Question

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(1-y^2))/y]-sqrt(1-y^2) satisfies the differential equation you found to prove this equation generates the curve.

Answer 1

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

Answer 2

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...

Answer 3

well, y^2 + x^2 = 1 is a circle of radius 1

x = +/- (1-y^2)^.5

y = +/- (1-x^2)^.5

dy/dx = +/- .5 (1-x^2)^-.5 (-2x)

dy/dx = +/- x (1-x^2)^-.5
but we know x = +/- sqrt(1-y^2)
dy/dx = +/- sqrt(1-y^2)/y^2