View Full Version : Second Implicit Derivatives
Razanir
2013-05-12, 10:07 AM
How would I find the second derivative with respect to t of x²+y²=r²
Douglas
2013-05-12, 10:17 AM
Without knowing how x, y, and r relate to t, you can't.
warty goblin
2013-05-12, 10:17 AM
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
2013-05-12, 11:26 AM
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
2013-05-12, 11:29 AM
Then solve for either dx/dt or dy/dt, depending on what you're looking for, and derive again.
Mando Knight
2013-05-12, 11:43 AM
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
2013-05-12, 11:44 AM
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.
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