# Homework Help: Radius of an Arc Inside an Arc

1. Jul 3, 2010

### IBY

1. The problem statement, all variables and given/known data
Say there are two arcs which are part of the circumference of a circle within a circle with the same angles, but since one arc belongs to the circle outside, that arc is longer than the arc inside it. If the first arc has lenght s, then the second arc has length $$s + \Delta s$$. The distance between the two arcs are $$\Delta r$$. Supposing that we don't know what the lenght of the inner circle is, and the information we know is the angle of the arcs, the difference of lenght between the two arcs, and the distance between them, what is the radius of the inner circle, supposing that the outer circle is $$r+\Delta r$$?

This is a problem I made up, so there is no hurry in answering it. Anyways, I want someone to check the logic of my procedure because when I plugged in some numbers, they did not compute.
2. Relevant equations

$$\theta=\frac{s}{r}=\frac{s+ \Delta s}{r+ \Delta r}$$

$$\Delta r=\frac{s+\Delta s}{\theta}-\frac{s}{\theta}$$
(the second one is the one where I start from, basically the distance between the two arcs is the radius of the outside minus the radius of the inside)

3. The attempt at a solution

$$\Delta r=\frac{s+\Delta s}{\theta}-\frac{s}{\theta}$$

Since s over theta is the radius:

$$\Delta r=\frac{s+\Delta s}{\theta}-r$$

And since theta is s over r:

$$\Delta r=(s+\Delta s)\frac{r}{s}-r$$

$$\Delta r=(1+\frac{\Delta s}{s})*r-r$$

Factoring out the r:

$$\Delta r=r*((1+\frac{\Delta s}{s})-1)$$

$$\Delta r=r*\frac{\Delta s}{s}$$

Looking for r, I get:

$$r=\Delta r \frac{s}{\Delta s}$$ ***

Now, using equation 1 of the relevant equations:

$$\theta=\frac{s+ \Delta s}{r+ \Delta r}$$

I solve for s, which is:

$$\theta*r+\theta*\Delta r-\Delta s$$

Now, using the equation I asterisked, I substitute for s:

$$r=\Delta r*\frac{\theta*r+\theta*\Delta r-\Delta s}{\Delta s}$$

I simplify:

$$r=\frac{\theta*r*\Delta r}{\Delta s}+\frac{\theta*\Delta r^2}{\Delta s}-\Delta r$$

Substracting both sides with the term at the right with the r in it, I get:

$$r-\frac{\theta*r*\Delta r}{\Delta s}=\frac{\theta*\Delta r^2}{\Delta s}-\Delta r$$

Factoring out r, I get:

$$r(1-\frac{\theta*\Delta r}{\Delta s})=\frac{\theta*\Delta r^2}{\Delta s}-\Delta r$$

Solving for r:

$$r=\frac{\frac{\theta*\Delta r^2}{\Delta s}-\Delta r}{1-\frac{\theta*\Delta r}{\Delta s}}$$

2. Jul 3, 2010

### ehild

$$\Delta r=\frac{s}{\theta}+\frac{\Delta s}{\theta}-\frac{s}{\theta}=\frac{\Delta s}{\theta}$$

Δs/Δr= θ, they are not independent quantities.

ehild

3. Jul 3, 2010

### IBY

I am sorry to tell you that I don't know what you mean by that. Perhaps you could clarify that part to me a bit more.

4. Jul 3, 2010

### ehild

I mean that you can not choose Δs and Δr and θ arbitrarily, and you can not find r from them. Your resulting formula for r is equivalent to r=0/0.

5. Jul 3, 2010

### IBY

Does this mean the problem is unsolveable?

6. Jul 3, 2010

Yes.

ehild

7. Jul 3, 2010

Ok, thanks!

8. Jul 3, 2010

### Char. Limit

Wait, but it isn't. You're thinking of differentials. Since he has no limits in his equation, he's not dealing with differentials. So, there's no 0/0 involved.

9. Jul 4, 2010

### ehild

Sorry, I do not understand. What do you speak about?

ehild

10. Jul 4, 2010

### Unit

A diagram explaining your problem would really help. From your explanation, this is what I'm imagining:

So, to define our variables:
• θ is the angle AOB
• r = AO = BO
• Δr = AC = BD
• s = lesser arc AB
• s + Δs = lesser arc CD
Is that what you had in mind?

I've also attached file, in case the image above goes down in the future.

#### Attached Files:

• ###### IBY.bmp
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Last edited: Jul 4, 2010
11. Jul 4, 2010

### HallsofIvy

Yes, this is correct.

Why did you go back to r? Just do the subtraction:
$$\Delta r= \frac{s}{\theta}+ \frac{\Delta s}{\theta}- \frac{s}{\theta}= \frac{\Delta s}{\theta}[/itex] Isn't that what you want? 12. Jul 5, 2010 ### Mentallic It is unsolvable because the values of [tex]\Delta r, \Delta s, \theta$$ don't depend on the inner circle's radius. The only relationship you need (and can deduce for that fact) is $$\Delta s=\frac{2\pi \Delta r}{\theta}$$ which can be easily shown.

13. Jul 5, 2010

### ehild

This problem was solved long ago in this thread, and it is really very simple, I do not understand why was it started again.
The length of arc is angle times radius, just like mass is density times volume. If you have two pieces of gold, and you know the density and you know that one piece has 2 g more mass than the other, you can find the difference of the volumes, but there is no way to find out the individual volumes or masses.

ehild

Last edited: Jul 5, 2010