Accuracy of a smith's crafted metal disc

[mcastillo356]

TL;DR

A metal worker must make a circular metal disc with a certain accuracy. Got a solved example, but what if the disc's radius is not that of the example?

The area of a circular disc with radius $r$ is $A=\pi r^2\;\mbox{cm}^2$. A smith must make a circular metal disc of $400\pi\;\mbox{cm}^2$ with an accuracy of $\pm{5}\;\mbox{cm}^2$. Which accuracy range must have a $20\;\mbox{cm}$ disc?
Answer The metal worker wants to obtain $|\pi r^2-400\pi|<5$, this is

$$400\pi-5<\pi r^2<400\pi+5$$

or​

$$\sqrt{400-(5/\pi)}<r<\sqrt{400+(5/\pi)}$$
$$19.96017<r<20.03975$$
Therefore, the smith needs $|r-20|<0.03975$; so he must check the radius of the disc is less than $0.04\;\mbox{cm}$ the value of $20\;\mbox{cm}$.

The question: what about if the radius is $30$ or, for example, $5\;\mbox{cm}$?

PS: I'm not native. Forgive my english

 

[PeroK]

What's the question? How to do the same problem with different numbers?

 

[mcastillo356]

Yes, I can't manage to know which should be the answer if the question is the same, but the disc was a different radius.
See you later! My greyhound requires her walk...

 

[PeroK]

Yes, I can't manage to know which should be the answer if the question is the same, but the disc was a different radius.
See you later! My greyhound requires her walk...

You just follow the same steps with different numbers.

 

[hutchphd]

The usual way to do this is to use the fact that the percent accuracy is a small number. In this case an error of $5cm^2$ for a $5cm$ radius means you could hire a Beaver to do the cutting!

In general you just use the derivative. In this case $$\Delta A \approx \frac {dA} {dr} \Delta r \\= 2\pi r \Delta r$$ or for fractional change $$\frac {\Delta A} A\approx 2\frac { \Delta r} r$$

 

[mcastillo356]

For example, if the radius is $5\;\mbox{cm}^2$:
$$|\pi r^2-25\pi|<5$$
$$25\pi-5<\pi r^2<25\pi+5$$
$$\sqrt{25-(5/\pi)}<r<\sqrt{25+(5/\pi)}$$
$$4.83823<r<5.15671$$

therefore,​

$$|r-5|<0.15671$$
So he must check the radius is less than $0.15671$. Definitely, he will need a beaver.
Conclusion: bad example to introduce the concept of limit formally.
But in the next paragraph repares: "actually means it is possible to warrant that the gap $|f(x)-L|$ will be as small as any admissible accuracy, no matter how small is it".

 

[hutchphd]

Oh that's not too bad ...I somehow misread the number. So that's about 3% accuracy in the radius. So is your original question answered?

 

[mcastillo356]

Well ...No. How do you obtain that percentage?

 

[mcastillo356]

Well, good morning PF
$$|r-5|<0.15671$$
$$\dfrac{0.15671}{5}=0.031342\approx{3\%}$$. Compared with accuracy at the first post: $0.198...\approx{0.2\%}$, the difference between the first and the second is very big. I cannot coclude anything. I still think the example of the book is unfortunate. What do you think?

 

[mcastillo356]

Thread revisited. I was wrong. The example is perfect.
Thanks!

 

[Mark44]

For example, if the radius is $5\;\mbox{cm}^2$:
$$|\pi r^2-25\pi|<5$$
$$25\pi-5<\pi r^2<25\pi+5$$
$$\sqrt{25-(5/\pi)}<r<\sqrt{25+(5/\pi)}$$
$$4.83823<r<5.15671$$

therefore,​

$$|r-5|<0.15671$$
So he must check the radius is less than $0.15671$. Definitely, he will need a beaver.
Conclusion: bad example to introduce the concept of limit formally.

I don't believe the intent of this example is to introduce the limit concept. Instead, it is more likely to introduce estimating some value by the use of the derivative.

For example, in the context of this problem, if we have a disk of radius r, by how much does the area A change ($\Delta A$) when there is a small change in r of $\Delta r$?

So we have $A + \Delta A = \pi(r + \Delta r)^2 = \pi(r^2 + 2r\Delta r + (\Delta r)^2 \approx \pi r^2 + 2\pi r \Delta r$. If $\Delta r$ is small relative to r, then $\Delta r^2$ will be very much smaller, so ignoring it doesn't change our result by much.
From the above, we have $A + \Delta A \approx \pi r^2 + 2\pi r \Delta r = A + 2\pi r \Delta r$, hence $\Delta A \approx 2\pi r \Delta r = \frac {dA}{dr} \Delta r$.

This is what the example is trying to get across, I believe.

 

[hutchphd]

Often written as

In general you just use the derivative. In this case $$\Delta A \approx \frac {dA} {dr} \Delta r \\= 2\pi r \Delta r$$ or for fractional change $$\frac {\Delta A} A\approx 2\frac { \Delta r} r$$

 

[mcastillo356]

I definitively don't trust this book, but I will quote:

1.5 Formal definition of limit

(...). The more precise formal definition is based on the idea of assuring that the $x$ entry of a function $f$ is such that the $f(x)$ outcome will be in a certain interval

Sorry, it might be a bad translation

 

[Mark44]

1.5 Formal definition of limit

(...). The more precise formal definition is based on the idea of assuring that the x entry of a function f is such that the f(x) outcome will be in a certain interval

Right. I think I misunderstood what the textbook was trying to do, which appears to be this.
You want $|f(r) - L| < \epsilon$ (the conclusion of the formal definition of the derivative), so the idea is so solve this inequality for an interval around some $r_0$.

In the example of post #1, the inequality is $|\pi r^2 - 400\pi| < 5$, and after working with this inequality, they end up with the interval 19.96017 <r <20.03975. Note that here the function I referred to earlier is $f(r) = \pi r^2$, and since the ideal area of the disk is $400\pi$, the radius would be $r_0 = 20$. The interval that the textbook showed is roughly $20 \pm 0.04 \text{cm}$.