Pendulum motion and pythagorean theorem

[sunnyday01]

Homework Statement

Show that the relation between the horizontal and vertical components of the ball's position is given by the equation: y = L - [(L^2 - x^2)^1/2]

http://www.flickr.com/photos/94066958@N08/8553595522/in/photostream/

Homework Equations

y = L - [(L^2 - x^2)^1/2]

The Attempt at a Solution

I know the solution must involve Pythagorean theorem and drawing a second triangle.
The first triangle has hypotenuse of length L, and other sides are L-y and x. That identity is given by L^2 = (L-y)^2 - x^2.
Drawing a second triangle, the sides are x and y but I don't know what the identity of the hypotenuse is. that equation would be x^2 + y^2 = hypotenuse^2

I don't know where to go from there...

 

[SammyS]

Homework Statement

Show that the relation between the horizontal and vertical components of the ball's position is given by the equation: y = L - [(L^2 - x^2)^1/2]

http://www.flickr.com/photos/94066958@N08/8553595522/in/photostream/

Homework Equations

y = L - [(L^2 - x^2)^1/2]

The Attempt at a Solution

I know the solution must involve Pythagorean theorem and drawing a second triangle.
The first triangle has hypotenuse of length L, and other sides are L-y and x. That identity is given by L^2 = (L-y)^2 - x^2.
Drawing a second triangle, the sides are x and y but I don't know what the identity of the hypotenuse is. that equation would be x^2 + y^2 = hypotenuse^2

I don't know where to go from there...

Hello sunnyday01. Welcome to PF !

If a & b are legs of a right triangle with hypotenuse, c, then

c² = a² + b²​

In your equation, L^2 = (L-y)^2 - x^2, why do you have the sign between. (L-y)² and x² as a minus sign ?

 

[sunnyday01]

formula correction

It should be:
L² = (L-y)² + x²

 

[SammyS]

It should be:
L² = (L-y)² + x²

Yes.

... and that should be all you need to get the desired result.

 

[sunnyday01]

I think I figured it out!

so, I think:
L² = (L-y)² + x²
L²-x² = (L-y)²
(L²-x²)^1/2 = L - y
-(L²-x²)^1/2 = - L + y
L -(L²-x²)^1/2 = y

I think that's it! Having to crop a picture for this question made me focus only on the variables I needed in the diagram.

I see the solution now is really simple, I feel silly for not seeing it sooner. Thank you for being so nice about it!