What are the signs of \frac{dx}{d\theta} and \frac{dy}{d\theta} at the orange point P?

\frac{dx}{d\theta} is negative and \frac{dy}{d\theta} is positive.

At the ringed point at the top of the circle, the arrow is horizontal and points left. What is true there?

\frac{dy}{d\theta} = 0 and \frac{dx}{d\theta} < 0

\frac{dx}{d\theta} = 0 and \frac{dy}{d\theta} < 0

\frac{dx}{d\theta} > 0 and \frac{dy}{d\theta} = 0

Topic 9.7 · BC

Differentiate the Cartesian coordinates

Builds on converting between polar and Cartesian coordinates. 7 minutes.

A point winding outward

The spiral \(r = \theta\) leaves the pole and winds outward. As \(\theta\) increases, the orange point P moves along the spiral, so \(x\) and \(y\) are functions of \(\theta\).

The orange arrow shows how P moves. Its horizontal part is the rate \(\frac{dx}{d\theta}\), and its vertical part is the rate \(\frac{dy}{d\theta}\). A part is green if its rate is positive, so it points to the right or up. It is red if its rate is negative.

At \(\theta = \frac{\pi}{2}\), P is at \((0, \frac{\pi}{2})\). The horizontal rate is \(-\frac{\pi}{2}\) and the vertical rate is 1, so P moves left and up. The horizontal part is red, and the vertical part is green.

To find these rates from an equation \(r = f(\theta)\), you differentiate \(x\) and \(y\) with respect to \(\theta\).

Two moves at once

P is now on a general curve \(r = f(\theta)\). When \(\theta\) increases, two changes move P.

The radius changes at the rate \(f'\), which moves P along its ray. The ray also turns, which moves P perpendicular to the ray. A point at distance \(f\) from the pole travels an arc of length \(f\) per radian of turn, so this move has the rate \(f\).

The faint arrows show these two moves. As you drag either slider, the orange arrow stays the sum of the two faint arrows.

The horizontal parts of the two faint arrows add up to \(\frac{dx}{d\theta}\), and their vertical parts add up to \(\frac{dy}{d\theta}\).

Notation

Because \(x = r\cos\theta\) and \(y = r\sin\theta\), each coordinate of the curve \(r = f(\theta)\) is a function of \(\theta\). A polar curve is a parametric curve with parameter \(\theta\).
\[x = f(\theta)\cos\theta, \qquad y = f(\theta)\sin\theta\]
The coordinate \(x\) is a product of two functions of \(\theta\), so use the product rule.
\[\cy{\dfrac{dx}{d\theta}} = f'(\theta)\cos\theta - f(\theta)\sin\theta\]
The first term is the horizontal part of the move along the ray. The second term is the horizontal part of the turn.
The coordinate \(y\) is also a product.
\[\cy{\dfrac{dy}{d\theta}} = f'(\theta)\sin\theta + f(\theta)\cos\theta\]
The two terms are the vertical parts of the same two moves.
If \(f\) contains another function of \(\theta\), find \(f'\) first with the chain rule.
\[f(\theta) = \sin 3\theta \;\Rightarrow\; f'(\theta) = 3\cos 3\theta\]
Both rates are derivatives with respect to \(\theta\), the parameter. The next skill divides them to find the slope \(\frac{dy}{dx}\).

Example 1: a curve with two terms

Find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\) for \(r = 2 + \cos\theta\). Check \(\frac{dx}{d\theta}\) with a second method.

Here \(f(\theta) = 2 + \cos\theta\), so \(f'(\theta) = -\sin\theta\).
Put \(f\) and \(f'\) into the formula for \(\frac{dx}{d\theta}\), and factor.
\[\begin{aligned} \dfrac{dx}{d\theta} &= -\sin\theta\cos\theta \\ &\quad - (2 + \cos\theta)\sin\theta \\ &= -2\sin\theta\,(1 + \cos\theta) \end{aligned}\]
Put them into the formula for \(\frac{dy}{d\theta}\). Use \(\cos^2\theta - \sin^2\theta = \cos 2\theta\).
\[\begin{aligned} \dfrac{dy}{d\theta} &= -\sin^2\theta + (2 + \cos\theta)\cos\theta \\ &= 2\cos\theta + \cos 2\theta \end{aligned}\]
Check \(\frac{dx}{d\theta}\) another way. Expand \(x\) with a double-angle formula and differentiate. The result equals the factored form, because \(\sin 2\theta = 2\sin\theta\cos\theta\).
\[\begin{aligned} x &= 2\cos\theta + \cos^2\theta \\ &= 2\cos\theta + \tfrac12(1 + \cos 2\theta) \\ \dfrac{dx}{d\theta} &= -2\sin\theta - \sin 2\theta \end{aligned}\]
At \(\theta = \frac{\pi}{2}\), \(\frac{dx}{d\theta} = -2\) and \(\frac{dy}{d\theta} = -1\). P is at \((0, 2)\) and moves left and down, as the two red parts of the arrow show.

