Find a polar tangent slope
The slope of a moving point
The point P moves along the curve \(r = 1 + 2\cos\theta\) as \(\theta\) increases. The orange arrow shows the direction in which P moves, and the orange tangent line at P runs along it.
The horizontal part of the arrow is the horizontal rate \(\frac{dx}{d\theta}\), and the vertical part is the vertical rate \(\frac{dy}{d\theta}\). Each part is green when it points to the right or up, and red when it points left or down.
At \(\theta = \frac{\pi}{2}\), P is at \((0, 1)\), with \(\frac{dx}{d\theta} = -1\) and \(\frac{dy}{d\theta} = -2\), so for every 1 unit that P moves left, it moves 2 units down. The slope is the rise divided by the run, so here it is \(\frac{-2}{-1} = 2\).
Drag \(\theta\). The tangent line turns, and at each position its slope is the vertical rate divided by the horizontal rate.
A ratio of two rates
Without numbers, the horizontal part of the arrow, \(\frac{dx}{d\theta}\), is the run, and the vertical part, \(\frac{dy}{d\theta}\), is the rise.
Press the first button to double the arrow. Both parts double, and the tangent line does not move, because the slope depends on the direction of the arrow and not on its length.
The slope of the tangent is the rise over the run, \(\frac{dy}{d\theta}\) divided by \(\frac{dx}{d\theta}\). This is the slope rule for a parametric curve, with \(\theta\) as the parameter.
You cannot divide by 0, so the rule needs \(\frac{dx}{d\theta} \ne 0\).
Notation
The slope is the vertical rate divided by the horizontal rate.
\[\cy{\dfrac{dy}{dx}} = \dfrac{dy/d\theta}{dx/d\theta}, \qquad \dfrac{dx}{d\theta} \ne 0\]Substitute the two rates from the last skill.
\[\cy{\dfrac{dy}{dx}} = \dfrac{f'(\theta)\sin\theta + f(\theta)\cos\theta}{f'(\theta)\cos\theta - f(\theta)\sin\theta}\]The slope needs both \(f\) and \(f'\). It is not \(f'(\theta)\), which is only \(\frac{dr}{d\theta}\), the rate of the radius.
To find the slope at one angle, calculate \(f\) and \(f'\) at that angle. Then calculate the top and the bottom, and divide.
At the pole, \(r = f(\theta_0) = 0\), so the terms with \(f\) vanish and \(f'(\theta_0)\) cancels from the top and the bottom.
\[\cy{\dfrac{dy}{dx}} = \dfrac{f'(\theta_0)\sin\theta_0}{f'(\theta_0)\cos\theta_0} = \tan\theta_0\]This holds only when \(f(\theta_0) = 0\) and \(f'(\theta_0) \ne 0\). If \(f'(\theta_0) = 0\) also, the quotient is \(\frac00\). A curve can pass through the pole at several angles, and each angle has its own tangent.
When it fails
At \(\theta = 0\), P is at \((3, 0)\). Here \(\frac{dx}{d\theta} = 0\) and \(\frac{dy}{d\theta} = 3\), so the arrow points straight up and the tangent line is vertical.
The formula gives \(\frac{3}{0}\), which is undefined. A vertical line has no slope, so \(\frac{dy}{dx}\) does not exist at this point, although the tangent line still exists.
Example 1: a circle with a check
Find the slope of the tangent to \(r = 6\sin\theta\) at \(\theta = \frac{\pi}{6}\). Check the answer with the circle.
Here \(f = 6\sin\theta\) and \(f' = 6\cos\theta\). Find both at \(\theta = \frac{\pi}{6}\).
\[f(\tfrac{\pi}{6}) = 3, \qquad f'(\tfrac{\pi}{6}) = 3\sqrt{3}\]Calculate the top, \(f'\sin\theta + f\cos\theta\), and the bottom, \(f'\cos\theta - f\sin\theta\), of the formula.
\[\begin{aligned} \text{top} &= 3\sqrt{3}\cdot\tfrac12 + 3\cdot\tfrac{\sqrt{3}}{2} \\ &= 3\sqrt{3} \\ \text{bottom} &= 3\sqrt{3}\cdot\tfrac{\sqrt{3}}{2} - 3\cdot\tfrac12 \\ &= 3 \end{aligned}\]Divide the top by the bottom.
\[\dfrac{dy}{dx} = \dfrac{3\sqrt{3}}{3} = \cy{\sqrt{3}}\]Check with the circle. It has center \((0, 3)\) and radius 3, and P is at \((\frac{3\sqrt{3}}{2}, \frac32)\), so the radius from the center to P has slope \(-\frac{1}{\sqrt{3}}\). A tangent is perpendicular to the radius, so its slope is the negative reciprocal, \(\sqrt{3}\).
