Parametric and polar derivatives

Why these need their own method

Both of these describe curves that a single \(y = f(x)\) cannot. A parametric curve gives \(x\) and \(y\) separately in terms of a parameter \(t\); a polar curve gives the radius \(r\) as a function of the angle \(\theta\). In both cases the chain rule recovers \(dy/dx\).

They share a page because they share a method: differentiate each coordinate with respect to the parameter, then divide.

Parametric: dy/dx from x(t) and y(t)

The parametric formula, and the second-derivative trap

For a curve given by \(x = x(t)\) and \(y = y(t)\), the chain rule gives:

\[\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \frac{dx}{dt} \neq 0\]

Take \(x = t^{2}\), \(y = t^{3}\). Then \(dx/dt = 2t\) and \(dy/dt = 3t^{2}\), so \(\frac{dy}{dx} = \frac{3t^{2}}{2t} = \frac{3t}{2}\). The slope is expressed in the parameter, which is usually what you want — substitute the \(t\) of the point you care about.

Second derivatives

A common trap: \(\frac{d^{2}y}{dx^{2}}\) is not \(\frac{d^{2}y/dt^{2}}{d^{2}x/dt^{2}}\). Differentiate the first derivative with respect to \(t\), then divide by \(dx/dt\) again:

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

For the example above that gives \(\frac{3/2}{2t} = \frac{3}{4t}\).

Where the curve turns

• Horizontal tangent where \(dy/dt = 0\) and \(dx/dt \neq 0\).
• Vertical tangent where \(dx/dt = 0\) and \(dy/dt \neq 0\).
• If both vanish at once, the point may be a cusp — the formula says nothing, and you must look at the curve.

Polar: dy/dx from r(θ)

The polar formula

A polar curve \(r = r(\theta)\) becomes parametric through \(x = r\cos\theta\) and \(y = r\sin\theta\), with \(\theta\) as the parameter. Applying the product rule to each and dividing gives:

\[\frac{dy}{dx} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}\]

Take the cardioid \(r = 1 + \cos\theta\), so \(r' = -\sin\theta\). At \(\theta = \pi/2\) we have \(r = 1\), \(r' = -1\), and the formula gives \(\frac{-1 \cdot 1 + 0}{0 - 1} = 1\) — a slope of 45°.

The other derivative people mean

\(dr/d\theta\) is not the slope of the curve. It is how fast the radius grows as you sweep round, which is a different question. When a problem asks for "the slope of the tangent line", it wants \(dy/dx\) and the formula above.

Questions