Derivatives of Parametric and Polar Curves - Theory & Concepts

Derivatives of Parametric and Polar Curves

In many engineering scenarios, such as the analysis of projectile trajectories or the mapping of structural shapes, curves are not easily expressed as a simple function

y = f(x)

. Instead, they are expressed using parametric equations or polar coordinates. This section covers how to apply calculus to these alternative coordinate systems.

Derivatives of Parametric Equations

When a curve is defined by parametric equations

x = f(t)

and

y = g(t)

, we can find the derivative

\frac{dy}{dx}

without needing to eliminate the parameter

t

. This is done using the Chain Rule.

First Derivative of Parametric Equations

If $x = f(t)$ and $y = g(t)$ are differentiable functions of $t$ , and $\frac{dx}{dt} \neq 0$ , then the derivative $\frac{dy}{dx}$ is given by:

First Derivative of Parametric Equations

Finding dy/dx without eliminating the parameter t.

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{y'(t)}{x'(t)}

Variables

Symbol	Description	Unit
$\frac{dy}{dx}$	Derivative of y with respect to x	-
$\frac{dx}{dt}, \frac{dy}{dt}$	Derivatives of x and y with respect to parameter t	-

Second Derivative of Parametric Equations

Finding the second derivative

\frac{d^2y}{dx^2}

requires careful application of the chain rule. It is not simply the ratio of the second derivatives.

Second Derivative Formula

To find the second derivative $\frac{d^2y}{dx^2}$ , differentiate the first derivative $\frac{dy}{dx}$ with respect to $t$ , and then divide by $\frac{dx}{dt}$ :

Second Derivative of Parametric Equations

Finding the second derivative.

\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}

Variables

Symbol	Description	Unit
$\frac{d^2y}{dx^2}$	Second derivative of y with respect to x	-
$\frac{dy}{dx}$	First derivative	-
$\frac{dx}{dt}$	Derivative of x with respect to parameter t	-

Interact with the simulation below to explore a physical application of parametric derivatives in projectile ballistics, analyzing horizontal and vertical rates of change and trajectory concavity.

Parametric Trajectory: Projectile Motion

Explore how parametric derivatives govern vertical and horizontal velocity rates, constructing the instantaneous tangent slope vector $dy/dx$ .

Launch Velocity (

v_0

): 30 m/s

Launch Angle (

\theta

): 45°

Time (

t

): 2.16 / 4.32 s

0.00sPeak: 2.16sImpact: 4.32s

Parametric Derivatives

x(t) position:45.82 m

y(t) position:22.94 m

dx/dt (Horizontal rate):21.21 m/s

dy/dt (Vertical rate):0.02 m/s

Tangent Slope dy/dx:0.0011

Concavity d²y/dx²:-0.02180 m⁻¹

Dynamic Motion Path & Tangent Vector

Observation: Horizontal velocity

v_x = dx/dt

is constant throughout flight (ignoring air drag). Vertical velocity

v_y = dy/dt

decreases linearly from positive to negative due to gravity. The combined velocity vector is always exactly tangent to the curve, represented by the parametric derivative

dy/dx

Derivatives of Polar Curves

In polar coordinates, a curve is defined by an equation

r = f(\theta)

, where

r

is the distance from the origin and

\theta

is the angle from the positive x-axis. To find the slope

\frac{dy}{dx}

of a polar curve, we first convert it into parametric equations using

\theta

as the parameter.

Parametric Form of Polar Equations

Using the standard conversion formulas between polar and Cartesian coordinates: $x = r \cos \theta$ and $y = r \sin \theta$ .
Substituting $r = f(\theta)$ , we get the parametric equations:
$x(\theta) = f(\theta) \cos \theta$
$y(\theta) = f(\theta) \sin \theta$

Slope of a Polar Curve

By applying the product rule to the parametric forms, the slope $\frac{dy}{dx}$ of the polar curve $r = f(\theta)$ is:

Slope of a Polar Curve

Finding dy/dx for polar curves.

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\frac{dr}{d\theta} \sin \theta + r \cos \theta}{\frac{dr}{d\theta} \cos \theta - r \sin \theta}

Variables

Symbol	Description	Unit
$\frac{dy}{dx}$	Slope of the tangent line	-
$r$	Polar radius	-
$\theta$	Polar angle	-

Interact with the simulation below to explore parametric and polar coordinates and their derivatives interactively.

Parametric & Polar Tangent Explorer

Curve Type

Parameter (θ) = 1.05

The tangent line slope is given by:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

For polar, substitute x=r cos(θ) and y=r sin(θ) first.

Angle Between the Radius Vector and the Tangent Line

In analyzing polar curves, it is often useful to find the angle

\psi

(psi) between the radial line (the position vector extending from the origin to the point on the curve) and the tangent line to the curve at that point. This angle characterizes the spiral nature of the curve.

Angle of Tangency in Polar Coordinates

Let $r = f(\theta)$ be a polar curve. The angle $\psi$ between the extended radius vector and the tangent line at a point $(r, \theta)$ is given by the formula:

Angle of Tangency in Polar Coordinates

Angle between the radius vector and the tangent line.

\tan \psi = \frac{r}{\frac{dr}{d\theta}}

Variables

Symbol	Description	Unit
$\psi$	Angle between radius vector and tangent line	-
$r$	Radius vector	-
$\frac{dr}{d\theta}$	Derivative of r with respect to \theta	-

This formula provides an elegant way to find the direction of a tangent line purely in polar terms without converting to Cartesian slope

\frac{dy}{dx}

. For example, in a logarithmic spiral

r = ae^{b\theta}

, the angle

\psi

is remarkably constant, since

\tan \psi = \frac{ae^{b\theta}}{abe^{b\theta}} = \frac{1}{b}

Key Takeaways

The first derivative of a parametric curve is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ .
The second derivative is found by dividing the $t$ -derivative of $\frac{dy}{dx}$ by $\frac{dx}{dt}$ . It is not the ratio of second partials.
To find the Cartesian slope of a polar curve, treat $\theta$ as a parameter and convert to Cartesian coordinates using $x = r \cos \theta$ and $y = r \sin \theta$ .
Use the product rule to find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$ , then apply the parametric derivative formula $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ .
The angle $\psi$ between the radius vector and the tangent line is given by $\tan \psi = r / \frac{dr}{d\theta}$ .

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