(y')*y+(x')*x=0
where both derivatives are with respect to t, alternatively written as
y(dy/dt) + x(dx/dt) = 0
Can't see a solution at all, but may just be having a mind blank. Tried searching google but nothing came up.
Not even sure if there is a solution, went wrong trying to find a solution to Euler's angular motion equations, somehow ended up with that, and in the true spirit of procrastination have spent more time working this out than the original problem. Any help much appreciated.
where both derivatives are with respect to t, alternatively written as
y(dy/dt) + x(dx/dt) = 0
Can't see a solution at all, but may just be having a mind blank. Tried searching google but nothing came up.
Not even sure if there is a solution, went wrong trying to find a solution to Euler's angular motion equations, somehow ended up with that, and in the true spirit of procrastination have spent more time working this out than the original problem. Any help much appreciated.
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You already know that (y^2)' = 2yy'. Therefore yy' + xx' = 0 implies that (y^2)' + (x^2)' = 0. As (y^2)' + (x^2)' = (y^2 + x^2)' = 0. This means that y^2 + x^2 is a constant function of t. This should remind you of a familiar family of functions!
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Sometimes (cos(t),sin(t)) is called a "parametrization" of the circle. Maybe you could say something like
"the solutions x(t), y(t) constitute a parametrization of a circle." You can do the same for the hyperbola (or your favorite graph): x^2 - y^2 = 1, so xx' - yy' = 0, and so on.
"the solutions x(t), y(t) constitute a parametrization of a circle." You can do the same for the hyperbola (or your favorite graph): x^2 - y^2 = 1, so xx' - yy' = 0, and so on.
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