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Homework Statement
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 length s, then the second arc has length [tex] s + \Delta s [/tex]. The distance between the two arcs are [tex]\Delta r[/tex]. Supposing that we don't know what the length of the inner circle is, and the information we know is the angle of the arcs, the difference of length between the two arcs, and the distance between them, what is the radius of the inner circle, supposing that the outer circle is [tex]r+\Delta r[/tex]?
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.
Homework Equations
[tex]\theta=\frac{s}{r}=\frac{s+ \Delta s}{r+ \Delta r}[/tex]
[tex]\Delta r=\frac{s+\Delta s}{\theta}-\frac{s}{\theta}[/tex]
(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)
The Attempt at a Solution
[tex]\Delta r=\frac{s+\Delta s}{\theta}-\frac{s}{\theta}[/tex]
Since s over theta is the radius:
[tex]\Delta r=\frac{s+\Delta s}{\theta}-r[/tex]
And since theta is s over r:
[tex]\Delta r=(s+\Delta s)\frac{r}{s}-r[/tex]
[tex]\Delta r=(1+\frac{\Delta s}{s})*r-r[/tex]
Factoring out the r:
[tex]\Delta r=r*((1+\frac{\Delta s}{s})-1)[/tex]
[tex]\Delta r=r*\frac{\Delta s}{s}[/tex]
Looking for r, I get:
[tex]r=\Delta r \frac{s}{\Delta s}[/tex] ***
Now, using equation 1 of the relevant equations:
[tex]\theta=\frac{s+ \Delta s}{r+ \Delta r}[/tex]
I solve for s, which is:
[tex]\theta*r+\theta*\Delta r-\Delta s[/tex]
Now, using the equation I asterisked, I substitute for s:
[tex]r=\Delta r*\frac{\theta*r+\theta*\Delta r-\Delta s}{\Delta s}[/tex]
I simplify:
[tex]r=\frac{\theta*r*\Delta r}{\Delta s}+\frac{\theta*\Delta r^2}{\Delta s}-\Delta r[/tex]
Substracting both sides with the term at the right with the r in it, I get:
[tex]r-\frac{\theta*r*\Delta r}{\Delta s}=\frac{\theta*\Delta r^2}{\Delta s}-\Delta r[/tex]
Factoring out r, I get:
[tex]r(1-\frac{\theta*\Delta r}{\Delta s})=\frac{\theta*\Delta r^2}{\Delta s}-\Delta r[/tex]
Solving for r:
[tex]r=\frac{\frac{\theta*\Delta r^2}{\Delta s}-\Delta r}{1-\frac{\theta*\Delta r}{\Delta s}}[/tex]