Calculus can be applied to equations expressed in polar coordinates. The angular coordinate
φ is expressed in radians throughout this section, which is the conventional choice when doing calculus.
Differential calculus Using and , 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 derivatives) or \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} Hence, we have the following formula: \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} Using the inverse coordinates transformation, an analogous reciprocal relationship can be derived between the derivatives. Given a function
u(
r,
φ), it follows that \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} or \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} Hence, we have the following formulae: \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} 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. \begin{align} x &= r(\varphi)\cos\varphi \\ y &= r(\varphi)\sin\varphi \end{align}
Differentiating both equations with respect to
φ yields \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} Dividing the second equation by the first yields the Cartesian slope of the tangent line to the curve at the point (r(\varphi), \varphi) :\frac{dy}{dx} = \frac{r'(\varphi)\sin\varphi + r(\varphi)\cos\varphi}{r'(\varphi)\cos\varphi-r(\varphi)\sin\varphi}. For other useful formulas including divergence, gradient, and Laplacian in polar coordinates, see
curvilinear coordinates.
Integral calculus Arc length The arc length (length of a line segment) defined by a polar function is found by the integration over the curve
r(
φ). Let
L denote this length along the curve starting from points
A through to point
B, where these points correspond to
φ =
a and
φ =
b such that . The length of
L is given by the following integral L = \int_a^b \sqrt{ \left[r(\varphi)\right]^2 + \left[ {\tfrac{dr(\varphi) }{ d\varphi }} \right] ^2 } d\varphi
Area Let
R denote the region enclosed by a curve
r(
φ) and the rays
φ =
a and
φ =
b, where . Then, the area of
R is \frac12\int_a^b \left[r(\varphi)\right]^2\, d\varphi. This result can be found as follows. First, the interval is divided into
n subintervals, where
n is some positive integer. Thus Δ
φ, the angle measure of each subinterval, is equal to (the total angle measure of the interval), divided by
n, the number of subintervals. For each subinterval
i = 1, 2, ...,
n, let
φi be the midpoint of the subinterval, and construct a
sector with the center at the pole, radius
r(
φi),
central angle Δ
φ and arc length
r(
φi)Δ
φ. The area of each constructed sector is therefore equal to \pi \left[r(\varphi_i)\right]^2 \cdot \frac{\Delta \varphi}{2\pi} = \frac{1}{2}\left[r(\varphi_i)\right]^2 \Delta \varphi. Hence, the total area of all of the sectors is \sum_{i=1}^n \tfrac12r(\varphi_i)^2\,\Delta\varphi. As the number of subintervals
n is increased, the approximation of the area improves. Taking , the sum becomes the
Riemann sum for the above integral. , which mechanically computes polar integrals A mechanical device that computes area integrals is the
planimeter, which measures the area of plane figures by tracing them out: this replicates integration in polar coordinates by adding a joint so that the 2-element
linkage effects
Green's theorem, converting the quadratic polar integral to a linear integral.
Area generalization Using
Cartesian coordinates, an infinitesimal area element can be calculated as
dA =
dx dy. The
substitution rule for multiple integrals states that, when using other coordinates, the
Jacobian determinant of the coordinate conversion formula has to be considered: J = \det \frac{\partial(x, y)}{\partial(r, \varphi)} = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \varphi} \\[2pt] \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \varphi} \end{vmatrix} = \begin{vmatrix} \cos\varphi & -r\sin\varphi \\ \sin\varphi & r\cos\varphi \end{vmatrix} = r\cos^2\varphi + r\sin^2\varphi = r. Hence, an area element in polar coordinates can be written as dA = dx\,dy\ = J\,dr\,d\varphi = r\,dr\,d\varphi. Now, a function, that is given in polar coordinates, can be integrated as follows: \iint_R f(x, y)\, dA = \int_a^b \int_{r_1(\varphi_1)}^{r_2(\varphi_2)} f(r, \varphi)\,r\,dr\,d\varphi. Here,
R is the same region as above, namely, the region enclosed by a curve
r(
φ) and the rays
φ =
a and
φ =
b. The formula for the area of
R is retrieved by taking
f identically equal to 1. A more surprising application of this result yields the
Gaussian integral: \int_{-\infty}^\infty e^{-x^2} \, dx = \sqrt\pi.
