The discussion focuses on parametrizing the equation $(x^2+y^2)^2 = r^2 (x^2 - y^2)$ using polar coordinates. Participants confirm that the polar transformations $x = r\cos{\theta}$ and $y = r\sin{\theta}$ are valid, leading to the equation $r^4 = r^4(\cos(2\theta) - \sin(2\theta))$. The parametrization is established as $0 \leq r < \infty$ along the lines $\theta = 0$ or $\theta = \pi$, which corresponds to Cartesian coordinates $(x(t), y(t)) = (t, 0)$ for any real number t. The discussion also highlights the need to evaluate a line integral involving the curve.
PREREQUISITESStudents and educators in mathematics, particularly those studying calculus and polar coordinate systems, as well as anyone involved in evaluating line integrals in polar coordinates.
)jakey said:
tiny-tim said:hi jakey!
(have a theta: θ and try using the X2 icon just above the Reply box
yes it does …
show us how far you've got
jakey said:
if the question says they're equal, then they're equal! tiny-tim said:hi jakey!
d'oh!
the equation has become r4 = r4(cos2θ - sin2θ) …
what are the solutions to that, and what curve does it represent?
tiny-tim said:solve it anyway
jakey said:
tiny-tim said:and what curve is that?
jakey said:
tiny-tim said:well that's the parametrisation, isn't it? …
0 ≤ r < ∞, along the line θ = 0 or π …
in Cartesian coordinates: (x(t),y(t)) = (t, 0) for any t
tiny-tim said:period?
what's the question?
jakey said:
)tiny-tim said:(btw, if you don't use polar coordinates, the original equation becomes 2y2(x2+y2) = 0, or y = 0
∫ |y| ds ? …
well that's 0