Second Implicit Derivatives

[Razanir]

How would I find the second derivative with respect to t of x²+y²=r²

[Douglas]

Without knowing how x, y, and r relate to t, you can't.

[warty goblin]

I'm assuming there's functions relating x, and y to t. That is you could write

r² = (fx(t))²+(fy(t))² for some fx(t), fy(t).

In which case, chain rule for the first derivative, followed by product rule for the second.

[Razanir]

Glad people figured out r is a constant. I needed a random function, so I just used a circle.

So how you can find a first derivative:
x²+y²=r²
2x dx/dt + 2y dy/dt = 0

How would you find the second derivative?

[AttilaTheGeek]

Then solve for either dx/dt or dy/dt, depending on what you're looking for, and derive again.

[Mando Knight]

Or just use the d²x/dt² and d²y/dt² terms, for the general form?

2(dx/dt)² + 2x*d²x/dt² + 2(dy/dt)² + 2y*(d²y/dt²) = 0

[warty goblin]

You find the second derivative by taking another derivative. It gets slightly messy, because both the 2x*x' and 2y*y' terms require the product rule, but the theory is simple

0 = 2x*x' + 2y*y'

d/dt(0) = d/dt(2x*x' +2y*y')

0 = 2(x'*x' + x*x'' + y'*y' + y*y'')

0 = (x')^2 + (y')^2 + x*x'' + y*y''

Writing x' = dx/dt
Without having fx(t), fy(t) I can't really say any more.