Defining Polar Coordinates and Differentiating in Polar Form - AP Calculus BC Study Guide

The Core Idea: Defining Polar Coordinates and Differentiating in Polar Form

The polar coordinate system describes points in a plane using a distance from a central point (the pole) and an angle from a reference direction. This topic introduces the fundamental connection between the polar coordinate system $(r,\theta )$ and the more familiar rectangular (Cartesian) coordinate system $(x,y)$. While polar equations like $r=f(\theta )$ are excellent for describing certain types of curves (like circles, cardioids, and spirals), calculus operations such as finding the slope of a tangent line are defined in terms of $x$ and $y$ coordinates ($dy/dx$).

The central problem this topic solves is how to find the slope of a tangent line to a polar curve. The solution is to treat the polar equation as a set of parametric equations where the angle $\theta$ serves as the parameter. By expressing $x$ and $y$ in terms of $\theta$, we can leverage the established rules for differentiating parametric functions to find $dy/dx$, the slope of the curve in the $xy$-plane. This allows us to apply the tools of differential calculus to the unique geometry of polar curves.

Key Formulas

The entire process of differentiating in polar form is built upon the conversion between coordinate systems and the rule for parametric derivatives.

Coordinate Conversion

The relationship between polar coordinates $(r,\theta )$ and rectangular coordinates $(x,y)$ is defined by the following equations:

$$x=r\cos (\theta )$$

$$y=r\sin (\theta )$$

Parametric Form of a Polar Curve

For a polar curve defined by the function $r=f(\theta )$, we can substitute $f(\theta )$ for $r$ in the conversion formulas to express $x$ and $y$ as functions of the parameter $\theta$:

$$x(\theta )=f(\theta )\cos (\theta )$$

$$y(\theta )=f(\theta )\sin (\theta )$$

The Derivative $dy/dx$ in Polar Form

To find the slope of the tangent line to the polar curve in the $xy$-plane, we use the formula for the derivative of parametric equations, where $\theta$ is the parameter:

$$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}$$

To find the derivatives $dy/d\theta$ and $dx/d\theta$, we must apply the product rule to the parametric forms of $x$ and $y$. Let $r^{\prime }=dr/d\theta =f^{\prime }(\theta )$.

• Derivative of y with respect to \theta:

  $$\frac{dy}{d\theta }=\frac{d}{d\theta }[r\sin (\theta )]=\frac{dr}{d\theta }\sin (\theta )+r\cos (\theta )$$

• Derivative of x with respect to \theta:

  $$\frac{dx}{d\theta }=\frac{d}{d\theta }[r\cos (\theta )]=\frac{dr}{d\theta }\cos (\theta )-r\sin (\theta )$$

Combining these results gives the complete formula for the slope of a polar curve:

$$\frac{dy}{dx}=\frac{r^{\prime }\sin (\theta )+r\cos (\theta )}{r^{\prime }\cos (\theta )-r\sin (\theta )}$$

Understanding the Connection to Parametric Equations

The most critical conceptual leap in this topic is recognizing that a polar equation is simply a special case of a parametric equation. The Essential Knowledge statements guide us to treat $\theta$ as the parameter, just as $t$ is often used in general parametric contexts.

By defining $x=r\cos (\theta )$ and $y=r\sin (\theta )$, and knowing that $r$ itself is a function of $\theta$ ($r=f(\theta )$), we have successfully defined both $x$ and $y$ in terms of a single variable, $\theta$. This is the exact definition of a set of parametric equations.

Therefore, we do not need to learn a "new" rule for polar derivatives. Instead, we apply the existing parametric derivative rule:

$$\frac{dy}{dx}=\frac{\text{derivative of the y-component}}{\text{derivative of the x-component}}$$

The only new work is correctly applying the product rule to find the derivatives of the specific component functions $x(\theta )=f(\theta )\cos (\theta )$ and $y(\theta )=f(\theta )\sin (\theta )$. This insight simplifies the topic by connecting it directly to prior knowledge of parametric calculus.

Core Concepts & Rules

• Coordinate Conversion is Key: The foundation of polar calculus rests on the conversion formulas $x=r\cos (\theta )$ and $y=r\sin (\theta )$.

• Polar Curves are Parametric: Any polar curve $r=f(\theta )$ can be rewritten as a pair of parametric equations with parameter $\theta$: $x(\theta )=f(\theta )\cos (\theta )$ and $y(\theta )=f(\theta )\sin (\theta )$.

• Slope is $dy/dx$: The slope of the tangent line to a polar curve is $dy/dx$, not $dr/d\theta$. The term $dr/d\theta$ represents the rate of change of the radius with respect to the angle.

• Use the Parametric Derivative Formula: The slope $dy/dx$ is calculated as the ratio of the derivatives of the component functions: $(dy/d\theta )/(dx/d\theta )$.

• Product Rule is Essential: Calculating $dy/d\theta$ and $dx/d\theta$ requires the correct application of the product rule, since $r$ is a function of $\theta$.

• Horizontal Tangents: A polar curve has a horizontal tangent line when $dy/d\theta =0$ and $dx/d\theta \ne 0$.

