The Rate of Change of a Polar Function

Distance that changes with direction

On the cardioid $r=1+\cos \theta$r = 1 + cos θ, the distance from the pole depends on the direction. It is 2 along the initial line, 1 straight up at $\theta =\frac{\pi }{2}$θ = π/2, and 0 at $\theta =\pi$θ = π, where the curve reaches the pole.

So as $\theta$θ turns, r changes, and the curve moves in toward the pole or out away from it. The rate at which r changes as $\theta$θ turns is $\frac{dr}{d\theta }$dr/dθ.

Plot r against $\theta$θ

The polar sketch shows where the curve goes, but its horizontal direction is not $\theta$θ, so a rate cannot be read from it. Plot r against $\theta$θ on ordinary axes instead, with $\theta$θ across and r up.

For $r=1+\cos \theta$r = 1 + cos θ the graph starts at 2 when $\theta =0$θ = 0, falls to 1 at $\frac{\pi }{2}$π/2 and to 0 at $\pi$π, then rises through 1 at $\frac{3\pi }{2}$3π/2 back to 2 at $2\pi$2π. It is the cosine graph lifted up by 1.

Differentiate r

r is an ordinary function of $\theta$θ, so differentiate it in the ordinary way. The constant 1 gives 0 and $\cos \theta$cos θ gives $-\sin \theta$−sin θ, so $\frac{dr}{d\theta }=-\sin \theta$dr/dθ = −sin θ.

At $\theta =\frac{\pi }{6}$θ = π/6, $\frac{dr}{d\theta }=-\frac{1}{2}$dr/dθ = −½: r is falling at half a unit for each radian turned. At $\theta =\frac{\pi }{2}$θ = π/2, $\frac{dr}{d\theta }=-1$dr/dθ = −1. At $\theta =\frac{3\pi }{2}$θ = 3π/2, $\frac{dr}{d\theta }=1$dr/dθ = 1, and r is rising.

Moving in, moving out

For $0<\theta <\pi$0 < θ < π, $\sin \theta >0$sin θ > 0, so $\frac{dr}{d\theta }=-\sin \theta <0$dr/dθ = −sin θ < 0. r is falling, and the curve moves in toward the pole along the whole upper half.

For $\pi <\theta <2\pi$π < θ < 2π, $\sin \theta <0$sin θ < 0, so $\frac{dr}{d\theta }>0$dr/dθ > 0. r is rising, and the curve moves back out along the lower half.

Where r stops changing

$\frac{dr}{d\theta }=-\sin \theta$dr/dθ = −sin θ is zero at $\theta =0$θ = 0 and $\theta =\pi$θ = π. At $\theta =0$θ = 0, r = 2: the derivative changes from positive just before a full turn to negative just after, so r has a maximum, the point farthest from the pole.

At $\theta =\pi$θ = π, r = 0: the derivative changes from negative to positive, so r has a minimum, and the curve is at the pole.

The average rate

For an average rate of change across a stretch of $\theta$θ, divide the change in r by the change in $\theta$θ. From $\theta =0$θ = 0 to $\theta =\frac{\pi }{2}$θ = π/2, r goes from 2 to 1, so the average rate is $\left(1-2\right)/\left(\frac{\pi }{2}-0\right)=-\frac{2}{\pi }\approx -0.637$(1 − 2) / (π/2 − 0) = −2/π ≈ −0.637.

That lies between the rates at the two ends, 0 at $\theta =0$θ = 0 and −1 at $\theta =\frac{\pi }{2}$θ = π/2. On the graph of r against $\theta$θ it is the gradient of the chord between the two points.

$r=1+\cos \theta$r = 1 + cos θ plotted against $\theta$θ from 0 to $2\pi$2π, falling to 0 at $\pi$π and rising back to 2. The gold line is the tangent at $\theta =\frac{\pi }{2}$θ = π/2, with gradient $\frac{dr}{d\theta }=-1$dr/dθ = −1. The dashed line is the chord from $\theta =0$θ = 0 to $\frac{\pi }{2}$π/2, with gradient $-\frac{2}{\pi }$−2/π, the average rate over that stretch.

$\frac{dr}{d\theta }$dr/dθ is not the gradient of the curve

$\frac{dr}{d\theta }$dr/dθ says how fast the distance from the pole changes. The gradient of the curve on the page is $\frac{dy}{dx}$dy/dx, which is something else. To find it, treat $x=r\cos \theta$x = r cos θ and $y=r\sin \theta$y = r sin θ as parametric equations in $\theta$θ.

