Chapter 26 — Quiz

10 questions covering conversion, the polar curve gallery, slope, area (the sector formula), area between curves, arc length, and orbits. Try each before opening the answer. Section references point back to index.md.

1. Which pair of formulas converts a polar point $(r,\theta)$ to Cartesian $(x,y)$?

  • A) $x = r\sin\theta,\quad y = r\cos\theta$
  • B) $x = r\cos\theta,\quad y = r\sin\theta$
  • C) $x = r + \theta,\quad y = r - \theta$
  • D) $x = \sqrt{r^2+\theta^2},\quad y = \arctan(\theta/r)$
Answer**B.** Reading the right triangle with hypotenuse $r$ at angle $\theta$: horizontal leg $x = r\cos\theta$, vertical leg $y = r\sin\theta$. (*Section 26.1.*)

2. A student converts the Cartesian point $(-1,-1)$ to polar and writes $\theta = \arctan\!\frac{-1}{-1} = \arctan(1) = \pi/4$. What went wrong, and what is the correct angle (with $r>0$)?

  • A) Nothing is wrong; $\theta = \pi/4$.
  • B) The point is in the third quadrant; $\arctan$ lost the sign information, so $\theta = 5\pi/4$.
  • C) The radius should be negative; $\theta = \pi/4$ with $r = -\sqrt2$.
  • D) The point is on the $y$-axis; $\theta = 3\pi/2$.
Answer**B.** Both $(1,1)$ and $(-1,-1)$ give the ratio $y/x = 1$, but they sit in opposite quadrants. Since $(-1,-1)$ is in the third quadrant, $\theta = \pi + \pi/4 = 5\pi/4$. Using `atan2(y,x)` avoids this trap. (*Section 26.1; Common Pitfall.*)

3. The polar equation $r = 2\cos\theta$ describes which curve?

  • A) A cardioid pointing right
  • B) A circle of radius $1$ centered at $(1,0)$, passing through the pole
  • C) A line through the origin
  • D) A two-petal rose
Answer**B.** Multiply by $r$: $r^2 = 2r\cos\theta \Rightarrow x^2+y^2 = 2x \Rightarrow (x-1)^2 + y^2 = 1$. A circle of radius $1$ centered at $(1,0)$. (*Section 26.1 / 26.2.*)

4. How many petals does the rose $r = \cos(4\theta)$ have?

  • A) 2
  • B) 4
  • C) 8
  • D) 16
Answer**C — 8.** The rule is $n$ petals when $n$ is odd, but $2n$ petals when $n$ is even. Here $n=4$ is even, so $2n = 8$. (The tempting wrong answer is 4.) (*Section 26.2; petal-counting rule.*)

5. A polar curve $r = f(\theta)$ passes through the pole at $\theta = \theta_0$ (so $f(\theta_0)=0$). In what direction does the tangent line point there?

  • A) Always horizontal
  • B) Along the ray $\theta = \theta_0$
  • C) Perpendicular to the ray $\theta = \theta_0$
  • D) Undefined in every case
Answer**B.** When $r=0$, the slope formula collapses to $\dfrac{dy}{dx} = \dfrac{r'\sin\theta_0}{r'\cos\theta_0} = \tan\theta_0$, so the tangent points along $\theta = \theta_0$. This is how you find the directions a rose's petals leave the origin. (*Section 26.3.*)

6. The area enclosed by a polar curve $r = f(\theta)$ swept from $\theta = \alpha$ to $\theta = \beta$ is:

  • A) $\displaystyle\int_\alpha^\beta r\,d\theta$
  • B) $\displaystyle\int_\alpha^\beta r^2\,d\theta$
  • C) $\dfrac{1}{2}\displaystyle\int_\alpha^\beta r^2\,d\theta$
  • D) $\dfrac{1}{2}\displaystyle\int_\alpha^\beta r\,d\theta$
Answer**C.** The natural slice is a thin circular sector of area $\tfrac12 r^2\,d\theta$, not a rectangle. The factor $\tfrac12$ and the $r^2$ *are* the sector area. Forgetting the $\tfrac12$ is the most common polar-area error. (*Section 26.5.*)

7. Find the area enclosed by the cardioid $r = 1 + \cos\theta$.

  • A) $\pi$
  • B) $\dfrac{3\pi}{2}$
  • C) $2\pi$
  • D) $\dfrac{\pi}{2}$
Answer**B — $3\pi/2$.** $A = \tfrac12\int_0^{2\pi}(1+\cos\theta)^2\,d\theta = \tfrac12\int_0^{2\pi}(1 + 2\cos\theta + \cos^2\theta)\,d\theta = \tfrac12(2\pi + 0 + \pi) = \tfrac{3\pi}{2}$, using $\int_0^{2\pi}\cos^2\theta\,d\theta = \pi$. (*Section 26.5, Worked Example 26.5.1.*)

8. The arc length of a polar curve $r = f(\theta)$ is $\displaystyle\int_\alpha^\beta \sqrt{r^2 + (dr/d\theta)^2}\,d\theta$. The two terms under the root represent, respectively:

  • A) horizontal and vertical motion
  • B) tangential (swing around the pole) and radial (toward/away from the pole) motion
  • C) area and perimeter
  • D) the cosine and sine components of $r$
Answer**B.** The $r^2$ term is tangential motion (arc $r\,d\theta$ around the pole); the $(r')^2$ term is radial motion ($dr$ along the spoke). These are perpendicular, so Pythagoras combines them into speed $\sqrt{r^2+(r')^2}$. (*Section 26.7.*)

9. Orbits under gravity are conics in polar form $r = \dfrac{p}{1 + e\cos\theta}$ with the attractor at one focus. The orbit is an ellipse (a bound planet) precisely when:

  • A) $e = 0$
  • B) $0 < e < 1$
  • C) $e = 1$
  • D) $e > 1$
Answer**B.** $e=0$ is a circle, $01$ a hyperbola (unbound flyby). The full geometry of these conics is studied in Chapter 27. (*Section 26.8.*)

10. Kepler's Second Law states that a planet sweeps out equal areas in equal times. In the language of this chapter, the swept-area rate is $\dfrac{dA}{dt} = \tfrac12 r^2\dfrac{d\theta}{dt}$. The law says this quantity is constant, which is equivalent to the conservation of:

  • A) energy
  • B) angular momentum
  • C) mass
  • D) the eccentricity
Answer**B — angular momentum.** $\tfrac12 r^2\dot\theta = $ const is exactly $r^2\dot\theta = $ const, the conservation of angular momentum. Kepler's empirical "equal areas" law and Newton's conservation law are the *same equation*, bridged by the polar area element $\tfrac12 r^2\,d\theta$. (*Section 26.8.*)

Scoring Guide

Score Interpretation
9–10 Excellent. You command conversion, the curve gallery, and the calculus of polar regions. Move on to Chapter 27 (conics).
7–8 Solid. Re-read the section flagged by any miss — most likely the petal rule (Q4), the $\tfrac12$ in area (Q6/Q7), or the arc-length geometry (Q8).
5–6 Shaky. Revisit Sections 26.2 (curve gallery), 26.5 (sector area), and 26.7 (arc length), then redo the worked examples by hand.
0–4 Restart with Sections 26.1–26.2. Practice converting points and sketching the standard curves until the gallery is automatic; the calculus rests on it.