A polar point is written \((r,\theta)\): start facing along the positive x-axis, rotate by the angle \(\theta\), then walk a distance \(r\) outward. That's the location.
To move between the two systems, drop a right triangle from the point. The conversions are just the legs of that triangle:
A polar curve \(r=f(\theta)\) gives a rule: for each angle, how far out to go. As the angle sweeps around, the point traces a shape. The magic is that shapes which need messy equations in \(x\) and \(y\) — circles off-center, loops, petals — often have beautifully simple polar forms.
This is the whole idea in one picture. The amber arm is the current angle \(\theta\); its length is \(r=f(\theta)\) for that angle. As \(\theta\) sweeps from \(0\) around, the tip of the arm traces the cyan curve. Pick a curve and hit play — or scrub the angle by hand and watch the arm swing and stretch.
• Two views, one motion. Left is the polar curve; right is the same \(r\) plotted
the ordinary way — \(r\) up, \(\theta\) across. The dots move together, so you can see exactly which
angle produces which point.
• The negative-r reveal. Watch the right graph dip into the shaded band below the
\(r=0\) line. The instant \(r\) goes negative, the polar point (and its arm) flips to the
opposite side — that's the whole secret of how roses grow
petals and limaçons grow loops, shown directly.
• The rose's k. Pick the rose and drag the k slider — petals appear and vanish in
real time, and the label shows the odd/even rule as you go.
When \(f(\theta)\) is negative, you don't go forward along the angle — you plot the point in the opposite direction, \(180°\) around. So \((-1,\tfrac{\pi}{3})\) is the same point as \((1,\tfrac{4\pi}{3})\). This is exactly how a rose grows extra petals and a limaçon grows its inner loop — the curve swings to the other side whenever \(r\lt 0\).
A handful of forms cover almost everything the exam throws at you. Learn to recognize them at a glance from the equation:
For a rose \(r=a\cos(k\theta)\) or \(r=a\sin(k\theta)\): if \(k\) is odd, you get \(k\) petals; if \(k\) is even, you get \(2k\) petals. (The odd case looks like it should give \(2k\), but negative \(r\) makes the second set of petals land exactly on top of the first.)
"What's \(\tfrac{dy}{dx}\) at this point?" still makes sense for a polar curve — it's the slope of the tangent line in the ordinary xy-plane. The trick: convert to parametric form using \(\theta\) as the parameter, then use the parametric slope rule from the start of this unit.
Now \(x\) and \(y\) are both functions of \(\theta\) (remember \(r=f(\theta)\), so each is a product). Differentiate with the product rule and divide, exactly like parametric:
Since \(x=f(\theta)\cos\theta\), the product rule gives \(\tfrac{dx}{d\theta}=f'(\theta)\cos\theta-f(\theta)\sin\theta=r'\cos\theta-r\sin\theta\). Likewise \(\tfrac{dy}{d\theta}=r'\sin\theta+r\cos\theta\). You only need \(r\) and its derivative \(r'\).
See it on a curve. Drag \(\theta\) around this cardioid \(r=1+\cos\theta\) and watch the tangent line tilt. The slope \(\tfrac{dy}{dx}\) updates live — notice it reads horizontal (slope 0) and vertical (undefined) at exactly the spots your eye expects.
A natural mistake is to report \(\tfrac{dr}{d\theta}\) as the slope. But \(\tfrac{dr}{d\theta}\) only measures how fast the radius grows as you sweep the angle — it says nothing about the tilt of the curve in the plane. The actual tangent slope is \(\tfrac{dy}{dx}\), so you always have to go through \(x\) and \(y\) first.
Convert the polar point \((4,\tfrac{2\pi}{3})\) to rectangular coordinates.
Use \(x=r\cos\theta,\ y=r\sin\theta\). (−2, 2√3)
What shape is \(r=4\cos(2\theta)\), and how many petals does it have?
Recognize the form. It's a rose, \(r=a\cos(k\theta)\), with \(a=4\), \(k=2\).
Apply the petal rule. \(k=2\) is even, so the rose has \(2k=4\) petals, each reaching out to a maximum radius of \(4\). 4-petal rose, radius 4
Find \(\tfrac{dy}{dx}\) for \(r=2\cos\theta\) at \(\theta=\tfrac{\pi}{4}\).
Set up. Here \(r=2\cos\theta\) and \(r'=-2\sin\theta\). At \(\theta=\tfrac{\pi}{4}\): \(r=\sqrt2\), \(r'=-\sqrt2\).
Build the two derivatives.
Divide. \(\tfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}=\dfrac{0}{-2}=0\) — a horizontal tangent, which makes sense: \(\theta=\tfrac{\pi}{4}\) lands at the top of this circle. dy/dx = 0
Five questions on conversions, recognizing curves, the petal rule, negative r, and polar slope.