Tangents to Polar Curves

Finding the gradient of a polar curve, and in particular locating its horizontal and vertical tangents, is a recurring AQA Further Mathematics skill that ties polar coordinates back to the differentiation you already know. The trick is to treat $\theta$θ as a parameter: write $x$x and $y$y as functions of $\theta$θ, differentiate each, and form $\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy/\mathrm d\theta}{\mathrm dx/\mathrm d\theta}$dxdy​=dx/dθdy/dθ​. This lesson derives the two key derivative formulae, applies them to cardioids, circles and spirals, handles the special case of tangents at the pole, and introduces the elegant radius–tangent angle $\psi$ψ.

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1 Spec mapping

Tangents to polar curves are compulsory pure content for Papers 1 and 2 of AQA 7367. The work is a blend of AO1 (carry out the parametric differentiation) and AO2 (interpret "horizontal", "vertical" and "tangent at the pole" correctly, and communicate the reasoning). It draws directly on the parametric differentiation and product/chain rule techniques from A-Level Maths, applied to the polar parametrisation $x=r(\theta)\cos\theta,\ y=r(\theta)\sin\theta$x=r(θ)cosθ, y=r(θ)sinθ. In an exam this topic tends to appear in two guises: "find where the tangent is horizontal/vertical" (a differentiate-and-solve exercise) and "find the equation of the tangent at this point" (a point-plus-gradient exercise). Both reduce to the same two derivative formulae derived below, so mastering those formulae and the discipline of computing them separately is the whole game.

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2 Core theory — gradient of a polar curve

A polar curve $r=f(\theta)$r=f(θ) is really a parametric curve with parameter $\theta$θ:

$x = r\cos\theta = f(\theta)\cos\theta, \qquad y = r\sin\theta = f(\theta)\sin\theta.$x=rcosθ=f(θ)cosθ,y=rsinθ=f(θ)sinθ.

The crucial conceptual shift is that although the equation is $r=f(\theta)$r=f(θ), the gradient we want — the slope of the tangent line drawn on ordinary $xy$xy-axes — is still $\tfrac{\mathrm dy}{\mathrm dx}$dxdy​, a Cartesian quantity. We cannot read it off the polar equation directly; we have to route through the parameter $\theta$θ. This is precisely the parametric-differentiation situation from A-Level Maths, where $\tfrac{\mathrm dy}{\mathrm dx}=\tfrac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt}$dxdy​=dx/dtdy/dt​; here the parameter just happens to be the polar angle. So the plan is: express $x$x and $y$y as functions of $\theta$θ, differentiate each, and divide.

Differentiate each with respect to $\theta$θ by the product rule — both $x$x and $y$y are products of $r(\theta)$r(θ) with a trig function of $\theta$θ — writing $r'=\dfrac{\mathrm dr}{\mathrm d\theta}$r′=dθdr​:

$\begin{aligned} \frac{\mathrm dx}{\mathrm d\theta} &= r'\cos\theta - r\sin\theta, \\[2pt] \frac{\mathrm dy}{\mathrm d\theta} &= r'\sin\theta + r\cos\theta. \end{aligned}$dθdx​dθdy​​=r′cosθ−rsinθ,=r′sinθ+rcosθ.​

The gradient of the curve in Cartesian terms is then the quotient of these two:

$\boxed{\ \frac{\mathrm dy}{\mathrm dx} = \frac{\dfrac{\mathrm dy}{\mathrm d\theta}}{\dfrac{\mathrm dx}{\mathrm d\theta}} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}\ }$ dxdy​=dθdx​dθdy​​=r′cosθ−rsinθr′sinθ+rcosθ​ ​

Method tip. Always compute $\dfrac{\mathrm dy}{\mathrm d\theta}$dθdy​ and $\dfrac{\mathrm dx}{\mathrm d\theta}$dθdx​ separately and fully before dividing. Trying to differentiate a single messy quotient is error-prone; the two-step approach keeps the algebra clean and makes the horizontal/vertical conditions transparent.

