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Numerade Educator



Problem 11 Medium Difficulty

A machinist is required to manufacture a circular metal disk with area 1000 $ cm^2 $.
(a) What radius produces such a disk?
(b) If the machinist is allowed an error tolerance of $ \pm 5 cm^2 $ in the area of the disk, how close to the ideal radius in part (a) must the machinist control the radius?
(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) $r \approx 17.8412 cm$
(b) Radius must be within $0.0445 cm$
(c) $x$ is the radius, $f(x)$ is the area, $a$ is the target radius given in part (a), $L$ is the target area $(1000), e$ is the tolerance in the
$\operatorname{arca}(5),$ and $\delta$ is the tolerance in the radius given in part (b).

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Video Transcript

in this exercise, the machinist is making circular metal disc that is 1000 centimeters squared in area and He's allowed tolerance or plus plus or -5 cm. And we are initially asked what is the radius necessary to make 1000 centimeters squared area disk? Well recall geometry. The formula for area of a circle. His pi times the radius squared. If I want an area of 1000 cube square centimeters, that's going to be given to me by pi times whatever the previous is squared, dividing both sides by pi and taking the square root. Kiss me. Yeah. Uh The radius Is equal to the square root of 1000 over pie. So this is the Specific radius that will give me 1000 square centimeter disk. The second question asks if this machinist is allowed an error of plaster five plus or minus five square centimetres. How close does the radius have to be? So -5 cm is going to be 955 square centimeters. That's the lower limit of the area. The area. His pi times the radius squared. And the upper limit for the Tolerance of the area is 1005 square centimeters solving this inequality, dividing everything by pi. That means back The radius squared has to be greater than or equal to 955 divided by Pi. Okay, and it has to be less than or equal to 1005 divided by pi. Which means that the radius can be anywhere between the square break of 955 divided by pi. And let's square root of 1005 divided by pi. So if the actual radius that's necessary is 1000 Square root of 1000 divided by Pi. And the lower limit is 955 divided by Pi than the difference is The square root of 1000 Over the square root of pi minus the square root of 955 Over the square root of pi. Yeah come on. No. Which is equal to the square root of 1000 minus square root Of 955. All over the square root of pi. Going to go ahead and evaluate that. Oh it didn't like that. Last level modification. If I calculate that as a decimal I get .406. So this can be right 0.406 less than The radius of the squared of 1000 divided by PI or Yeah, I'm going to move down a little bit. Yeah. Mhm Okay, it can be 1005 minus the square rail of 1000. Over the square root of pi. Greater than The radius of 1000 over Pi Square root. Yeah. And if I evaluate this square root of one 1000. Oops, I can do that. A square root of 1005 minus the square root of 1000. No, I didn't like that. Yeah, let me try. Redrawing squared of 1005 minus the square root of 1000. Mhm Right, That evaluates to negative .445. Oh yeah That's 0.046. Yeah. Yeah. No. Uh huh. Yeah I'm sorry I'm double checking my mouth. So this is .406. Less than 4.0445. Greater than the So for part B the radius can be anywhere from the square root 1000 over pie minus 0.406. Yeah let me rewrite that. Mhm. Yes. Yeah. Yeah. The square root of 1000 over a pie. Yeah So the radius will have to be 17.84. It can be 17 8.8 for -106. So 17 okay for one minus 0.406 is the smallest that this can be. I hope it didn't like that. Mhm. Mhm. Yeah. Yeah. Okay. Peterson, So the radius can be anywhere between 17.435 And 17.885. Yeah. Which means that the radius must be within .0445 cm. In order for the tolerance of the area to be within five square centimetres. And then part C. Is asking us about the delta epsilon definition of limits. So a delta epsilon definition of limits says I'm going to go ahead and type this. No okay we are looking at but make sure you write this down correctly. Okay a limit and X approaches some value A uh my function after of Acts is equal to no and the delta epsilon definition says that if my limit varies by some delta that the input can vary by some absalon I can choose a delta around my limit pill by choosing an appropriate absalon difference around my um target input value K. And then this asks, what does X mean in this situation? Well in this situation Acts is my input. So X is my radius I'm going to type Yeah. And my function F of X Is the area that I am trying to achieve? 1000 years area A So fX is the function that gives me the area. The input is the radius. And what is a hey is my target radius. That will give me the correct Area and that is 17.8. 4 books. Yeah. Or one who take for one cm and L is a target area I just 1000 square cm. And then delta is the furthest we can get away from our target input. So furthest we can get away from our target and put paint and still be close enough to 1000 cm is 0.04455 17 years. And that allows us to stay within our salon of us. Four minus five square centimetres of our target of 1000. So our area is 1000. That's the l limit that we're trying to get to our absalon. The tolerance is plus or minus five square centimetres. And in order to stay within that tolerance of plus or minus five square centimetres I have to stay within plus or minus 0.4455 centimeters of the radius. I can go technically, Oops, sorry about that. I can technically go uh huh. Above or I can technically subtract .4 cm and still be within the tolerance of the area. But my absalon has to be the amount I can go in either direction, so I'm going to choose the smaller value for my delta. Yeah. Yeah, yep. Yes. Mhm.