9.7 Defining Polar Coordinates and Differentiating in Polar Form

Polar coordinates locate a point using its distance from the origin $r$ and its angle $\theta$ from the positive x-axis. To differentiate a polar curve $r = f(\theta)$, convert to $x = r\cos\theta$ and $y = r\sin\theta$, then use the chain rule to find $\frac{dy}{dx}$. For AP Calculus BC, treat $\theta$ like the parameter when finding slopes and tangent lines.

Why This Matters for the AP Calculus Exam

Polar differentiation only appears on the AP Calculus BC exam. Unit 9 carries about 11 to 12 percent of the BC exam weight, and polar topics are a recurring part of that. This section sets up the area problems in 9.8 and 9.9, so getting comfortable with polar derivatives now pays off later.

The big skill here is recognizing that polar curves are really a special case of parametric equations with $\theta$ as the parameter. Once you see that, you can reuse the differentiation procedures you already know. On the exam you may need to find $\frac{dr}{d\theta}$, find $\frac{dy}{dx}$, locate horizontal or vertical tangent lines, or write a tangent line equation. Precise notation and careful chain rule work matter for clear exam work, especially when you switch between coordinate systems.

Key Takeaways

• Polar coordinates use distance $r$ and angle $\theta$ instead of $x$ and $y$.
• Convert with $x = r\cos\theta$, $y = r\sin\theta$, and $r = \sqrt{x^2 + y^2}$.
• $\frac{dr}{d\theta}$ tells you where $r$ reaches relative max or min distance from the origin, not the slope of the tangent line.
• The slope of the tangent line is $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$, which you can derive with the chain rule.
• Treat $\theta$ like the parameter $t$ from parametric equations; the procedures carry over.
• Watch your trig identities, radian values, and unit circle work, since small errors there cause most mistakes.

What Are Polar Functions?

Polar functions are graphed in a polar coordinate system, which uses a distance $r$ from a fixed point called the pole and an angle $\theta$ measured counter-clockwise from the positive x-axis. These functions show up in physics and engineering to model things like waves, orbits, and fields.

Differentiating polar functions directly is awkward because they are defined in terms of $r$ and $\theta$, not $x$ and $y$. Our usual idea of a derivative, the slope of the tangent line at a point, does not translate cleanly in polar form.

To work around this, convert polar equations to Cartesian form using these relations:

$$x=r\cos\theta\\y=r\sin\theta$$

Another useful conversion is:

$$r=\sqrt{x^2+y^2}$$

Converting Polar to Cartesian: Example 1

Convert the following polar function to a Cartesian function:

$$r=4\sin\theta$$

Rearrange the equation:

$$\sin\theta=\frac{r}{4}$$

Now bring in $y = r\sin\theta$, rearrange it, and set it equal to the equation above:

$$y=r\sin\theta \Rightarrow \sin\theta = \frac{y}{r}=\frac{r}{4}\\r^2=4y$$

Use $r = \sqrt{x^2 + y^2}$:

$$(\sqrt{x^2+y^2})^2=4y\\x^2+y^2=4y$$

Complete the square to get the final Cartesian function:

$$x^2+(y^2-4y)=0\\x^2+(y^2-4y+4)=4\\x^2+(y-2)^2=4$$

This is a circle. Converting between polar and Cartesian comes down to familiarity with the three relations connecting $x$, $y$, $r$, and $\theta$.

Converting Polar to Cartesian: Example 2

Find the values of $\theta$ on $r = 2 + 3\sin\theta$ where $x = 2$.

You have two pieces here: the $r = \dots$ equation and the $x = 2$ equation. Connect them. Plug $x = r\cos\theta$ into $x = 2$:

$$x=2\\r\cos\theta=2$$

Substitute the $r = \dots$ equation:

$$r=2+3\sin\theta$$ $$(2+3\sin\theta)\cos\theta=2$$ $$2\cos\theta+3\sin\theta \cos\theta=2$$

Using a calculator solver, you get:

$$\theta=0\\\theta=1.133$$

The general approach stays the same: use the relations to connect $x$, $y$, $r$, and $\theta$.

Derivatives of Polar Functions

When you take $\frac{dr}{d\theta}$, you find the angles where $r$ reaches its relative maximum or minimum distance from the origin. You find $\frac{dr}{d\theta}$ the same way you find any derivative: differentiate the polar function.

Polar Function Derivative Walkthrough

Find the points closest and furthest from the origin for $r = \theta + \cos(2\theta)$ on $[0, \pi)$.

Differentiate $r$ with respect to $\theta$, then set $\frac{dr}{d\theta} = 0$:

$$r=\theta+\cos(2\theta)\\\frac{dr}{d\theta}=1-\sin(2\theta)\cdot 2\\0=1-2\sin(2\theta)\\\frac{1}{2}=\sin(2\theta)$$

Using the unit circle, where does $\sin$ equal $\frac{1}{2}$?

$$2\theta=\frac{\pi}{6},\frac{5\pi}{6}\\$$ $$\theta=\frac{\pi}{12},\frac{5\pi}{12}$$

The question asks for points, not angles, so plug these $\theta$ values back into the original equation to get $r$:

$$r\left(\frac{\pi}{12}\right)=\frac{\pi}{12}+\cos\left(2\cdot\frac{\pi}{12}\right)=1.128\\$$ $$r\left(\frac{5\pi}{12}\right)=\frac{5\pi}{12}+\cos\left(2\cdot\frac{5\pi}{12}\right)=0.443$$

Now check the endpoints, $0$ and $\pi$:

$$r(0)=1\\r(\pi)=\pi$$

The point closest to the origin has $r = 0.443$, and the point furthest from the origin has $r = \pi$ (about 3.141). The values $1$ and $1.128$ fall in between.