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3 Horizontal and vertical tangents

The geometry is exactly as in Cartesian parametric work. A tangent is horizontal when $y$y is momentarily stationary as $\theta$θ varies but $x$x is still changing, and vertical in the opposite case:

$\begin{aligned} \textbf{horizontal tangent:}\quad & \frac{\mathrm dy}{\mathrm d\theta}=0 \ \text{ and }\ \frac{\mathrm dx}{\mathrm d\theta}\ne0, \\[2pt] \textbf{vertical tangent:}\quad & \frac{\mathrm dx}{\mathrm d\theta}=0 \ \text{ and }\ \frac{\mathrm dy}{\mathrm d\theta}\ne0. \end{aligned}$horizontal tangent:vertical tangent:​dθdy​=0  and  dθdx​=0,dθdx​=0  and  dθdy​=0.​

The "and $\ne0$=0" condition is not pedantry — it is what rules out the degenerate case. If both derivatives vanish at the same $\theta$θ, which typically happens at the pole, then $\tfrac{\mathrm dy}{\mathrm dx}$dxdy​ is the indeterminate form $\tfrac00$00​ and tells you nothing; that point needs the separate pole analysis of §5. A very common exam structure is to find several $\theta$θ satisfying $\tfrac{\mathrm dy}{\mathrm d\theta}=0$dθdy​=0, one of which is secretly the pole, and to expect you to weed it out — so always check the other derivative at each candidate before declaring it a genuine horizontal or vertical tangent.

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4 Worked examples with mark scheme

Worked Example 1 — tangents to a cardioid

Find the values of $\theta$θ at which $r=a(1+\cos\theta)$r=a(1+cosθ) has horizontal and vertical tangents.

Here $r'=-a\sin\theta$r′=−asinθ. Build the two derivatives:

$\begin{aligned} \frac{\mathrm dy}{\mathrm d\theta} &= -a\sin\theta\sin\theta + a(1+\cos\theta)\cos\theta = a\big(\cos\theta + \cos^2\theta - \sin^2\theta\big) \\ &= a\big(\cos\theta + \cos 2\theta\big) = 2a\cos\tfrac{3\theta}{2}\cos\tfrac{\theta}{2}, \end{aligned}$dθdy​​=−asinθsinθ+a(1+cosθ)cosθ=a(cosθ+cos2θ−sin2θ)=a(cosθ+cos2θ)=2acos23θ​cos2θ​,​

using $\cos^2\theta-\sin^2\theta=\cos2\theta$cos2θ−sin2θ=cos2θ and then the sum-to-product identity $\cos A+\cos B=2\cos\tfrac{A+B}{2}\cos\tfrac{A-B}{2}$cosA+cosB=2cos2A+B​cos2A−B​ with $A=2\theta,\ B=\theta$A=2θ, B=θ. The factorised form is the key to the whole question: a product is zero exactly when one of its factors is zero, so the messy-looking condition $\cos\theta+\cos2\theta=0$cosθ+cos2θ=0 breaks into two simple sub-cases. Reaching for the sum-to-product identity here is the single technique that turns this from an intractable trig equation into a routine one. Similarly,

$\frac{\mathrm dx}{\mathrm d\theta} = -a\sin\theta\cos\theta - a(1+\cos\theta)\sin\theta = -a\sin\theta\,(1+2\cos\theta).$dθdx​=−asinθcosθ−a(1+cosθ)sinθ=−asinθ(1+2cosθ).

Horizontal tangents ($\tfrac{\mathrm dy}{\mathrm d\theta}=0$dθdy​=0): $\cos\tfrac{3\theta}{2}=0\Rightarrow\theta=\tfrac{\pi}{3},\pi,\tfrac{5\pi}{3}$cos23θ​=0⇒θ=3π​,π,35π​ (on $[0,2\pi)$[0,2π)). At $\theta=\pi$θ=π the denominator also vanishes (the pole — treat separately), so the genuine horizontal tangents are at $\theta=\tfrac{\pi}{3}$θ=3π​ and $\theta=\tfrac{5\pi}{3}$θ=35π​, where $r=\tfrac{3a}{2}$r=23a​.

Vertical tangents ($\tfrac{\mathrm dx}{\mathrm d\theta}=0$dθdx​=0): the factorised form $-a\sin\theta(1+2\cos\theta)=0$−asinθ(1+2cosθ)=0 splits cleanly. Either $\sin\theta=0\Rightarrow\theta=0\ (r=2a)$sinθ=0⇒θ=0 (r=2a) or $\theta=\pi\ (\text{pole, exclude})$θ=π (pole, exclude); or $1+2\cos\theta=0\Rightarrow\cos\theta=-\tfrac12\Rightarrow\theta=\tfrac{2\pi}{3},\tfrac{4\pi}{3}$1+2cosθ=0⇒cosθ=−21​⇒θ=32π​,34π​. So genuine vertical tangents occur at $\theta=0,\tfrac{2\pi}{3},\tfrac{4\pi}{3}$θ=0,32π​,34π​. Again the only trap is $\theta=\pi$θ=π, where both derivatives vanish — the cusp at the pole.

Mark scheme: M1 form $\tfrac{\mathrm dy}{\mathrm d\theta}$dθdy​; M1 form $\tfrac{\mathrm dx}{\mathrm d\theta}$dθdx​; A1 factorise each; A1 horizontal $\theta=\tfrac{\pi}{3},\tfrac{5\pi}{3}$θ=3π​,35π​; A1 vertical $\theta=0,\tfrac{2\pi}{3},\tfrac{4\pi}{3}$θ=0,32π​,34π​; B1 identify $\theta=\pi$θ=π as the pole.

Worked Example 2 — gradient at a point on a circle

Find $\dfrac{\mathrm dy}{\mathrm dx}$dxdy​ for $r=2\cos\theta$r=2cosθ at $\theta=\tfrac{\pi}{4}$θ=4π​.

$r'=-2\sin\theta$r′=−2sinθ, so

$\frac{\mathrm dy}{\mathrm d\theta} = -2\sin\theta\sin\theta + 2\cos\theta\cos\theta = 2\cos 2\theta, \qquad \frac{\mathrm dx}{\mathrm d\theta} = -2\sin\theta\cos\theta - 2\cos\theta\sin\theta = -2\sin 2\theta.$dθdy​=−2sinθsinθ+2cosθcosθ=2cos2θ,dθdx​=−2sinθcosθ−2cosθsinθ=−2sin2θ.

$\therefore\ \frac{\mathrm dy}{\mathrm dx} = \frac{2\cos 2\theta}{-2\sin 2\theta} = -\cot 2\theta, \qquad \text{at } \theta=\tfrac{\pi}{4}:\ -\cot\tfrac{\pi}{2}=0.$∴ dxdy​=−2sin2θ2cos2θ​=−cot2θ,at θ=4π​: −cot2π​=0.

The tangent is horizontal at $\theta=\tfrac{\pi}{4}$θ=4π​ — geometrically the top of this circle, which makes sense: $r=2\cos\theta$r=2cosθ is the circle of diameter $2$2 on the initial line, and its highest point is reached at $\theta=\tfrac{\pi}{4}$θ=4π​. Notice how the two derivatives simplified to single double-angle terms, making the quotient $-\cot2\theta$−cot2θ extremely clean; recognising $\sin\theta\cos\theta=\tfrac12\sin2\theta$sinθcosθ=21​sin2θ and $\cos^2\theta-\sin^2\theta=\cos2\theta$cos2θ−sin2θ=cos2θ as you go is what produces such tidy forms.

Mark scheme: M1 both derivatives; A1 $\tfrac{\mathrm dy}{\mathrm dx}=-\cot2\theta$dxdy​=−cot2θ; A1 value $0$0 with interpretation.

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5 Tangent lines at the pole

There is a beautiful shortcut for the tangent to a curve at the pole. If $f(\theta_0)=0$f(θ0​)=0 (so the curve passes through the pole at angle $\theta_0$θ0​) and $f'(\theta_0)\ne0$f′(θ0​)=0, then the tangent to the curve at the pole is simply the line

$\theta=\theta_0.$θ=θ0​.

The reason is geometric: as the curve approaches the pole the radius shrinks to zero along the direction $\theta_0$θ0​, so the point arrives at the origin travelling along that ray, and the limiting tangent line is the ray itself. This converts what looks like a hard calculus problem (the gradient there is $\tfrac00$00​) into a piece of trivial algebra: just solve $r=0$r=0. It is one of the most marks-per-effort results in the whole topic, and it is exactly why curve-sketching and tangents are taught together — the angles where a petal or loop touches the pole are simultaneously the tangents at the pole. One subtlety: each line through the pole corresponds to two $\theta$θ-values differing by $\pi$π, so after solving $r=0$r=0 you must collapse pairs like $\theta=0$θ=0 and $\theta=\pi$θ=π into the single line they describe.

Worked Example 3 — tangents at the pole for a rose

Find the tangent lines at the pole for $r=\sin 2\theta$r=sin2θ.

No differentiation is required — the pole-tangent shortcut reduces this to solving $r=0$r=0. Solve $\sin 2\theta=0\Rightarrow 2\theta=0,\pi,2\pi,3\pi\Rightarrow\theta=0,\tfrac{\pi}{2},\pi,\tfrac{3\pi}{2}$sin2θ=0⇒2θ=0,π,2π,3π⇒θ=0,2π​,π,23π​. These four angles are the directions in which the four petals of this rose leave the pole. As lines through the pole, however, $\theta=0$θ=0 and $\theta=\pi$θ=π describe the same line (the $x$x-axis), and $\theta=\tfrac{\pi}{2}$θ=2π​ and $\theta=\tfrac{3\pi}{2}$θ=23π​ the same line (the $y$y-axis). So the distinct tangent lines at the pole are the $x$x-axis and the $y$y-axis — even though four petals emanate from there, each pair shares a tangent line.

Mark scheme: M1 solve $r=0$r=0; A1 the four $\theta$θ-values; A1 identify the two distinct lines.

Worked Example 4 — equation of a tangent

Find the equation of the tangent to $r=1+2\cos\theta$r=1+2cosθ at $\theta=\tfrac{\pi}{3}$θ=3π​.

Finding the equation of a tangent needs two ingredients: a point it passes through, and its gradient. The point is found by converting the polar coordinates of the point of contact to Cartesian; the gradient comes from the quotient formula. At $\theta=\tfrac{\pi}{3}$θ=3π​: $r=1+2\cdot\tfrac12=2$r=1+2⋅21​=2, so $x=2\cos\tfrac{\pi}{3}=1$x=2cos3π​=1, $y=2\sin\tfrac{\pi}{3}=\sqrt3$y=2sin3π​=3​. With $r'=-2\sin\theta$r′=−2sinθ,

$\begin{aligned} \frac{\mathrm dy}{\mathrm d\theta}\Big|_{\pi/3} &= -2\sin\tfrac{\pi}{3}\sin\tfrac{\pi}{3}+2\cos\tfrac{\pi}{3} = -2\cdot\tfrac34 + 2\cdot\tfrac12 = -\tfrac32+1 = -\tfrac12, \\ \frac{\mathrm dx}{\mathrm d\theta}\Big|_{\pi/3} &= -2\sin\tfrac{\pi}{3}\cos\tfrac{\pi}{3}-2\sin\tfrac{\pi}{3} = -\tfrac{\sqrt3}{2}-\sqrt3 = -\tfrac{3\sqrt3}{2}. \end{aligned}$dθdy​​π/3​dθdx​​π/3​​=−2sin3π​sin3π​+2cos3π​=−2⋅43​+2⋅21​=−23​+1=−21​,=−2sin3π​cos3π​−2sin3π​=−23​​−3​=−233​​.​

$\frac{\mathrm dy}{\mathrm dx} = \frac{-\tfrac12}{-\tfrac{3\sqrt3}{2}} = \frac{1}{3\sqrt3} = \frac{\sqrt3}{9}.$dxdy​=−233​​−21​​=33​1​=93​​.

Tangent through $(1,\sqrt3)$(1,3​) with gradient $\tfrac{\sqrt3}{9}$93​​:

$y - \sqrt3 = \tfrac{\sqrt3}{9}(x-1) \ \implies\ y = \tfrac{\sqrt3}{9}x + \tfrac{8\sqrt3}{9}.$y−3​=93​​(x−1) ⟹ y=93​​x+983​​.

Mark scheme: M1 find $(x,y)$(x,y) at the point; M1 both derivatives; A1 $\tfrac{\mathrm dy}{\mathrm dx}=\tfrac{\sqrt3}{9}$dxdy​=93​​; A1 correct tangent equation.

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6 The angle $\psi$ψ between radius and tangent

So far we have measured tangent direction relative to the fixed $x$x-axis, via $\tfrac{\mathrm dy}{\mathrm dx}$dxdy​. But in polar problems it is often more natural to measure direction relative to the radius — after all, the radius is the curve's own built-in reference line at each point. A second, often slicker, measure of direction is therefore the angle $\psi$ψ (psi) between the radius vector $OP$OP and the tangent to the curve at $P$P. It satisfies the remarkably compact formula

$\boxed{\ \tan\psi = \frac{r}{\,\mathrm dr/\mathrm d\theta\,} = \frac{r}{r'}\ }$ tanψ=dr/dθr​=r′r​ ​