Example 2: slope is not the rate of the radius
Find the slope of the tangent to \(r = e^{\theta}\) at \(\theta = \frac{\pi}{2}\).
Here \(f = f' = e^{\theta}\). At \(\theta = \frac{\pi}{2}\), \(\sin\theta = 1\) and \(\cos\theta = 0\), so P is at \((0, e^{\pi/2})\).
A tempting wrong move: The radius grows at the rate \(\frac{dr}{d\theta} = e^{\theta}\), so take the slope to be \(e^{\pi/2}\), which is about 4.8.
\[\dfrac{dy}{dx} = \dfrac{dr}{d\theta} = \cr{e^{\pi/2}}\]The red line with slope 4.8 rises steeply, but the spiral falls to the right at P, so the tangent slope there is negative. The rate \(\frac{dr}{d\theta}\) measures how fast the distance from the pole grows, not a rise over a run.
Use the slope formula with \(f = f' = e^{\pi/2}\).
\[\begin{aligned} \dfrac{dy}{dx} &= \dfrac{e^{\pi/2}\cdot 1 + e^{\pi/2}\cdot 0}{e^{\pi/2}\cdot 0 - e^{\pi/2}\cdot 1} \\ &= \dfrac{e^{\pi/2}}{-e^{\pi/2}} = \cy{-1} \end{aligned}\]The slope is −1, so the tangent line at P falls at 45 degrees.
Practice
The curve \(r = 2\sin 3\theta\) passes through the pole for \(0 \le \theta < \pi\). Find the slope of each tangent line at the pole.
At the pole, \(r = 0\), so solve \(2\sin 3\theta = 0\). Check that \(f'\) is not zero at each solution. Then use \(\tan\theta_0\).
Solution
The point is at the pole when \(2\sin 3\theta = 0\), so \(3\theta\) is a multiple of \(\pi\).
\[\theta_0 = 0, \ \tfrac{\pi}{3}, \ \tfrac{2\pi}{3}\]Check \(f'\) at each angle. It is not zero, so the rule for the pole applies.
\[f'(\theta) = 6\cos 3\theta = 6, \ -6, \ 6\]The slopes are \(\tan\theta_0\).
\[\begin{aligned} \tan 0 &= 0 \\ \tan\tfrac{\pi}{3} &= \sqrt{3} \\ \tan\tfrac{2\pi}{3} &= -\sqrt{3} \end{aligned}\]
Questions
The figure shows the curve \(r = 3 - 2\cos\theta\). The orange arrow at the orange point shows \(\frac{dx}{d\theta}\) as its horizontal part and \(\frac{dy}{d\theta}\) as its vertical part. A part is green if it points to the right or up, and red if it points left or down. The ring marks a second point, B.
What is the slope of the tangent line at the orange point?
- \(-\tfrac23\)
- \(-\tfrac32\)
- \(\tfrac23\)
Answer
\(-\tfrac23\)
What is true of the tangent line at the ringed point B?
- It is vertical, and \(\frac{dy}{dx}\) does not exist.
- Its slope is \(-5\).
- Its slope is 0.
Answer
It is vertical, and \(\frac{dy}{dx}\) does not exist.
The curve \(r = 4\cos 2\theta\) passes through the pole at \(\theta = \frac{\pi}{4}\). What is the slope of the tangent line there?
- 1
- 0
- −8
Answer
1
Summary
The tangent slope of a polar curve is \(\frac{dy}{d\theta}\) divided by \(\frac{dx}{d\theta}\), because the tangent line follows the direction of motion.
At the pole, where \(f(\theta_0) = 0\) and \(f'(\theta_0) \ne 0\), this slope is \(\tan\theta_0\).
If \(\frac{dx}{d\theta} = 0\) and \(\frac{dy}{d\theta} \ne 0\), the tangent line is vertical and has no slope.
This is one of four skills in topic 9.7 of Which & Why: AP Calculus.