Rates of Change in Polar Functions - AP PreCalculus Study Guide

The Core Idea: Rates of Change in Polar Functions

In the polar coordinate system, a point's location is described by its distance from the origin (the pole), $r$, and its angle relative to the positive x-axis, $\theta$. The function $r=f(\theta )$ defines a curve by specifying this directed distance $r$ for any given angle $\theta$. The central concept of this topic is to analyze how this distance from the pole changes as the angle $\theta$ increases.

We are not concerned with the overall speed of a point moving along the curve, but specifically with the rate at which it is moving directly toward or away from the pole. This is determined by finding the derivative of the distance function $r$ with respect to the angle $\theta$. This derivative, $dr/d\theta$, provides a precise measure of this rate of change, allowing us to determine if the point is getting closer to, farther from, or is at a maximum or minimum distance from the pole at any given angle.

Key Formulas/Rules/Theorems

The primary rule for analyzing the rate of change in polar functions is based on the derivative of the polar equation with respect to the angle $\theta$.

Rate of Change of Directed Distance from the Pole

Given a polar function defined by $r=f(\theta )$, the rate of change of the directed distance from the pole with respect to the angle $\theta$ is given by its derivative:

$$\frac{dr}{d\theta }=f^{\prime }(\theta )$$

• $r$: The directed distance from the pole.

• $\theta$: The angle in radians.

• $dr/d\theta$: The instantaneous rate at which the distance $r$ is changing as $\theta$ increases.

Understanding the Sign of the Derivative

The sign of the derivative $dr/d\theta$ provides a crucial interpretation of the point's motion relative to the pole as the angle $\theta$ increases. It does not describe movement in the x or y direction, but exclusively movement toward or away from the origin.

• If $dr/d\theta >0$: The rate of change of the distance is positive. This means the distance $r$ from the pole is increasing. Therefore, the point on the curve is moving away from the pole.

• If $dr/d\theta <0$: The rate of change of the distance is negative. This means the distance $r$ from the pole is decreasing. Therefore, the point on the curve is moving toward the pole.

• If $dr/d\theta =0$: The rate of change of the distance is zero. This means the distance $r$ is momentarily not changing. These points are candidates for where the curve is closest to or farthest from the pole. To determine the absolute maximum or minimum distance on a given interval, one must evaluate $r$ at the angles where $dr/d\theta =0$ and at the endpoints of the interval.

Core Concepts & Rules

• Distance Function: The equation $r=f(\theta )$ explicitly defines the directed distance ($r$) of a point from the pole for any angle $\theta$.

• Rate of Change: The derivative, $dr/d\theta =f^{\prime }(\theta )$, measures the instantaneous rate of change of the distance from the pole with respect to the angle $\theta$.

• Moving Away: A positive value for $dr/d\theta$ indicates the point on the curve is moving away from the pole as $\theta$ increases.

• Moving Toward: A negative value for $dr/d\theta$ indicates the point on the curve is moving toward the pole as $\theta$ increases.

• Finding Extrema: The angles $\theta$ at which a point on the curve is farthest from or closest to the pole can be determined by analyzing the derivative $dr/d\theta$. These extrema often occur where $dr/d\theta =0$.

Step-by-Step Example 1: Interpreting the Rate of Change

Problem: Consider the polar curve given by the equation $r=4\cos (\theta )$. At the angle $\theta =\pi /6$, is the point on the curve moving toward or away from the pole? Justify your answer.

Step 1: Identify the function and find its derivative.

The function describing the distance from the pole is $r=f(\theta )=4\cos (\theta )$. We need to find the rate of change, $dr/d\theta$.

$$\frac{dr}{d\theta }=\frac{d}{d\theta }(4\cos (\theta ))$$

Using the derivative rule for cosine, we get:

$$\frac{dr}{d\theta }=-4\sin (\theta )$$

Step 2: Evaluate the derivative at the specified angle.

The problem asks for the behavior at $\theta =\pi /6$. We substitute this value into our derivative.

$${\frac{dr}{d\theta }|}_{\theta =\pi /6}=-4\sin \left(\frac{\pi }{6}\right)$$

We know that $sin(\pi /6)=1/2$.

$${\frac{dr}{d\theta }|}_{\theta =\pi /6}=-4\left(\frac{1}{2}\right)=-2$$

Step 3: Interpret the sign of the derivative.

The value of the derivative at $\theta =\pi /6$ is $-2$, which is negative ($dr/d\theta <0$).

Conclusion:

Because $dr/d\theta =-2<0$ at $\theta =\pi /6$, the distance from the pole is decreasing at this angle. Therefore, the point on the curve is moving toward the pole.

Step-by-Step Example 2: Finding Maximum Distance from the Pole

Problem: For the polar curve $r=2+3\sin (\theta )$ on the interval $0\le \theta \le 2\pi$, find the angle $\theta$ at which the point is farthest from the pole and state this maximum distance.

Step 1: Find the derivative of the distance function.

The distance from the pole is given by $r=f(\theta )=2+3\sin (\theta )$. To find where this distance might be a maximum or minimum, we first find the derivative, $dr/d\theta$.

$$\frac{dr}{d\theta }=\frac{d}{d\theta }(2+3\sin (\theta ))=3\cos (\theta )$$

Step 2: Find the critical points by setting the derivative to zero.

The distance $r$ can be maximized or minimized when its rate of change is zero. We set $dr/d\theta =0$ and solve for $\theta$ within the interval $[0,2\pi ]$.

$$3\cos (\theta )=0$$

$$\cos (\theta )=0$$

This equation is true for $\theta =\pi /2$ and $\theta =3\pi /2$. These are our critical angles.

Step 3: Evaluate the distance $r$ at the critical points and the endpoints of the interval.

To find the absolute maximum distance on the closed interval $[0,2\pi ]$, we must test the critical angles we found ($\pi /2$, $3\pi /2$) and the endpoints of the interval ($0$, $2\pi$). We substitute these values back into the original distance function, $r=2+3\sin (\theta )$.

• At endpoint $\theta =0$:

  $r(0)=2+3\sin (0)=2+3(0)=2$

• At critical point $\theta =\pi /2$:

  $r(\pi /2)=2+3\sin (\pi /2)=2+3(1)=5$

• At critical point $\theta =3\pi /2$:

  $r(3\pi /2)=2+3\sin (3\pi /2)=2+3(-1)=-1$

• At endpoint $\theta =2\pi$:

  $r(2\pi )=2+3\sin (2\pi )=2+3(0)=2$

Note: A negative $r$ value means the point is located in the opposite quadrant from the angle, but its distance from the pole is the absolute value, $|r|$. The directed distance is $-1$, but the actual distance is $1$.

Step 4: Compare the values and state the conclusion.

The calculated distances are 2, 5, 1 (from $|-1|$), and 2. The largest value is 5.

Conclusion:

The point is farthest from the pole at the angle $\theta =\pi /2$. The maximum distance from the pole is 5.

Using Your Calculator

A graphing calculator is a powerful tool for analyzing rates of change in polar functions, especially for equations where finding the derivative or solving for $\theta$ is difficult by hand. Ensure your calculator is always in Radian Mode.

Problem: For the polar curve $r=1+\sqrt{\theta }\cos (\theta )$ on $0<\theta \le 4\pi$, find the first angle $\theta$ where the point is moving away from the pole at a rate of 0.5.

Step 1: Input the function and its derivative.

• In your calculator's function editor, enter the polar function r(\theta) into Y1. Use`X$ for $\theta$.

  Y1 = 1 + √(X) * cos(X) - To analyze the rate of change, you need $dr/d\theta. Use the calculator's numerical derivative function (often called nDerivord/dx) to graph the derivative in Y2`.

  Y2 = nDeriv(Y1, X, X)

  This tells the calculator to take the derivative of Y1 with respect to $X$, evaluated at $X$.

