Re: polar and cartesian

On 11 Sep 2006 21:06:58 -0700, "vicky" <vikash.vks123@xxxxxxxxx>
wrote:

Friends
As we knowrelation between poler and cartesian co-ordinates are
x=r*cos(t) -------(1) // sorry i
use t for angle theta//
y=r*sin(t) --------(2)
r ^2= x^2+y^2 -------(3)
t= tan^-1(y/x) -------(4)

By equation (1) x_r = cos(t) ----(5)
By equation (3) differentiating w.r.to r we get
x_r= sec(t) -------(6)

why?

I don't see how you came up with sec(t). Remember that the x and y on
the right side of (3) depend on r so you need the chain rule if you
differentiate with respect to r:

2r = 2x * x_r + 2y * y_r
= 2x * cos(t) + 2y * sin(t)
= 2* r * cos(t) * cos(t) + 2 * r * sin(t) * sin(t)
= 2r, not very enlightening but at least consistent, since x^2 + y^2
= r^2 after all.

--Lynn
.

Relevant Pages

  • polar and cartesian
    ... As we knowrelation between poler and cartesian co-ordinates are ... use t for angle theta// ... By equation differentiating w.r.to r we get ...
    (sci.math)
  • Re: polar and cartesian
    ... As we knowrelation between poler and cartesian co-ordinates are ... use t for angle theta// ...
    (sci.math)