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Formula:

(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 d²y/dx².

After much simplification, we get [3] d²y/dx² = [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?

Calculus

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