• Vertical Tangents: A polar curve has a vertical tangent line when $dx/d\theta =0$ and $dy/d\theta \ne 0$.

Step-by-Step Example 1: Finding the Slope at a Point

Problem: Find the slope of the tangent line to the cardioid $r=1+\cos (\theta )$ at the point where $\theta =\pi /3$.

Step 1: Identify the function and find $dr/d\theta$.

The polar function is $r=1+\cos (\theta )$.

First, we find the derivative of $r$ with respect to $\theta$:

$$\frac{dr}{d\theta }=-\sin (\theta )$$

Step 2: Write the general expressions for $dy/d\theta$ and $dx/d\theta$.

Using the formulas derived from the product rule:

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

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

Step 3: Evaluate $r$, $dr/d\theta$, $dy/d\theta$, and $dx/d\theta$ at $\theta =\pi /3$.

First, evaluate the trigonometric functions and $r$ at $\theta =\pi /3$:

• $\sin (\pi /3)=\sqrt{3}/2$

• $\cos (\pi /3)=1/2$

• $r=1+\cos (\pi /3)=1+1/2=3/2$

• $dr/d\theta =-\sin (\pi /3)=-\sqrt{3}/2$

Now, substitute these values into the expressions for $dy/d\theta$ and $dx/d\theta$:

$$\frac{dy}{d\theta }{|}_{\theta =\pi /3}=(-\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2})+(\frac{3}{2})(\frac{1}{2})=-\frac{3}{4}+\frac{3}{4}=0$$

$$\frac{dx}{d\theta }{|}_{\theta =\pi /3}=(-\frac{\sqrt{3}}{2})(\frac{1}{2})-(\frac{3}{2})(\frac{\sqrt{3}}{2})=-\frac{\sqrt{3}}{4}-\frac{3\sqrt{3}}{4}=-\frac{4\sqrt{3}}{4}=-\sqrt{3}$$

Step 4: Calculate $dy/dx$.

Form the ratio to find the slope:

$$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{0}{-\sqrt{3}}=0$$

Conclusion: The slope of the tangent line to $r=1+\cos (\theta )$ at $\theta =\pi /3$ is 0. This indicates a horizontal tangent line at that point.

Step-by-Step Example 2: Exam-Style Application

Problem: Find all values of $\theta$ in the interval $[0,\pi ]$ for which the polar curve $r=2\cos (\theta )+1$ has a horizontal or vertical tangent line.

Step 1: Find expressions for $dy/d\theta$ and $dx/d\theta$.

Given $r=2\cos (\theta )+1$, we first find $dr/d\theta$:

$$\frac{dr}{d\theta }=-2\sin (\theta )$$

Now, we use the general formulas for $dy/d\theta$ and $dx/d\theta$:

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

$$=-2{\sin }^2(\theta )+2{\cos }^2(\theta )+\cos (\theta )$$

Using the identity ${\sin }^2(\theta )=1-{\cos }^2(\theta )$:

$$=-2(1-{\cos }^2(\theta ))+2{\cos }^2(\theta )+\cos (\theta )=4{\cos }^2(\theta )+\cos (\theta )-2$$

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

$$=-2\sin (\theta )\cos (\theta )-2\sin (\theta )\cos (\theta )-\sin (\theta )$$

$$=-4\sin (\theta )\cos (\theta )-\sin (\theta )=-\sin (\theta )(4\cos (\theta )+1)$$

Step 2: Find $\theta$ values for horizontal tangents.

Set $dy/d\theta =0$:

$$4{\cos }^2(\theta )+\cos (\theta )-2=0$$

This is a quadratic equation in terms of $\cos (\theta )$. Using the quadratic formula where $a=4,b=1,c=-2$:

$$\cos (\theta )=\frac{-1\pm \sqrt{1^2-4(4)(-2)}}{2(4)}=\frac{-1\pm \sqrt{1+32}}{8}=\frac{-1\pm \sqrt{33}}{8}$$

We have two possible values for $\cos (\theta )$:

• $\cos (\theta )=\frac{-1+\sqrt{33}}{8}\approx 0.593$. Since this value is between -1 and 1, there is a solution in $[0,\pi ]$. $\theta =\arccos \left(\frac{-1+\sqrt{33}}{8}\right)$.

• $\cos (\theta )=\frac{-1-\sqrt{33}}{8}\approx -0.843$. Since this value is between -1 and 1, there is a solution in $[0,\pi ]$. $\theta =\arccos \left(\frac{-1-\sqrt{33}}{8}\right)$.

At these $\theta$ values, $dx/d\theta$ is not zero, so they correspond to horizontal tangents.

Step 3: Find $\theta$ values for vertical tangents.

Set $dx/d\theta =0$:

$$-\sin (\theta )(4\cos (\theta )+1)=0$$

This gives two possibilities:

• $\sin (\theta )=0⟹\theta =0,\pi$

• $4\cos (\theta )+1=0⟹\cos (\theta )=-1/4$. This gives a solution $\theta =\arccos (-1/4)$ in the interval $[0,\pi ]$.

