Unit 6, lesson 87

The Slope of Polar Curves

How to find the slope of a polar curve by treating it as a parametric curve with the angle as parameter, where its tangents are horizontal or vertical, and how to find the tangent lines at the pole.

  • Theory part 173
  • Practice part 174
  • Calculus of Polar Curves

The derivative dr/dθdr/d\theta tells you how fast a polar curve moves away from the pole. It is not the slope of the curve. A circle around the pole has dr/dθ=0dr/d\theta = 0 everywhere, yet its slope takes every value. To get the actual slope you go back to xx and yy and let θ\theta play the role of the parameter.

Key terms

A polar curve as a parametric curve

Definition The polar curve r=f(θ)r = f(\theta) is the parametric curve

x=f(θ)cos⁡θ,y=f(θ)sin⁡θ,x = f(\theta)\cos\theta, \qquad y = f(\theta)\sin\theta,

with parameter θ\theta. If ff is differentiable, the product rule gives

dxdθ=f′(θ)cos⁡θ−f(θ)sin⁡θ,dydθ=f′(θ)sin⁡θ+f(θ)cos⁡θ.\frac{dx}{d\theta} = f'(\theta)\cos\theta - f(\theta)\sin\theta, \qquad \frac{dy}{d\theta} = f'(\theta)\sin\theta + f(\theta)\cos\theta.

In plain words Substitute r=f(θ)r = f(\theta) into x=rcos⁡θx = r\cos\theta and y=rsin⁡θy = r\sin\theta. Now both coordinates are functions of one variable, and everything known about parametric curves applies.

Example For the spiral r=θr = \theta we get x=θcos⁡θx = \theta\cos\theta and y=θsin⁡θy = \theta\sin\theta. Then

dxdθ=cos⁡θ−θsin⁡θ,dydθ=sin⁡θ+θcos⁡θ.\frac{dx}{d\theta} = \cos\theta - \theta\sin\theta, \qquad \frac{dy}{d\theta} = \sin\theta + \theta\cos\theta.

The derivative ratio

Definition Let ff be differentiable at θ\theta and suppose dx/dθ≠0dx/d\theta \neq 0 there. Then the slope of the tangent line to r=f(θ)r = f(\theta) at the point with parameter θ\theta is

dydx=dydθdxdθ=f′(θ)sin⁡θ+f(θ)cos⁡θf′(θ)cos⁡θ−f(θ)sin⁡θ.\frac{dy}{dx} = \frac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}} = \frac{f'(\theta)\sin\theta + f(\theta)\cos\theta}{f'(\theta)\cos\theta - f(\theta)\sin\theta}.

In plain words Slope is the rate of change of yy divided by the rate of change of xx, both measured per unit of θ\theta.

Example For the cardioid r=1+sin⁡θr = 1 + \sin\theta we have f′(θ)=cos⁡θf'(\theta) = \cos\theta, so

dydx=cos⁡θsin⁡θ+(1+sin⁡θ)cos⁡θcos⁡2θ−(1+sin⁡θ)sin⁡θ=cos⁡θ (1+2sin⁡θ)(1+sin⁡θ)(1−2sin⁡θ).\frac{dy}{dx} = \frac{\cos\theta\sin\theta + (1 + \sin\theta)\cos\theta}{\cos^2\theta - (1 + \sin\theta)\sin\theta} = \frac{\cos\theta\,(1 + 2\sin\theta)}{(1 + \sin\theta)(1 - 2\sin\theta)}.

At θ=π3\theta = \frac{\pi}{3} the slope is

12(1+3)(1+32)(1−3)=1+3−1−3=−1.\frac{\frac{1}{2}(1 + \sqrt{3})}{\left(1 + \frac{\sqrt{3}}{2}\right)(1 - \sqrt{3})} = \frac{1 + \sqrt{3}}{-1 - \sqrt{3}} = -1.

Tangent direction

Definition If ff is differentiable at θ\theta, the vector

⟨dxdθ,dydθ⟩=⟨f′cos⁡θ−fsin⁡θ, f′sin⁡θ+fcos⁡θ⟩\left\langle \frac{dx}{d\theta}, \frac{dy}{d\theta} \right\rangle = \left\langle f'\cos\theta - f\sin\theta,\ f'\sin\theta + f\cos\theta \right\rangle

points along the curve in the direction of increasing θ\theta. Its length is f(θ)2+f′(θ)2\sqrt{f(\theta)^2 + f'(\theta)^2}, so it is nonzero unless f(θ)=f′(θ)=0f(\theta) = f'(\theta) = 0. Where it is nonzero, the tangent line is the line through the point in this direction, and this works even when dx/dθ=0dx/d\theta = 0.

In plain words The pair of derivatives is a direction arrow. It gives the tangent line in every case except when the point momentarily stops, which can only happen at the pole.

Example On the circle r=2r = 2 we have f′=0f' = 0, so the tangent vector is ⟨−2sin⁡θ,2cos⁡θ⟩\langle -2\sin\theta, 2\cos\theta \rangle. Its dot product with the position vector ⟨2cos⁡θ,2sin⁡θ⟩\langle 2\cos\theta, 2\sin\theta \rangle is 00, so the tangent is perpendicular to the radius, as it should be for a circle.

Horizontal and vertical tangents

Definition Let ff be differentiable at θ0\theta_0. The curve has a horizontal tangent at the point with parameter θ0\theta_0 if

dydθ(θ0)=0anddxdθ(θ0)≠0,\frac{dy}{d\theta}(\theta_0) = 0 \quad \text{and} \quad \frac{dx}{d\theta}(\theta_0) \neq 0,

and a vertical tangent if

dxdθ(θ0)=0anddydθ(θ0)≠0.\frac{dx}{d\theta}(\theta_0) = 0 \quad \text{and} \quad \frac{dy}{d\theta}(\theta_0) \neq 0.

