Editing Polar coordinate system (section)

===Differential calculus===
Using {{math|1=''x'' = ''r'' cos ''φ'' }} and {{math|1=''y'' = ''r'' sin ''φ'' }}, one can derive a relationship between derivatives in Cartesian and polar coordinates. For a given function, ''u''(''x'',''y''), it follows that (by computing its [[total derivative]]s)
or
<math display="block">\begin{align}
  r \frac{du}{dr}
    &= r \frac{\partial u}{\partial x} \cos\varphi  +
       r \frac{\partial u}{\partial y} \sin\varphi
     = x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}, \\[2pt]
  \frac{du}{d\varphi}
    &= - \frac{\partial u}{\partial x} r \sin\varphi +
         \frac{\partial u}{\partial y} r \cos\varphi
     = -y \frac{\partial u}{\partial x} + x \frac{\partial u}{\partial y}.
\end{align}</math>

Hence, we have the following formula:
<math display="block">\begin{align}
      r \frac{d}{dr} &=  x \frac{\partial}{\partial x} + y \frac{\partial}{\partial y} \\[2pt]
  \frac{d}{d\varphi} &= -y \frac{\partial}{\partial x} + x \frac{\partial}{\partial y}.
\end{align}</math>

Using the inverse coordinates transformation, an analogous reciprocal relationship can be derived between the derivatives. Given a function ''u''(''r'',''φ''), it follows that
<math display="block">\begin{align}
  \frac{du}{dx}
    &= \frac{\partial u}{\partial r}\frac{\partial r}{\partial x} +
       \frac{\partial u}{\partial \varphi}\frac{\partial \varphi}{\partial x}, \\[2pt]
  \frac{du}{dy}
    &= \frac{\partial u}{\partial r}\frac{\partial r}{\partial y} +
       \frac{\partial u}{\partial \varphi}\frac{\partial \varphi}{\partial y},
\end{align}</math>
or
<math display="block">\begin{align}
  \frac{du}{dx}
    &= \frac{\partial u}{\partial r}\frac{x}{\sqrt{x^2+y^2}} -
       \frac{\partial u}{\partial \varphi}\frac{y}{x^2+y^2} \\[2pt]
    &= \cos \varphi \frac{\partial u}{\partial r} -
       \frac{1}{r} \sin\varphi \frac{\partial u}{\partial \varphi}, \\[2pt]
  \frac{du}{dy}
    &= \frac{\partial u}{\partial r}\frac{y}{\sqrt{x^2+y^2}} +
       \frac{\partial u}{\partial \varphi}\frac{x}{x^2+y^2} \\[2pt]
    &= \sin\varphi \frac{\partial u}{\partial r} +
       \frac{1}{r} \cos\varphi \frac{\partial u}{\partial \varphi}.
\end{align}</math>

Hence, we have the following formulae:
<math display="block">\begin{align}
  \frac{d}{dx}
    &= \cos \varphi \frac{\partial}{\partial r} -
       \frac{1}{r} \sin\varphi \frac{\partial}{\partial \varphi} \\[2pt]
  \frac{d}{dy}
    &= \sin \varphi \frac{\partial}{\partial r} +
       \frac{1}{r} \cos\varphi \frac{\partial}{\partial \varphi}.
\end{align}</math>

To find the Cartesian slope of the tangent line to a polar curve ''r''(''φ'') at any given point, the curve is first expressed as a system of [[parametric equations]].
<math display="block">\begin{align}
  x &= r(\varphi)\cos\varphi \\
  y &= r(\varphi)\sin\varphi
\end{align}</math>

[[Derivative|Differentiating]] both equations with respect to ''φ'' yields
<math display="block">\begin{align}
  \frac{dx}{d\varphi} &= r'(\varphi)\cos\varphi - r(\varphi)\sin\varphi \\[2pt]
  \frac{dy}{d\varphi} &= r'(\varphi)\sin\varphi + r(\varphi)\cos\varphi.
\end{align}</math>

Dividing the second equation by the first yields the Cartesian slope of the tangent line to the curve at the point {{nowrap|(''r''(''φ''),&nbsp;''φ'')}}:
<math display="block">\frac{dy}{dx} = \frac{r'(\varphi)\sin\varphi + r(\varphi)\cos\varphi}{r'(\varphi)\cos\varphi-r(\varphi)\sin\varphi}.</math>

For other useful formulas including divergence, gradient, and Laplacian in polar coordinates, see [[curvilinear coordinates]].