Question

Find the radius of curvature for √x/a - √y/b =1 at the pt where it touches the coordinale axis 9​

Answer 1

Answer:

This is a tough one!

The curvature of y = f(x) is

K = abs(y’’(x)) / [1 + (y’(x))^2]^(3/2) and the radius of curvature at a point (x1,y1) is

R = 1/K at that point.

We have √(x/a) + √(y/b) = 1

Rearranging and squaring, y = b[1 - 2√(x/a) + x/a]

So, for y = b[1 - 2√(x/a) + x/a] = b – 2b√(x/a) + bx/a

y’ = b/a – b/√(ax)

y’’ = (b√a)/2x^(3/2)

So, K = [(b√a)/2x^(3/2)] / {1 + [b/a – b/√(ax)]^2)}^(3/2)

And this can be evaluated at a point (x1, y1)

Absent a point (x1, y1) and the values of a and b, this is as far as we can go.

Step-by-step explanation: