Convert the polar coordinates \( (r,\theta) = (4, \pi/3) \) to Cartesian coordinates.
\( x = r\cos\theta = 4\cos(\pi/3) = 4\cdot\dfrac12 = 2 \)
\( y = r\sin\theta = 4\sin(\pi/3) = 4\cdot\dfrac{\sqrt{3}}{2} = 2\sqrt{3} \)
Point: \( (2, 2\sqrt{3}) \).
Polar Coordinates
12 exam-style questions on polar curves, conversion, area and tangents. Tap Show answer for full working.
4 Easy4 Medium4 Hard
Convert the Cartesian equation \( x^2 + y^2 = 9 \) into polar form.
\( r^2 = 9 \) ⇒ \( r = 3 \) (taking \( r \ge 0 \)).
Sketch the curve \( r = 2\cos\theta \) for \( -\pi/2 \le \theta \le \pi/2 \) and state its Cartesian equation.
Circle of radius 1 centred at \( (1,0) \). Cartesian: \( r = 2\cos\theta \) ⇒ \( r^2 = 2r\cos\theta \) ⇒ \( x^2 + y^2 = 2x \) ⇒ \( (x-1)^2 + y^2 = 1 \).
Find the area enclosed by the curve \( r = 2 \) from \( \theta = 0 \) to \( \theta = \pi/2 \).
Area \( = \dfrac12\int_0^{\pi/2} r^2\,d\theta = \dfrac12\int_0^{\pi/2} 4\,d\theta = 2\cdot\dfrac{\pi}{2} = \pi \).
Find the polar equation of the line \( y = x + 1 \).
\( r\sin\theta = r\cos\theta + 1 \) ⇒ \( r(\sin\theta - \cos\theta) = 1 \) ⇒ \( r = \dfrac{1}{\sin\theta - \cos\theta} \).
Find the area of one loop of the curve \( r = \sin 2\theta \).
One loop occurs for \( 0 \le \theta \le \pi/2 \). Area \( = \dfrac12\int_0^{\pi/2} \sin^2 2\theta\,d\theta = \dfrac12\int_0^{\pi/2} \dfrac{1-\cos 4\theta}{2}\,d\theta = \dfrac14\Big[\theta - \dfrac14\sin 4\theta\Big]_0^{\pi/2} = \dfrac{\pi}{8} \).
For the curve \( r = 1 + \cos\theta \), find \( \dfrac{dy}{dx} \) at the point where \( \theta = \pi/2 \).
\( x = r\cos\theta = (1+\cos\theta)\cos\theta \), \( y = (1+\cos\theta)\sin\theta \).
\( \dfrac{dy}{d\theta} = \cos\theta + \cos 2\theta \), \( \dfrac{dx}{d\theta} = -\sin\theta - \sin 2\theta \) (standard differentiation).
At \( \theta=\pi/2 \): \( \dfrac{dy}{d\theta} = 0 + (-1) = -1 \), \( \dfrac{dx}{d\theta} = -1 - 0 = -1 \).
\( \dfrac{dy}{dx} = 1 \).
Convert \( r = \dfrac{2}{1 + \cos\theta} \) to Cartesian form.
\( r(1+\cos\theta) = 2 \) ⇒ \( r + x = 2 \) (since \( x = r\cos\theta \)) ⇒ \( r = 2 - x \).
Square: \( x^2 + y^2 = (2-x)^2 = 4 - 4x + x^2 \) ⇒ \( y^2 = 4 - 4x \) ⇒ \( y^2 = 4(1-x) \). Parabola.
Find the area common to the circles \( r = 2\cos\theta \) and \( r = 2\sin\theta \).
Intersect when \( 2\cos\theta = 2\sin\theta \) ⇒ \( \theta = \pi/4 \) (in first quadrant).
Area = \( 2\cdot\dfrac12\int_0^{\pi/4} (2\sin\theta)^2\,d\theta = 2\int_0^{\pi/4} 2\sin^2\theta\,d\theta = 4\int_0^{\pi/4}\dfrac{1-\cos 2\theta}{2}\,d\theta = 2\Big[\theta - \dfrac12\sin 2\theta\Big]_0^{\pi/4} = 2\Big(\dfrac{\pi}{4} - \dfrac12\Big) = \dfrac{\pi}{2} - 1 \).
Find the equation of the tangent to \( r = a(1+\cos\theta) \) at the point where \( \theta = \pi/3 \).
At \( \theta=\pi/3 \), \( r = a(1+\tfrac12) = \tfrac32 a \). Point: \( x = \tfrac32 a\cdot\tfrac12 = \tfrac34 a \), \( y = \tfrac32 a\cdot\tfrac{\sqrt{3}}{2} = \tfrac{3\sqrt{3}}{4}a \).
Using \( \dfrac{dy}{dx} = \dfrac{(dr/d\theta)\sin\theta + r\cos\theta}{(dr/d\theta)\cos\theta - r\sin\theta} \) with \( dr/d\theta = -a\sin\theta \).
At \( \theta=\pi/3 \): numerator \( = (-a\sqrt{3}/2)(\sqrt{3}/2) + (3a/2)(1/2) = -3a/4 + 3a/4 = 0 \).
Denominator ≠ 0. Thus horizontal tangent: \( y = \dfrac{3\sqrt{3}}{4}a \).
Show that the area of one petal of the curve \( r = a\cos 2\theta \) is \( \dfrac{\pi a^2}{8} \).
One petal: \( -\pi/4 \le \theta \le \pi/4 \). Area \( = \dfrac12\int_{-\pi/4}^{\pi/4} a^2\cos^2 2\theta\,d\theta = a^2\int_0^{\pi/4}\dfrac{1+\cos 4\theta}{2}\,d\theta = \dfrac{a^2}{2}\Big[\theta + \dfrac14\sin 4\theta\Big]_0^{\pi/4} = \dfrac{a^2}{2}\cdot\dfrac{\pi}{4} = \dfrac{\pi a^2}{8} \).
Find the points of intersection of \( r = 1 + \sin\theta \) and \( r = 1 - \sin\theta \) and the area common to both curves.
Set \( 1+\sin\theta = 1-\sin\theta \) ⇒ \( \sin\theta = 0 \) ⇒ \( \theta = 0, \pi \).
At \( \theta=0 \): \( r=1 \), point \( (1,0) \). At \( \theta=\pi \): \( r=1 \), point \( (-1,0) \).
By symmetry the common area is \( \displaystyle\int_0^{\pi}(1-\sin\theta)^2\,d\theta = \int_0^{\pi}\big(1-2\sin\theta+\sin^2\theta\big)\,d\theta \).
\( = \Big[\theta + 2\cos\theta + \dfrac{\theta}{2} - \dfrac{\sin 2\theta}{4}\Big]_0^{\pi} = \big(\pi - 2 + \pi/2\big) - 2 = \dfrac{3\pi}{2} - 4 \).