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Calculus/Radius of curvature


(x/a)^m + (y/b)^n = 1      
Problem: Find radius of curvature for the following two conditions:      
Condition 1:      x = 0; y = b
Condition 2:      x = a; y = 0
Exponents m and n are greater than 0      

Put in another way:

For the following curve:

y = (1-(x/a)^m)^n*b

1. What is the radius of curvature when x = 0?
2. What is the radius of curvature when x = a?

For example:

a =   200
b =   100
m =   2.5
n =   0.6

y = (1-(x/200)^2.5)^.6*100

x   y
0   100
25   99.66817654
50   98.11310708
75   94.74034578
200   0

1. What is the rasius of curvature when x = 0?
1. What is the rasius of curvature when x = 100?

I am having problems with division by zero when using a standarg radius of curvature formula.

Thank you in advance,

I actually took time to review this.  I noticed that there was a mistake at the very start.
If we have (x/a)^m + (y/b)^n = 1, then [1] y = b(1 - (x/a)^m)^(1/n).
Note this is to the 1/n, not to the n.

Using [1], I can say that [2] dy/dx = [-m(1/x)^(m-1)]/[n(y/b)^(1-n)].

Using [2], we can find dy/dx.
After much simplification, we get [3] dy/dx = [A(x,y)/B(x,y)]/C(x,y) where
A(x,y) = m(x/a)^(m-2),
B(x,y) = n(y/b)^b, and
C(x,y) = (n-1)(x/a)(dy/dx) - (m-1)(y/b).

Does this look right to you?


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