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)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)x = f(t) and y=g(t)y = g(t), we can find the derivative dydx\frac{dy}{dx} without needing to eliminate the parameter tt. This is done using the Chain Rule.

First Derivative of Parametric Equations

  • If x=f(t)x = f(t) and y=g(t)y = g(t) are differentiable functions of tt, and dxdt≠0\frac{dx}{dt} \neq 0, then the derivative dydx\frac{dy}{dx} is given by:
dydx=dydtdxdt=y′(t)x′(t) \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{y'(t)}{x'(t)}

Second Derivative of Parametric Equations

Finding the second derivative d2ydx2\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 d2ydx2\frac{d^2y}{dx^2}, differentiate the first derivative dydx\frac{dy}{dx} with respect to tt, and then divide by dxdt\frac{dx}{dt}:
d2ydx2=ddt(dydx)dxdt \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}

Derivatives of Polar Curves

In polar coordinates, a curve is defined by an equation r=f(θ)r = f(\theta), where rr is the distance from the origin and θ\theta is the angle from the positive x-axis. To find the slope dydx\frac{dy}{dx} of a polar curve, we first convert it into parametric equations using θ\theta as the parameter.

Parametric Form of Polar Equations

  • Using the standard conversion formulas between polar and Cartesian coordinates: x=rcos⁡θx = r \cos \theta and y=rsin⁡θy = r \sin \theta.
  • Substituting r=f(θ)r = f(\theta), we get the parametric equations:
  • x(θ)=f(θ)cos⁡θx(\theta) = f(\theta) \cos \theta
  • y(θ)=f(θ)sin⁡θy(\theta) = f(\theta) \sin \theta

Slope of a Polar Curve

By applying the product rule to the parametric forms, the slope dydx\frac{dy}{dx} of the polar curve r=f(θ)r = f(\theta) is:
dydx=dydθdxdθ=drdθsin⁡θ+rcos⁡θdrdθcos⁡θ−rsin⁡θ \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}

Parametric & Polar Tangent Explorer

The tangent line slope is given by:

dydx=dy/dtdx/dt\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 ψ\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(θ)r = f(\theta) be a polar curve. The angle ψ\psi between the extended radius vector and the tangent line at a point (r,θ)(r, \theta) is given by the formula:
tan⁡ψ=rdrdθ \tan \psi = \frac{r}{\frac{dr}{d\theta}}
This formula provides an elegant way to find the direction of a tangent line purely in polar terms without converting to Cartesian slope dydx\frac{dy}{dx}. For example, in a logarithmic spiral r=aebθr = ae^{b\theta}, the angle ψ\psi is remarkably constant, since tan⁡ψ=aebθabebθ=1b\tan \psi = \frac{ae^{b\theta}}{abe^{b\theta}} = \frac{1}{b}.
Key Takeaways
  • The first derivative of a parametric curve is dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.
  • The second derivative is found by dividing the tt-derivative of dydx\frac{dy}{dx} by dxdt\frac{dx}{dt}. It is not the ratio of second partials.
  • To find the Cartesian slope of a polar curve, treat θ\theta as a parameter and convert to Cartesian coordinates using x=rcos⁡θx = r \cos \theta and y=rsin⁡θy = r \sin \theta.
  • Use the product rule to find dxdθ\frac{dx}{d\theta} and dydθ\frac{dy}{d\theta}, then apply the parametric derivative formula dydx=dy/dθdx/dθ\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}.
  • The angle ψ\psi between the radius vector and the tangent line is given by tan⁡ψ=r/drdθ\tan \psi = r / \frac{dr}{d\theta}.
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