 Squared minus four times. EH Let's 0.1. I'm C -180. All over two times. 0.1. Okay, so multiply This out, we get negative 2.155 Germany's the square root uh 76.644. All over 0.2. So this is -2.155, closer minus 8.7 55. All over 0.2. All right. Now, if I take the negative here, they don't have a negative number. Which doesn't make any sense. Right? So, I won't take that negative. I'll take the positive. Right? I'll take the plus. Don't give me a positive number. So that means I'm going to have 6.6 over 0.2, which is 33 want. So 33 wants of input. My relationship there gives us mm hmm gives us the 200 C. Now, the next one says if temperature varies by Plus or -1°C then we want to know what range of wanted is allowed for the input power. All right. So, yeah, we know that our relationship remember was This T. F.W. .1. Have you squared Times 2.155 W plus 20. So then doing the derivative right? D. T. Of W over T T. You can drill. I'm sorry, not t Bye. W Well, That means what would take the exponent down front and subtract one off the exponents. So this is 0.2 W Plus 2.155 20 goes away because there's no no W on it. So, the derivative is zero three. Ow is just one. So it's why it's 2.155. Okay. So then if we take then the change in T F W is are are derivative here. I'm still to tell you. Okay. And we do this at W equals 33 wants. So then this is 0.2 times. I'm 33 Plus 2.155. I'm still to w All right, well then this is 8.755 Delta W. Okay. And we're asked remember what the change in Dublin watts is going to be if Delta T is plus or -1. Okay, So if delta T is closer minus one, then I have a plus Vermont ist one over 8.755 is Delta W. And when I calculate this out I'll get Plus or zero 114. And this is in watts, isn't it? Okay, now, and so then part C were asked in terms of the epsilon delta definition of the limit as X approaches a uh, f of X equals L. Okay. Were asked what X is. Were asked what F of X is. Were asked what a is or asked what L is And were asked what value of epsilon is given? Were asked for the corresponding delta value. All right. So, were asked a series of questions here, aren't we that we can now work through? Well, first of all, X is the X. Going into our formula right into our F of X there, which is our input of power, which in our case was w Okay, F of X is T F W which is the temperature degree C. Okay, A is the input power in our case This is the 33 watts For the example, we just did L. Is the target temperature which in our case was 200° C. So then the value of epsilon that is given. Well remember by the epsilon delta definition, the epsilon is the the the absolute value of f of x minus L less than epsilon. So what does that mean? That means In our case epsilon was the value of the change in temperature. So this was posted modest 1°C, wasn't it? Which was the delta T. We had. Okay. And then the corresponding value of delta then remember in the epsilon uh delta definition, you have x minus C. Less than delta. So this means that delta is The plus or -0.1 14 114. What's wasn't it? Okay, so there we go. We got all of the different inputs and pieces for the epsilon delta definition to find

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