Locate horizontal and vertical tangents
A tangent that turns
On the heart-shaped curve \(r = 2 - 2\cos\theta\), the long orange line is the tangent at P and the short orange arrow shows the direction P moves. The arrow's parts \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\) are green when they point up or to the right and red otherwise. Faint rings mark five points with a horizontal or vertical tangent.
Drag \(\theta\). At \(\theta = \frac{2\pi}{3}\), P is at \((-\frac32, \frac{3\sqrt{3}}{2})\) and the arrow points straight left, so its vertical part is zero and the tangent is horizontal.
At \(\theta = \frac{\pi}{3}\), P is at \((\frac12, \frac{\sqrt{3}}{2})\) and the arrow points straight up, so its horizontal part is zero and the tangent is vertical.
At \(\theta = 0\), P is at the pole and both parts are zero, so the arrow has no direction and the tangent needs a closer look.
Which part of the arrow is zero
At every point, the tangent line points along the arrow, so the two parts of the arrow decide its direction.
If only the vertical part is zero, the arrow points sideways, so the tangent is horizontal.
If only the horizontal part is zero, the arrow points up or down, so the tangent is vertical.
If both parts are zero, the arrow has no length and no direction.
Notation
On a curve \(r = f(\theta)\), the slope is \(\frac{dy}{d\theta}\) divided by \(\frac{dx}{d\theta}\). A horizontal tangent has slope 0, so the top is zero and the bottom is not.
\[\cy{\dfrac{dy}{d\theta}} = 0 \quad\text{and}\quad \dfrac{dx}{d\theta} \ne 0\]A vertical tangent has no slope, which happens when the bottom is zero and the top is not.
\[\cy{\dfrac{dx}{d\theta}} = 0 \quad\text{and}\quad \dfrac{dy}{d\theta} \ne 0\]To find the angles, factor each rate and set each factor equal to zero. Then test the other rate at each solution.
If both rates are zero at an angle, the test does not apply. The tangent there needs the limit of \(\frac{dy}{dx}\).
To give a point, put each angle into \(x = f(\theta)\cos\theta\) and \(y = f(\theta)\sin\theta\).
When it fails
On \(r = 2 - 2\cos\theta\), the rates are \(\frac{dy}{d\theta} = 2(1 - \cos\theta)(2\cos\theta + 1)\) and \(\frac{dx}{d\theta} = 2\sin\theta\,(2\cos\theta - 1)\). Both are zero at \(\theta = 0\), so the test does not apply. The first button shows the arrow shrink to a point.
The pole rule \(\tan\theta_0\) does not apply either, because \(f'(0) = 0\). Instead, use \(\frac{1 - \cos\theta}{\sin\theta} = \tan\frac{\theta}{2}\) to write the ratio as \(\tan\frac{\theta}{2}\cdot\frac{2\cos\theta + 1}{2\cos\theta - 1}\). Its limit is 0 from both sides, so the tangent at the pole is horizontal, as the second button shows.
Example 1: a circle through the pole
Find the points of \(r = 4\cos\theta\), for \(0 \le \theta < \pi\), where the tangent is horizontal or vertical.
Here \(f = 4\cos\theta\) and \(f' = -4\sin\theta\). The double-angle formulas shorten both rates.
\[\begin{aligned} \dfrac{dx}{d\theta} &= -8\sin\theta\cos\theta \\ &= -4\sin 2\theta \\ \dfrac{dy}{d\theta} &= 4(\cos^2\theta - \sin^2\theta) \\ &= 4\cos 2\theta \end{aligned}\]Solve \(\frac{dy}{d\theta} = 0\). Then \(\cos 2\theta = 0\), so \(\theta = \frac{\pi}{4}\) or \(\frac{3\pi}{4}\). There \(\frac{dx}{d\theta}\) is \(-4\) or \(4\), which is not zero.
\[\theta = \tfrac{\pi}{4}, \ \tfrac{3\pi}{4}\]Solve \(\frac{dx}{d\theta} = 0\). Then \(\sin 2\theta = 0\), so \(\theta = 0\) or \(\frac{\pi}{2}\). There \(\frac{dy}{d\theta}\) is \(4\) or \(-4\), which is not zero.
\[\theta = 0, \ \tfrac{\pi}{2}\]At \(\frac{\pi}{4}\), \(r = 2\sqrt{2}\) and P is \((2, 2)\). At \(\frac{3\pi}{4}\), \(r = -2\sqrt{2}\) and P is \((2, -2)\). At 0, P is \((4, 0)\). At \(\frac{\pi}{2}\), \(r = 0\) and P is the pole.
On this circle with center (2, 0), the top and bottom have horizontal tangents, and the right end and the pole have vertical tangents.
Example 2: the highest point of a petal
Find the highest point of the petal of \(r = \sin 2\theta\) in the first quadrant, \(0 < \theta < \frac{\pi}{2}\).
