Further Maths/Lessons/Polar Coordinates/Tangents
Core Pure · Polar Coordinates · Year 2

Tangents to Polar Curves

Using the Cartesian parametric form of a polar curve to find where its tangent is horizontal or vertical.

8 min read 1 worked example

Tangents to polar curves

Method — tangents parallel or perpendicular to the initial line

Convert to Cartesian parametric form using $\theta$ as the parameter: $x=r\cos\theta=f(\theta)\cos\theta$, $y=r\sin\theta=f(\theta)\sin\theta$.

· The tangent is parallel to the initial line (horizontal) where $\dfrac{dy}{d\theta}=0$.
· The tangent is perpendicular to the initial line (vertical) where $\dfrac{dx}{d\theta}=0$.

The cardioid r=5(1+cos theta) with two horizontal dashed tangent lines touching its topmost and bottommost points at theta=pi/3 and theta=-pi/3, where dy/dtheta=0
$\dfrac{dy}{d\theta}=0$ has two solutions here: the tangent to $r=5(1+\cos\theta)$ is horizontal at $\theta=\dfrac\pi3$ (topmost point) and, by symmetry, at $\theta=-\dfrac\pi3$ (bottommost point).
The cardioid r=4(1+cos theta) with vertical dashed tangent lines touching the two dimple points at theta=2pi/3 and theta=-2pi/3, and the rightmost vertex at theta=0, where dx/dtheta=0
$\dfrac{dx}{d\theta}=0$ has three non-cusp solutions here: the tangent to $r=4(1+\cos\theta)$ is vertical at $\theta=\dfrac{2\pi}3$ and $\theta=-\dfrac{2\pi}3$, where the dimple touches the line $x=-1$, and also at $\theta=0$, the rightmost vertex $(8,0)$, where the tangent is $x=8$.
Example 1 · Tangents parallel to the initial line
Find the value of $\theta$, with $0<\theta<\dfrac\pi2$, at which the tangent to $r=5(1+\cos\theta)$ is parallel to the initial line, and find the corresponding value of $r$.

$y=r\sin\theta=5(1+\cos\theta)\sin\theta=5\sin\theta+5\sin\theta\cos\theta=5\sin\theta+\dfrac52\sin2\theta$.

$$\frac{dy}{d\theta}=5\cos\theta+5\cos2\theta=5\cos\theta+5(2\cos^2\theta-1)=10\cos^2\theta+5\cos\theta-5$$

Set to zero: $2\cos^2\theta+\cos\theta-1=0$, so $(2\cos\theta-1)(\cos\theta+1)=0$, giving $\cos\theta=\dfrac12$ or $\cos\theta=-1$.

Since $0<\theta<\dfrac\pi2$, we need $\cos\theta=\dfrac12$, so $\theta=\dfrac\pi3$.

Then $r=5\left(1+\dfrac12\right)=\dfrac{15}2$.

ProofThe dimple condition for r = a(p + q cosθ)
1
To justify the classification from the previous lesson, consider $r=a(p+q\cos\theta)$ with $p,q\gt0$ and examine where $\dfrac{dx}{d\theta}=0$ for $0\lt\theta\lt\pi$.
$x=r\cos\theta=a(p+q\cos\theta)\cos\theta=ap\cos\theta+aq\cos^2\theta$.
2
$\dfrac{dx}{d\theta}=-ap\sin\theta-2aq\cos\theta\sin\theta=-a\sin\theta(p+2q\cos\theta)$.
3
For $0\lt\theta\lt\pi$, $\sin\theta\neq0$, so stationary points of $x$ occur where $p+2q\cos\theta=0$, i.e. $\cos\theta=-\dfrac p{2q}$.
4
This equation has a solution with $-1\lt\cos\theta\lt1$ precisely when $-1\lt-\dfrac p{2q}\lt1$, i.e. when $p\lt2q$ (since $p,q\gt0$ already gives $-\dfrac p{2q}\lt0\lt1$). When such a $\theta$ exists strictly between $0$ and $\pi$, the boundary curve doubles back on itself in the $x$-direction there, creating the concave "dimple". When $p\geq2q$, no such interior stationary point exists and the curve remains convex.
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Exam-style questions

Question 1 7 marks

The curve $C$ has polar equation $r=2+\cos\theta$. Find the value of $\theta$, with $0<\theta<\pi$, at which the tangent to $C$ is perpendicular to the initial line, giving your answer to 3 significant figures if it is not exact.

$x=r\cos\theta=(2+\cos\theta)\cos\theta=2\cos\theta+\cos^2\theta$. (M1, A1)

$$\frac{dx}{d\theta}=-2\sin\theta-2\cos\theta\sin\theta=-2\sin\theta(1+\cos\theta)$$ (M1, A1)

Set to zero: $\sin\theta=0$ or $\cos\theta=-1$. For $0<\theta<\pi$, $\sin\theta\neq0$, and $\cos\theta=-1$ gives $\theta=\pi$, which is outside the open interval. (M1)

So there is no solution with $0<\theta<\pi$ — the tangent is never perpendicular to the initial line on this open interval (the curve is convex throughout, consistent with $p=2\geq2q=2$ in the dimple condition). (A1, A1)

Question 2 7 marks

The curve $C$ has polar equation $r=4(1+\cos\theta)$. Find the value of $\theta$, with $0<\theta<\pi$, at which the tangent to $C$ is perpendicular to the initial line, and find the corresponding value of $r$.

$x=r\cos\theta=4(1+\cos\theta)\cos\theta=4\cos\theta+4\cos^2\theta$. (M1, A1)

$$\frac{dx}{d\theta}=-4\sin\theta-8\cos\theta\sin\theta=-4\sin\theta(1+2\cos\theta)$$ (M1, A1)

Set to zero: $\sin\theta=0$ or $\cos\theta=-\dfrac12$. For $0<\theta<\pi$, $\sin\theta\neq0$, so $\cos\theta=-\dfrac12$, giving $\theta=\dfrac{2\pi}3$. (M1)

Then $r=4\left(1-\dfrac12\right)=2$. (A1, A1)

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