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The solution is -- so there's an unknown constant k (in this case, it corresponds to the "initial value"). If we tell what y equals at t=0, then we'll get the particular solution, (and that's often very important, that we get the exact solution and unique solution given the data).
We know that exponential functions have the property that the derivative is proportional to its function value (and is the special function for which the constant of proportionality is exactly 1).
or
The first doesn't involve y, and the second involves the second derivative of y.
A good introduction to parametric curves is given by ballistics. If we shoot a bullet into the air with speed v horizontally, and we neglect all forces but gravity, then the bullet will trace out a parabola:
Now how might we characterize the path of the bullet? The answer is a parametric curve, of the form C(t)=(x(t), y(t)).
or | ||
If we wish, we can solve for t in the equation for x and use that to eliminate the parameter t from the equation for y, hence getting an equation for the parabola traced out: | ||
Orbits of planets in the heavens, movements of ants on a hill, a robotic arm in an assembly plant: all these can be described by parametric curves.
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