Here \(f = \sin 2\theta\) and \(f' = 2\cos 2\theta\). The double-angle formulas let you factor \(\frac{dy}{d\theta}\).
\[\begin{aligned} \dfrac{dy}{d\theta} &= 2\cos 2\theta\sin\theta + \sin 2\theta\cos\theta \\ &= 2\sin\theta\,(2\cos^2\theta - \sin^2\theta) \end{aligned}\]A tempting wrong move: The highest point is where the radius is largest, so set \(\frac{dr}{d\theta} = 0\). This gives \(\theta = \frac{\pi}{4}\).
\[2\cos 2\theta = 0 \;\Rightarrow\; \theta = \cr{\tfrac{\pi}{4}}\]At \(\theta = \frac{\pi}{4}\), P is the tip of the petal, the point farthest from the pole. There \(\frac{dy}{d\theta} = \frac{\sqrt{2}}{2}\), not zero, and the orange tangent slopes down at 45 degrees instead of following the red horizontal line.
A horizontal tangent needs \(\frac{dy}{d\theta} = 0\), not \(\frac{dr}{d\theta} = 0\). On the open interval, \(\sin\theta \ne 0\), so the bracket must be zero.
\[\begin{aligned} 2\cos^2\theta - \sin^2\theta &= 0 \\ \tan^2\theta &= 2 \\ \tan\theta &= \sqrt{2} \end{aligned}\]In the same way, \(\frac{dx}{d\theta} = 2\cos\theta\,(\cos^2\theta - 2\sin^2\theta)\). With \(\tan^2\theta = 2\), the bracket is \(\cos^2\theta\,(1 - 4)\), so \(\frac{dx}{d\theta} \ne 0\) and the tangent is horizontal.
Here \(\cos^2\theta = \frac13\) and \(\sin\theta = \sqrt{\frac23}\), which go into \(x = 2\sin\theta\cos^2\theta\) and \(y = 2\sin^2\theta\cos\theta\). As a check, the point is about \((0.54, 0.77)\).
\[x = \cy{\tfrac{2\sqrt{6}}{9}}, \qquad y = \cy{\tfrac{4\sqrt{3}}{9}}\]
Practice
Find the points of \(r = \dfrac{4}{2 + \cos\theta}\) where the tangent is horizontal.
Find \(f'\) with the quotient rule. Then combine the fractions in \(\frac{dy}{d\theta}\) over \((2 + \cos\theta)^2\).
Solution
The quotient rule gives \(f'\). Then \(\frac{dy}{d\theta}\) simplifies, because \(\sin^2\theta + \cos^2\theta = 1\).
\[\begin{aligned} f' &= \dfrac{4\sin\theta}{(2 + \cos\theta)^2} \\ \dfrac{dy}{d\theta} &= \dfrac{4(1 + 2\cos\theta)}{(2 + \cos\theta)^2} \end{aligned}\]Set \(\frac{dy}{d\theta} = 0\). Then \(\cos\theta = -\frac12\), so \(\theta = \frac{2\pi}{3}\) or \(\frac{4\pi}{3}\). At both angles \(\sin\theta \ne 0\), so \(\frac{dx}{d\theta}\) is not zero.
\[\dfrac{dx}{d\theta} = \dfrac{-8\sin\theta}{(2 + \cos\theta)^2} \ne 0\]At both angles \(r = \frac{4}{2 - 1/2} = \frac83\). The points are \((-\frac43, \frac{4\sqrt{3}}{3})\) and \((-\frac43, -\frac{4\sqrt{3}}{3})\).
Questions
The figure shows the cardioid \(r = 2 + 2\sin\theta\) and the orange arrow of motion at a point. Its parts \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\) are green when they point up or to the right and red otherwise. Point A is at the top, and point B is the cusp, the sharp point at the pole.
What does the arrow at A show about the tangent line there?
- It is horizontal, because the vertical part is zero and the horizontal part is not.
- It is vertical, because the vertical part is zero.
- There is no tangent line, because one rate is zero.
Answer
It is horizontal, because the vertical part is zero and the horizontal part is not.
At B, the arrow has no length. What does this show?
- Both rates are zero, so the test does not apply and the tangent needs a limit.
- The tangent is horizontal, because \(\frac{dy}{d\theta} = 0\).
- The tangent is vertical, because \(\frac{dx}{d\theta} = 0\).
Answer
Both rates are zero, so the test does not apply and the tangent needs a limit.
The curve is \(r = 1 + \cos\theta\). At which angle \(\theta\) in \((0, \pi]\) is the tangent vertical?
- \(\frac{2\pi}{3}\)
- \(\frac{\pi}{3}\)
- \(\pi\)
- \(\frac{\pi}{2}\)
Answer
\(\frac{2\pi}{3}\)
Summary
You find horizontal tangents where \(\frac{dy}{d\theta} = 0\) and \(\frac{dx}{d\theta} \ne 0\), vertical tangents where \(\frac{dx}{d\theta} = 0\) and \(\frac{dy}{d\theta} \ne 0\), and you use a limit where both rates are zero.
It works because the slope \(\frac{dy/d\theta}{dx/d\theta}\) is zero when only its top is zero and has no value when only its bottom is zero.
You use it to find the highest, lowest, leftmost or rightmost points of a polar curve.
This is one of four skills in topic 9.7 of Which & Why: AP Calculus.