calculus
posted by Alex on .
Given: y^2+h^2=1
Find h as a function of y. If expression for dy/dx includes h, substitute the expression in terms of y for h. Write dy/dx in terms of y only.
Show that x=[(1+sqrt(1y^2))/y]sqrt(1y^2) satisfies the differential equation you found to prove this equation generates the curve.

You have not said anything about x in your problem statement so how do I know if dy/dx includes h?

This is the problem that I got. I thought of that too. I think that it might just be a type? x is probably suppose to be h...

well, y^2 + x^2 = 1 is a circle of radius 1
x = +/ (1y^2)^.5
y = +/ (1x^2)^.5
dy/dx = +/ .5 (1x^2)^.5 (2x)
dy/dx = +/ x (1x^2)^.5
but we know x = +/ sqrt(1y^2)
dy/dx = +/ sqrt(1y^2)/y^2