Polar / Rectangular Coordinates

Sparky_
Messages
227
Reaction score
5

Homework Statement



Convert into rectangular coordinates:

<br /> r = tan(theta)<br />



Homework Equations





The Attempt at a Solution



<br /> r = \frac {sin}{cos}<br />


I used
<br /> r = \sqrt{x^2+y^2}<br />
and

<br /> cos = \frac {x}{r}<br /> <br /> x = (r)(cos)<br /> sin = \frac {y}{r}<br /> <br /> y = (r)(sin)<br />

<br /> \sqrt {x^2+y^2} = \frac{\frac{y}{\sqrt {x^2+y^2}}}{\frac{x}{\sqrt {x^2+y^2}}}<br />

Then -

<br /> x\sqrt {x^2+y^2} -y =0<br />

The book gets
<br /> x^4 + (x^2)(y^2) - y^2 <br /> <br />


Can you help with the simplification to get to the book’s answer?

Thanks
-Sparky
 
Physics news on Phys.org
Take y to the other side, square and open brackets.
 
Thanks so much!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top