By the product rule, with r' standing for $\frac{dr}{d\theta }$dr/dθ, $\frac{dx}{d\theta }=r^{\prime }\cos \theta -r\sin \theta$dx/dθ = r' cos θ − r sin θ and $\frac{dy}{d\theta }=r^{\prime }\sin \theta +r\cos \theta$dy/dθ = r' sin θ + r cos θ. Dividing one by the other, $\frac{dy}{dx}=\left(r^{\prime }\sin \theta +r\cos \theta \right)/\left(r^{\prime }\cos \theta -r\sin \theta \right)$dy/dx = (r' sin θ + r cos θ) / (r' cos θ − r sin θ).

On the cardioid at $\theta =\frac{\pi }{2}$θ = π/2, r = 1 and r' = −1, so $\frac{dx}{d\theta }=-1\times 0-1\times 1=-1$dx/dθ = −1 × 0 − 1 × 1 = −1 and $\frac{dy}{d\theta }=-1\times 1+1\times 0=-1$dy/dθ = −1 × 1 + 1 × 0 = −1. So $\frac{dy}{dx}=1$dy/dx = 1: the tangent at (0, 1) rises at 45°, even though $\frac{dr}{d\theta }=-1$dr/dθ = −1 there.

The highest point of the cardioid is where $\frac{dy}{d\theta }=0$dy/dθ = 0. With $r^{\prime }=-\sin \theta$r' = −sin θ that is $-{\sin }^2\theta +\left(1+\cos \theta \right)\cos \theta =0$−sin²θ + (1 + cos θ)cos θ = 0, which becomes $2{\cos }^2\theta +\cos \theta -1=0$2cos²θ + cos θ − 1 = 0, or $\left(2\cos \theta -1\right)\left(\cos \theta +1\right)=0$(2cos θ − 1)(cos θ + 1) = 0. So $\cos \theta =\frac{1}{2}$cos θ = ½ and $\theta =\frac{\pi }{3}$θ = π/3, where $r=\frac{3}{2}$r = 3/2 and the point is $\left(\frac{3}{4},\frac{3\sqrt{3}}{4}\right)\approx \left(0.75,1.30\right)$(3/4, 3√3/4) ≈ (0.75, 1.30). There $\frac{dr}{d\theta }=-\frac{\sqrt{3}}{2}$dr/dθ = −√3/2, not 0: r is still falling at the top.

The cardioid $r=1+\cos \theta$r = 1 + cos θ. The gold line is its tangent at (0, 1), where $\theta =\frac{\pi }{2}$θ = π/2, with gradient $\frac{dy}{dx}=1$dy/dx = 1. The dashed line is the level tangent at the highest point, $\left(\frac{3}{4},\frac{3\sqrt{3}}{4}\right)$(3/4, 3√3/4), where $\theta =\frac{\pi }{3}$θ = π/3.

The usual mistakes

Losing the minus sign. The derivative of $\cos \theta$cos θ is $-\sin \theta$−sin θ, so $\frac{dr}{d\theta }=-\sin \theta$dr/dθ = −sin θ, and r falls where $\sin \theta$sin θ is positive.

Reading where r is positive or negative instead of where $\frac{dr}{d\theta }$dr/dθ is. For $r=2+\sin \theta$r = 2 + sin θ, r is always positive, but it falls where $\frac{dr}{d\theta }=\cos \theta$dr/dθ = cos θ is negative, for $\frac{\pi }{2}<\theta <\frac{3\pi }{2}$π/2 < θ < 3π/2.

Turning the average rate upside down, or stopping at the change in r. The average rate is the change in r divided by the change in $\theta$θ: $-1/\left(\frac{\pi }{2}\right)=-\frac{2}{\pi }$−1 / (π/2) = −2/π, not −1 and not $-\frac{\pi }{2}$−π/2.

Taking $\frac{dr}{d\theta }$dr/dθ as the gradient of the curve. The gradient is $\frac{dy}{dx}$dy/dx, which needs both $\frac{dx}{d\theta }$dx/dθ and $\frac{dy}{d\theta }$dy/dθ.

A camera following a sprinter

In the application below, a camera 30 meters from a straight track turns to follow a sprinter. The track is $r=30\sec \theta$r = 30 sec θ, so $\frac{dr}{d\theta }=30\sec \theta \tan \theta$dr/dθ = 30 sec θ tan θ, which is $60\sqrt{3}$60√3 at $\theta =\frac{\pi }{3}$θ = π/3. The camera turns at 0.08 radians per second, and the chain rule multiplies the two rates to give how fast r changes in time.