Note that $\frac{dr}{d\theta}$ tells you about relative max and min distances, but not the slope of the tangent line. For the tangent line slope, you need $\frac{dy}{dx}$ in the Cartesian system.

Slope of the Tangent Line of Polar Functions

The key formula for this section:

$$\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{\frac{d}{d\theta}(r\sin\theta)}{\frac{d}{d\theta}(r\cos\theta)}=\frac{r\cos\theta+\frac{dr}{d\theta}(\sin\theta)}{-r\sin\theta+\frac{dr}{d\theta}(\cos\theta)}$$

You can memorize this, but most students find it easier to derive it with the product rule and chain rule each time.

Practice: Finding the Tangent Line to a Polar Function

Find the equation of the line tangent to the polar curve $r = \theta + \cos(2\theta)$ at $\theta = \frac{\pi}{3}$.

This is the same polar function from the last example, but now you want $\frac{dy}{dx}$:

$$\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}\\$$ $$\frac{dy}{dx}=\frac{\frac{d}{d\theta}(\theta \sin\theta+\sin\theta \cos2\theta)}{\frac{d}{d\theta}(\theta \cos\theta+\cos\theta \cos2\theta)}=\frac{\sin\theta+\theta \cos\theta+\cos\theta \cos 2\theta-2\sin\theta \sin2\theta}{\cos\theta-\theta \sin\theta-\sin\theta \cos2\theta+2\cos\theta \sin 2\theta}$$

Substitute $\theta = \frac{\pi}{3}$:

$$\frac{\sin\frac{\pi}{3}+\frac{\pi}{3} \cos\frac{\pi}{3}+\cos\frac{\pi}{3} \cos \frac{2\pi}{3}-2\sin\frac{\pi}{3} \sin\frac{2\pi}{3}}{\cos\frac{\pi}{3}-\frac{\pi}{3} \sin\frac{\pi}{3}-\sin\frac{\pi}{3} \cos\frac{2\pi}{3}+2\cos\frac{\pi}{3} \sin \frac{2\pi}{3}}=0.429$$

Find the $x$ and $y$ coordinates of the point using $r$:

$$x=r\cos\theta=\left(\frac{\pi}{3}+\cos\frac{2\pi}{3}\right)\left(\cos\frac{\pi}{3}\right)=0.274\\$$ $$y=r\sin\theta=\left(\frac{\pi}{3}+\cos\frac{2\pi}{3}\right)\left(\sin\frac{\pi}{3}\right)=0.474$$

The tangent line in point-slope form is:

$$y-0.474=0.429(x-0.274)$$

For a polar curve $r = f(\theta)$, derivatives of $r$, $x$, and $y$ with respect to $\theta$, along with the first and second derivatives of $y$ with respect to $x$, all give information about the curve.

• $\frac{dr}{d\theta}$ describes how fast the distance from the origin changes as $\theta$ changes. Setting it to zero finds where $r$ is largest or smallest.
• $x = r\cos\theta$ and $y = r\sin\theta$ let you find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$, which describe how the point moves in the plane.
• Using the chain rule with these gives $\frac{dy}{dx}$, the slope of the tangent line, and you can extend to $\frac{d^2y}{dx^2}$ for concavity.

How to Use This on the AP Calculus Exam

Problem Solving

• When you see $r = f(\theta)$, set up $x = r\cos\theta$ and $y = r\sin\theta$ before differentiating anything for slope.
• For slope, find $\frac{dy}{d\theta}$ and $\frac{dx}{d\theta}$ separately, then divide. Apply the product rule carefully since both $r$ and the trig factor depend on $\theta$.
• Read the question. If it asks for points furthest or closest to the origin, use $\frac{dr}{d\theta} = 0$. If it asks for the tangent line, use $\frac{dy}{dx}$.
• Always convert angle answers back to $r$ or to $(x, y)$ points when the question asks for points, not angles.

Common Trap

• A horizontal tangent needs $\frac{dy}{d\theta} = 0$ while $\frac{dx}{d\theta} \neq 0$. A vertical tangent needs $\frac{dx}{d\theta} = 0$ while $\frac{dy}{d\theta} \neq 0$. Setting $\frac{dr}{d\theta} = 0$ does not give tangent line information.
• Keep your calculator in radian mode and remember to differentiate the inside of trig functions like $\cos(2\theta)$, which brings down a factor of $2$.

Common Misconceptions

• $\frac{dr}{d\theta}$ is not the slope of the tangent line. It tells you how the distance from the origin changes. The tangent slope is $\frac{dy}{dx}$.
• $\frac{dy}{dx} \neq \frac{dy/d\theta}{dx/d\theta}$ is wrong as written; the correct rule is $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$. Do not just divide $\frac{dy}{d\theta}$ by $\frac{dr}{d\theta}$.
• When you differentiate $r\sin\theta$ or $r\cos\theta$, you must use the product rule because $r$ depends on $\theta$. Forgetting this is a common error.
• Polar curves are a special case of parametric equations, not a separate system. The same chain rule ideas apply, just with $\theta$ instead of $t$.
• A point closest to the origin is found from $r$ values, not from the angle. Always plug the angle back in to get $r$ when the question asks for points.

Related AP Calculus Guides

• 9.1 Defining and Differentiating Parametric Equations
• 9.4 Defining and Differentiating Vector-Valued Functions
• Unit 9 Overview: Parametric Equations, Polar Coordinates, and Vector-Valued Functions
• 9.2 Second Derivatives of Parametric Equations
• 9.3 Finding Arc Lengths of Curves Given by Parametric Equations
• 9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve