# [SOLVED]Gradient

#### dwsmith

##### Well-known member
I have
$$u(r,\theta) = r\cos(\theta)\left[1 - \left(\frac{1}{r}\right)^2\right]$$
and the gradient is
$$1 + \frac{2 x^2}{(x^2 + y^2)^2} - \frac{1}{x^2 + y^2}, \frac{2 x y}{(x^2 + y^2)^2}$$
How was this obtained?

#### Sudharaka

##### Well-known member
MHB Math Helper
I have
$$u(r,\theta) = r\cos(\theta)\left[1 - \left(\frac{1}{r}\right)^2\right]$$
and the gradient is
$$1 + \frac{2 x^2}{(x^2 + y^2)^2} - \frac{1}{x^2 + y^2}, \frac{2 x y}{(x^2 + y^2)^2}$$
How was this obtained?
Hi dwsmith,

The gradient of a scalar field $$f$$ in spherical coordinates is given by,

$\nabla f(r, \theta, \phi) = \frac{\partial f}{\partial r}\mathbf{e}_r+\frac{1}{r}\frac{\partial f}{\partial \theta}\mathbf{e}_\theta+\frac{1}{r \sin\theta}\frac{\partial f}{\partial \phi}\mathbf{e}_\phi=\left(\frac{\partial f}{\partial r},\,\frac{1}{r}\frac{\partial f}{\partial \theta},\,\frac{1}{r \sin\theta}\frac{\partial f}{\partial \phi}\right)$

Therefore,

$\nabla u(r, \theta)=\left(\frac{\partial u}{\partial r},\,\frac{1}{r}\frac{\partial u}{\partial \theta},\,0\right)$

Calculate the partial derivatives and convert them into Cartesian coordinates. Hope you can continue.

Kind Regards,
Sudharaka.

#### Jester

##### Well-known member
MHB Math Helper
Probably the easiest way is to switch u in terms of x and y, $\it{ i.e.}$

$u = x \left( 1 - \dfrac{1}{x^2+y^2}\right)$,

then calculate the gradient the usual way.