# Math Help - Problem in Polar coordinates

1. ## Problem in Polar coordinates

1. The curve C has polar equation
r^2 cos 2Ѳ = a^2, -pi/4 < Ѳ < pi/4
Find in terms of the positive constant a
(a) the area of the region bounded by the half-lines Ѳ = + pi/6 and the arc of the curve C in the interval -pi/6 < Ѳ < pi/6
(b) the equation of the tangent to the curve which is perpendicular to the initial line, giving your answer in polar form.

2. The curve C has polar equation r = a sin Ѳ sin 2Ѳ, where a is a positive constant and 0 ≤ Ѳ ≤ pi/2.
(a) Find, in terms of a, the area of the region enclosed by C.
(b) Show that the tangent at the point (3/4a, pi/3) is parallel to the initial line.

2. Originally Posted by geton
1. The curve C has polar equation
r^2 cos 2Ѳ = a^2, -pi/4 < Ѳ < pi/4
Find in terms of the positive constant a
(a) the area of the region bounded by the half-lines Ѳ = + pi/6 and the arc of the curve C in the interval -pi/6 < Ѳ < pi/6
(b) the equation of the tangent to the curve which is perpendicular to the initial line, giving your answer in polar form.

2. The curve C has polar equation r = a sin Ѳ sin 2Ѳ, where a is a positive constant and 0 ≤ Ѳ ≤ pi/2.
(a) Find, in terms of a, the area of the region enclosed by C.
(b) Show that the tangent at the point (3/4a, pi/3) is parallel to the initial line.

1. (a) $A = \frac{1}{2} \int_{-\pi/6}^{\pi/6} \frac{a^2}{\cos (2\theta)} \, d \theta$.

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2. Area of single 'leaf': $A = \frac{1}{2} \int_{0}^{\pi/2} a \sin \theta \sin (2\theta) \, d \theta$.

Area of both 'leaves' (by symmetry): $A = \int_{0}^{\pi/2} a \sin \theta \sin (2\theta) \, d \theta$ and you should substitute $\sin \left( 2 \theta \right) = 2 \sin \theta \cos \theta$.

For (b) of both questions, I've got the same questions as in my other reply. What's meant by initial line etc.

3. Originally Posted by mr fantastic
I've got the same questions as in my other reply. What's meant by initial line etc.

Initial line is l, means x-axis. Thank you for your help. I’ve solved those questions. It is from polar coordinates chapter. And your area’s formula is absolutely right. Lately I understood that chapter. Thank you.