If both are zero, the test says nothing, and you study the limit of dy/dxdy/dx as θ→θ0\theta \to \theta_0.

In plain words Horizontal means yy stops changing while xx keeps moving. Vertical is the other way around.

Example For r=1+sin⁡θr = 1 + \sin\theta on [0,2π)[0, 2\pi) the two derivatives factor as

dydθ=cos⁡θ (1+2sin⁡θ),dxdθ=(1+sin⁡θ)(1−2sin⁡θ).\frac{dy}{d\theta} = \cos\theta\,(1 + 2\sin\theta), \qquad \frac{dx}{d\theta} = (1 + \sin\theta)(1 - 2\sin\theta).

Their zeros give this table.

θ\thetaPointTangent
π2\frac{\pi}{2}(0,2)(0, 2)horizontal
7π6\frac{7\pi}{6}(−34,−14)(-\frac{\sqrt{3}}{4}, -\frac{1}{4})horizontal
11π6\frac{11\pi}{6}(34,−14)(\frac{\sqrt{3}}{4}, -\frac{1}{4})horizontal
π6\frac{\pi}{6}(334,34)(\frac{3\sqrt{3}}{4}, \frac{3}{4})vertical
5π6\frac{5\pi}{6}(−334,34)(-\frac{3\sqrt{3}}{4}, \frac{3}{4})vertical

At θ=3π2\theta = \frac{3\pi}{2} both derivatives are 00, and that case is handled in the next section.

Tangents at the pole

Definition Let ff be differentiable at θ0\theta_0 and suppose f(θ0)=0f(\theta_0) = 0, so the curve passes through the pole at θ0\theta_0. If f′(θ0)≠0f'(\theta_0) \neq 0, the derivative ratio becomes

dydx=f′(θ0)sin⁡θ0f′(θ0)cos⁡θ0=tan⁡θ0(cos⁡θ0≠0),\frac{dy}{dx} = \frac{f'(\theta_0)\sin\theta_0}{f'(\theta_0)\cos\theta_0} = \tan\theta_0 \quad (\cos\theta_0 \neq 0),

so the tangent at the pole is the line θ=θ0\theta = \theta_0, which is vertical when cos⁡θ0=0\cos\theta_0 = 0. If f′(θ0)=0f'(\theta_0) = 0 as well, compute the limit of dy/dxdy/dx as θ→θ0\theta \to \theta_0 directly.

In plain words A curve passes through the pole along the ray at which rr becomes zero. Only when rr also has zero derivative there do you need a closer look.

Example For r=cos⁡2θr = \cos 2\theta at θ0=π4\theta_0 = \frac{\pi}{4} we have f=0f = 0 and f′=−2sin⁡π2=−2≠0f' = -2\sin\frac{\pi}{2} = -2 \neq 0, so the tangent at the pole is y=xy = x. For the cardioid r=1+sin⁡θr = 1 + \sin\theta at θ0=3π2\theta_0 = \frac{3\pi}{2} both ff and f′f' vanish. Near that angle the identity

cos⁡θ1+sin⁡θ=1−sin⁡θcos⁡θ\frac{\cos\theta}{1 + \sin\theta} = \frac{1 - \sin\theta}{\cos\theta}

turns the slope into

dydx=(1−sin⁡θ)(1+2sin⁡θ)cos⁡θ (1−2sin⁡θ),\frac{dy}{dx} = \frac{(1 - \sin\theta)(1 + 2\sin\theta)}{\cos\theta\,(1 - 2\sin\theta)},

and as θ→3π2\theta \to \frac{3\pi}{2} the numerator tends to −2-2 and the denominator to 00. The slope grows without bound, so the cusp has a vertical tangent.

Check yourself

  1. Find the slope of the spiral r=θr = \theta at θ=π2\theta = \frac{\pi}{2}.
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With f(θ)=θf(\theta) = \theta and f′(θ)=1f'(\theta) = 1 the slope is

dydx=sin⁡θ+θcos⁡θcos⁡θ−θsin⁡θ=1+00−π2=−2π.\frac{dy}{dx} = \frac{\sin\theta + \theta\cos\theta}{\cos\theta - \theta\sin\theta} = \frac{1 + 0}{0 - \frac{\pi}{2}} = -\frac{2}{\pi}.
  1. Find the tangent lines at the pole for the rose r=sin⁡3θr = \sin 3\theta.
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The radius sin⁡3θ\sin 3\theta is zero for θ=0,π3,2π3\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3} in [0,π)[0, \pi), and f′=3cos⁡3θ=±3≠0f' = 3\cos 3\theta = \pm 3 \neq 0 there. The tangents are y=0y = 0, y=3 xy = \sqrt{3}\,x and y=−3 xy = -\sqrt{3}\,x.

  1. Find the points where r=2cos⁡θr = 2\cos\theta has a horizontal tangent.
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In rectangular form y=2cos⁡θsin⁡θ=sin⁡2θy = 2\cos\theta\sin\theta = \sin 2\theta and x=2cos⁡2θ=1+cos⁡2θx = 2\cos^2\theta = 1 + \cos 2\theta. So

dydθ=2cos⁡2θ=0at θ=π4,3π4,\frac{dy}{d\theta} = 2\cos 2\theta = 0 \quad \text{at } \theta = \frac{\pi}{4}, \frac{3\pi}{4},

and there dx/dθ=−2sin⁡2θ=∓2≠0dx/d\theta = -2\sin 2\theta = \mp 2 \neq 0. The points are (1,1)(1, 1) and (1,−1)(1, -1), the top and bottom of the circle.

How this lesson connects