Step 2: Set up the equation to solve.

The problem asks for when the point is moving away from the pole at a rate of 0.5. This means we need to find where dr/d\theta = 0.5`. - In `Y3`, enter the target rate. `Y3 = 0.5` **Step 3: Graph and find the intersection.** - Set your viewing window appropriately for the interval. For $0 < \theta \le 4\pi, a window of $Xmin=0$, $Xmax=4\pi$ (approx. 12.57) is suitable. Adjust $Ymin$ and $Ymax‘toseetheintersectionof‘Y2‘and‘Y3‘.-Graph‘Y2‘(thederivative)and‘Y3‘(theconstantrate).-Usethecalculato{r}^{\prime }s"intersect"feature(oftenfoundinthe‘CALC‘menu)tofindthepointwherethetwographscross.-Thecalculatorwillaskyoutoselectthefirstcurve(‘Y2‘),thesecondcurve(‘Y3‘),andprovideaguess.Movethecursornearthefirstintersectionpointandpressenter.\ast \ast Step4:Statetheanswer.\ast \ast Thecalculatorwilldisplaythecoordinatesoftheintersection.Thex-coordinatecorrespondstotheangle$\theta$. For this problem, the first intersection occurs at approximately $\theta \approx 2.503$.

AP Exam Quick Hit

Common Question Types

• Interpreting the Sign of $dr/d\theta$: You will be given a polar function $r=f(\theta )$ and an angle $\theta =c$. You'll be asked to determine if the point is moving toward or away from the pole at that angle and to justify your answer by calculating and interpreting the sign of $f^{\prime }(c)$.

  • Example: "For the curve $r=3-3\sin (\theta )$, is the particle moving toward or away from the pole at $\theta =\pi$? Show the work that leads to your answer."

• Finding Maximum/Minimum Distance: You will be given a polar function $r=f(\theta )$ on a closed interval and asked to find the angle $\theta$ that corresponds to the point's maximum or minimum distance from the pole. This requires finding where $dr/d\theta =0$ and testing those critical points along with the interval endpoints.

  • Example: "Find the minimum distance from the pole for a point on the graph of $r=5+2\cos (\theta )$."

• Calculator-Based Rate Problems: In a calculator-active question, you may be given a more complex polar function and asked to find the value of $\theta$ where the rate of change $dr/d\theta$ is equal to a specific non-zero value.

  • Example: "A particle's position is given by $r=e^{0.2\theta }\sin (\theta )$. At what value of $\theta$ on the interval $[0,\pi ]$ is the particle moving toward the pole at a rate of 1 unit per radian?"

Common Mistakes

• Confusing $r$ with $dr/d\theta$: A common error is to evaluate $r$ at a given $\theta$ to determine motion. A large positive $r$ value does not mean the point is moving away from the pole. Motion is determined only by the sign of the derivative, $dr/d\theta$.

• Incorrect Interpretation of the Sign: Stating that $dr/d\theta >0$ means the point is "moving up" or that $dr/d\theta <0$ means it is "moving left." The sign of $dr/d\theta$only describes motion relative to the pole: away from the pole (positive) or toward the pole (negative).

• Forgetting to Check Endpoints: When finding the absolute maximum or minimum distance on a closed interval $[a,b]$, students frequently find the critical points where $dr/d\theta =0$ but forget to also evaluate $r$ at the endpoints $\theta =a$ and $\theta =b$. The absolute maximum or minimum could occur at an endpoint.

• Basic Derivative Errors: Simple mistakes in differentiating trigonometric functions, such as forgetting the negative sign when differentiating cosine ($d/d\theta (\cos \theta )=-\sin \theta$) or applying the product/chain rule incorrectly.

• Calculator in Degree Mode: All calculus operations in AP Precalculus, especially those involving trigonometric functions in polar coordinates, assume the angle $\theta$ is in radians. Performing calculations in Degree mode will produce incorrect answers.