At these $\theta$ values, $dy/d\theta$ is not zero, so they correspond to vertical tangents.

Conclusion:

• Horizontal tangents occur at $\theta =\arccos \left(\frac{-1+\sqrt{33}}{8}\right)$ and $\theta =\arccos \left(\frac{-1-\sqrt{33}}{8}\right)$.

• Vertical tangents occur at $\theta =0$, $\theta =\pi$, and $\theta =\arccos (-1/4)$.

Using Your Calculator

While the differentiation process is analytical, a graphing calculator is invaluable for verification and evaluation, especially on calculator-active sections of the AP Exam.

To find the slope $dy/dx$ of $r=f(\theta )$ at $\theta =A$:

1. Mode Setting: Ensure your calculator is in Polar and Radian mode.

2. Function Entry: In the Y= editor, enter your polar function as $r1=f(\theta )$.

3. Graph (Optional): Graph the function to visually inspect the behavior at the point of interest. This can help you confirm if the slope should be positive, negative, zero, or undefined.

4. Numerical Calculation: Use the numerical derivative feature (e.g., nDeriv on TI-84 or $d/d()$ from the math template) to calculate $dy/d\theta$ and $dx/d\theta$ at $\theta =A$.

   • Recall that $y=r\sin (\theta )$ and $x=r\cos (\theta )$.

   • To find $dy/d\theta$ at $\theta =A$, enter: nDeriv(r1*sin(\theta), \theta, A)` - To find $dx/d\theta at $\theta =A$, enter: nDeriv(r1*cos(\theta), \theta, A)` 5. **Compute the Ratio:** Divide the result for $dy/d\theta by the result for $dx/d\theta$ to get the value of $dy/dx$.

Example: For $r=1+cos(\theta )$ at $\theta =\pi /3$ from Example 1.

• $r1=1+cos(\theta )$

• $dy/d\theta$: $nDeriv(r1\ast sin(\theta ),\theta ,\pi /3)$ will return a value extremely close to $0$.

• $dx/d\theta$: $nDeriv(r1\ast cos(\theta ),\theta ,\pi /3)$ will return approximately $-1.732$ (which is $-\surd 3$).

• The ratio is $0/-1.732=0$.

AP Exam Quick Hit

Common Question Types

• Find the slope at a point: Given a polar function $r=f(\theta )$ and a specific angle ${\theta }_0$, you will be asked to find the value of $dy/dx$ at that angle. This is a direct application of the main formula.

  • Example: "Find the slope of the tangent line to the curve $r=3\theta$ at $\theta =\pi /2$."

• Write the equation of a tangent line: This is a two-step problem. First, find the slope $dy/dx$ at the given $\theta$. Second, find the rectangular coordinates $(x,y)$ of that point using $x=r\cos (\theta )$ and $y=r\sin (\theta )$. Finally, use the point-slope form $y-y_1=m(x-x_1)$.

  • Example: "Write an equation for the line tangent to the graph of $r=2-\sin (\theta )$ at the point where $\theta =\pi$."

• Find points of horizontal or vertical tangency: You will be given a polar function and asked to find the $\theta$ values (or the $(x,y)$ or $(r,\theta )$ coordinates) where the tangent line is horizontal or vertical. This involves setting $dy/d\theta =0$ or $dx/d\theta =0$ and solving for $\theta$.

  • Example: "For the polar curve $r=1+\sin (\theta )$, find the $\theta$ values on $[0,2\pi )$ for which the tangent line is vertical."

Common Mistakes

• Confusing $dy/dx$ with $dr/d\theta$: The most common conceptual error. Students calculate $dr/d\theta$ and state that it is the slope. Remember, $dy/dx$ is the slope in the Cartesian plane; $dr/d\theta$ is the rate at which the radius changes as the angle increases.

• Forgetting the Product Rule: When differentiating $x=r\cos (\theta )$ and $y=r\sin (\theta )$, students often forget that $r$ is a function of $\theta$ and fail to apply the product rule. They might incorrectly write $dx/d\theta =-r\sin (\theta )$, forgetting the $(dr/d\theta )\cos (\theta )$ term.

• Algebraic Errors in the Derivative: The expressions for $dy/d\theta$ and $dx/d\theta$ can become complex. Errors in distributing, combining terms, or using trigonometric identities are frequent. It is often safer to evaluate the components at the specific $\theta$ value before simplifying the general expression.

• Mixing up Horizontal and Vertical Conditions: Students may incorrectly set $dx/d\theta =0$ for horizontal tangents or $dy/d\theta =0$ for vertical tangents. Remember: horizontal lines have zero slope ($dy/dx=0$, so numerator $dy/d\theta =0$), and vertical lines have undefined slope ($dy/dx$ is undefined, so denominator $dx/d\theta =0$).

• Not Checking for $0/0$: When finding horizontal or vertical tangents, if a $\theta$ value makes both $dy/d\theta$ and $dx/d\theta$ equal to zero, the slope is indeterminate ($0/0$). This often occurs at the pole ($r=0$) and requires further analysis (like L'Hôpital's Rule on the ratio), though on the AP exam, it's most important to recognize that the standard conditions are not met.