Example 2: the chain rule inside the radius

Find \(\frac{dx}{d\theta}\) at \(\theta = \frac{\pi}{6}\) for \(r = \cos 2\theta\).

Here \(f(\theta) = \cos 2\theta\). The chain rule gives \(f'(\theta) = -2\sin 2\theta\).
A tempting wrong move: Differentiate each factor of \(x = \cos 2\theta\cos\theta\) and multiply the results.
\[\begin{aligned} \dfrac{dx}{d\theta} &= f'(\theta)\cdot(-\sin\theta) \\ &= \cr{2\sin 2\theta\sin\theta} \end{aligned}\]
At \(\theta = \frac{\pi}{6}\), this gives \(2\sin\frac{\pi}{3}\sin\frac{\pi}{6} = \frac{\sqrt{3}}{2}\), which is positive, so the arrow would point to the right. In the figure, P moves toward the pole, so the arrow points left.
The derivative of a product is not the product of the derivatives. The product rule adds two terms, one for each factor.
\[\begin{aligned} \dfrac{dx}{d\theta} &= f'\cos\theta - f\sin\theta \\ &= -2\sin 2\theta\cos\theta \\ &\quad - \cos 2\theta\sin\theta \end{aligned}\]
Put in \(\theta = \frac{\pi}{6}\). Here \(\sin 2\theta = \frac{\sqrt{3}}{2}\) and \(\cos 2\theta = \frac12\).
\[\begin{aligned} &-2\cdot\tfrac{\sqrt{3}}{2}\cdot\tfrac{\sqrt{3}}{2} - \tfrac12\cdot\tfrac12 \\ &= -\tfrac32 - \tfrac14 = \cr{-\tfrac74} \end{aligned}\]
The value is negative, so the horizontal part of the arrow is red and points left. The arrow is drawn at half size so that it fits.

Practice

Find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\) for \(r = \theta^2\) at \(\theta = \frac{\pi}{2}\).

Put \(f(\theta) = \theta^2\) and \(f'(\theta) = 2\theta\) into the two product-rule formulas. At \(\theta = \frac{\pi}{2}\), about 1.57 on the slider, P is on the positive y-axis, so \(\cos\theta = 0\) and \(\sin\theta = 1\).

Solution

The product rule gives both rates.
\[\begin{aligned} \dfrac{dx}{d\theta} &= 2\theta\cos\theta - \theta^2\sin\theta \\ \dfrac{dy}{d\theta} &= 2\theta\sin\theta + \theta^2\cos\theta \end{aligned}\]
At \(\theta = \frac{\pi}{2}\), \(\cos\theta = 0\) and \(\sin\theta = 1\).
\[\cr{\dfrac{dx}{d\theta} = -\dfrac{\pi^2}{4}}, \qquad \cg{\dfrac{dy}{d\theta} = \pi}\]
P moves left and up at this angle, so the horizontal part is red and the vertical part is green. The arrow is drawn at 60 percent of its size so that it fits.

Questions

The figure shows the circle \(r = 4\sin\theta\). The orange arrow at P has a horizontal part and a vertical part, drawn at half size. A part is green if it points to the right or up, and red if it points left or down. The ring marks the top of the circle.

What are the signs of \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\) at the orange point P?
- Both are negative.
- \(\frac{dx}{d\theta}\) is negative and \(\frac{dy}{d\theta}\) is positive.
- Both are positive.
Answer
Both are negative.
At the ringed point at the top of the circle, the arrow is horizontal and points left. What is true there?
- \(\frac{dy}{d\theta} = 0\) and \(\frac{dx}{d\theta} < 0\)
- \(\frac{dx}{d\theta} = 0\) and \(\frac{dy}{d\theta} < 0\)
- \(\frac{dx}{d\theta} > 0\) and \(\frac{dy}{d\theta} = 0\)
Answer
\(\frac{dy}{d\theta} = 0\) and \(\frac{dx}{d\theta} < 0\)
For \(r = 2\theta\), what is \(\frac{dy}{d\theta}\) at \(\theta = \pi\)?
- \(-2\pi\)
- 0
- \(-2\)
- \(2\pi\)
Answer
\(-2\pi\)

Summary

You find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\) by writing \(x = f(\theta)\cos\theta\) and \(y = f(\theta)\sin\theta\), and then using the product rule on each.

This works because a polar curve is a parametric curve in \(\theta\), and each coordinate is a product of two functions of \(\theta\).

You use these rates when you need the direction of motion along a polar curve, as for its tangent slope.

This is one of four skills in topic 9.7 of Which & Why: AP Calculus.