Vector calculus Vector calculus can also be applied to polar coordinates. For a planar motion, let \mathbf{r} be the position vector , with
r and depending on time
t. We define an
orthonormal basis with three unit vectors:
radial, transverse, and normal directions. The
radial direction is defined by normalizing \mathbf{r}: \hat{\mathbf{r}} = (\cos(\varphi), \sin(\varphi)) Radial and velocity directions span the
plane of motion, whose
normal direction is denoted \hat{\mathbf{k}}: \hat{\mathbf{k}} = \hat{\mathbf{v}} \times \hat{\mathbf{r}}. The
transverse direction is perpendicular to both radial and normal directions: \hat \boldsymbol \varphi = (-\sin(\varphi), \cos(\varphi)) = \hat{\mathbf{k}} \times \hat{\mathbf{r}} \ , Then, position and velocity can be expressed as follows \begin{align} \mathbf{r} &= (x,\ y) = r(\cos\varphi,\ \sin\varphi) = r \hat{\mathbf{r}}\ , \\[1.5ex] \dot{\mathbf{r}} &= \left(\dot{x},\ \dot{y}\right) = \dot{r}(\cos\varphi,\ \sin\varphi) + r\dot{\varphi}(-\sin\varphi,\ \cos\varphi) = \dot{r}\hat{\mathbf{r}} + r\dot{\varphi}\hat{\boldsymbol{\varphi}}\ ,\\[1.5ex] \end{align} with the
angular rate \dot{\varphi} necessarily in
radians per second (or multiples thereof), in degrees per second. Acceleration can be expressed as: \begin{align} \ddot{\mathbf{r}} &= \left(\ddot{x},\ \ddot{y}\right) \\[1ex] &= \ddot{r}(\cos\varphi,\ \sin\varphi) + 2\dot{r}\dot{\varphi}(-\sin\varphi,\ \cos\varphi) + r\ddot{\varphi}(-\sin\varphi,\ \cos\varphi) - r\dot{\varphi}^2(\cos\varphi,\ \sin\varphi) \\[1ex] &= \left(\ddot{r} - r\dot{\varphi}^2\right) \hat{\mathbf{r}} + \left(r\ddot{\varphi} + 2\dot{r}\dot{\varphi}\right) \hat{\boldsymbol{\varphi}} \\[1ex] &= \left(\ddot{r} - r\dot{\varphi}^2\right) \hat{\mathbf{r}} + \frac{1}{r}\; \frac{d}{dt} \left(r^2\dot{\varphi}\right) \hat{\boldsymbol{\varphi}}. \end{align} This equation can be obtained by taking the derivative of the function and derivatives of the unit basis vectors. For a curve in 2D where the parameter is \theta the previous equations simplify to: \begin{aligned} \mathbf{r} &= r(\theta) \hat \mathbf{e}_r \\[1ex] \frac {d\mathbf{r}}{d\theta} &= \frac {dr} {d\theta} \hat \mathbf{e}_r + r \hat \mathbf{e}_\theta\\[1ex] \frac {d^2\mathbf{r}}{d\theta^2} &= \left(\frac {d^2 r} {d\theta^2}-r\right) \hat \mathbf{e}_r + \frac {dr} {d\theta} \hat \mathbf{e}_\theta \end{aligned}
Centrifugal and Coriolis terms The term r\dot\varphi^2 is sometimes referred to as the
centripetal acceleration, and the term 2\dot r \dot\varphi as the
Coriolis acceleration. For example, see Shankar. These terms, which appear when acceleration is expressed in polar coordinates, are a mathematical consequence of differentiation; they appear whenever polar coordinates are used. In planar particle dynamics, these accelerations appear when setting up Newton's
second law of motion in a rotating frame of reference. Here, these extra terms are often called
fictitious forces; fictitious because they are simply a result of a change in coordinate frame. That does not mean they do not exist; rather, they exist only in the rotating frame.
Co-rotating frame For a particle in planar motion, one approach to attaching physical significance to these terms is based on the concept of an instantaneous
co-rotating frame of reference. To define a co-rotating
frame of reference, first an origin is selected from which the distance
r(
t) to the particle is defined. An
axis of rotation is set up that is perpendicular to the plane of motion of the particle, and passing through this origin. Then, at the selected moment
t, the
angular rate of the co-rotating frame Ω is made to match the rate of rotation of the particle about this axis,
dφ/
dt. Next, the terms in the acceleration in the inertial frame are related to those in the co-rotating frame. Let the location of the particle in the inertial frame be (
r(
t),
φ(
t)), and in the co-rotating frame be denoted primed (
r′(t),
φ′(t)). Because the co-rotating frame rotates at the same rate as the particle,
dφ′/
dt = 0. The fictitious centrifugal force in the co-rotating frame is
mrΩ2, radially outward. The velocity of the particle in the co-rotating frame also is radially outward, because
dφ′/
dt = 0. The
fictitious Coriolis force therefore has a value −2
m(
dr/
dt)Ω, pointed in the direction of increasing
φ only. Thus, using these forces in Newton's second law we find: \mathbf{F} + \mathbf{F}_\text{cf} + \mathbf{F}_\text{Cor} = m\ddot{\mathbf{r}} \, , where over dots represent derivatives with respect to time, and
F is the net real force (as opposed to the fictitious forces). In terms of components, this vector equation becomes: \begin{align} F_r + mr\Omega^2 &= m\ddot{r} \\ F_\varphi - 2m\dot{r}\Omega &= mr\ddot{\varphi} \ , \end{align} which can be compared to the equations for the inertial frame: \begin{align} F_r &= m\ddot{r} - mr\dot{\varphi}^2 \\ F_\varphi &= mr\ddot{\varphi} + 2m\dot{r}\dot{\varphi} \ . \end{align} This comparison, plus the recognition that by the definition of the co-rotating frame at time
t it has a rate of rotation Ω =
dφ/
dt, shows that we can interpret the terms in the acceleration (multiplied by the mass of the particle) as found in the inertial frame as the negative of the centrifugal and Coriolis forces that would be seen in the instantaneous, non-inertial co-rotating frame. For general motion of a particle (as opposed to simple circular motion), the centrifugal and Coriolis forces in a particle's frame of reference commonly are referred to the instantaneous
osculating circle of its motion, not to a fixed center of polar coordinates. For more detail, see
centripetal force. == Differential geometry ==