Polar differentiation AND integration?!?!?!

A curve is defined by the polar equation

r=1+\cos \theta, 0 \leqslant \theta < 2\pi

The tangents to the curve when r=1 meet at a single point.
Find the area of the region that lies inside the triangle formed by the tangents and the y axis, but outside the area bounded by the curve and the y axis.
I got the area as 3-3π/4
Is this correct??

Reply 1

TeeEm

19

Original post by Ano123

A curve is defined by the polar equation

r=1+\cos \theta, 0 \leqslant \theta < 2\pi

The tangents to the curve when r=1 meet at a single point.
Find the area of the region that lies inside the triangle formed by the tangents and the y axis, but outside the area bounded by the curve and the y axis.
I got the area as 3-3π/4
Is this correct??

@Zacken
@16Characters....
@EricPiphany

(edited 8 years ago)

Reply 2

16Characters....

13

Original post by Ano123

A curve is defined by the polar equation

r=1+\cos \theta, 0 \leqslant \theta < 2\pi

The tangents to the curve when r=1 meet at a single point.
Find the area of the region that lies inside the triangle formed by the tangents and the y axis, but outside the area bounded by the curve and the y axis.
I got the area as 3-3π/4
Is this correct??

I also got this.

Reply 3

Ano123

OP

15

Original post by 16Characters....

I also got this.

It's a nice question don't you think?

Reply 4

16Characters....

13

Original post by Ano123

It's a nice question don't you think?

It was a nice question. Tested quite a few different ideas but was not too long and it "flowed", unlike some questions which just force concepts together.

Reply 5

EricPiphany

17

got tagged here by teeEm. Finally maths I can actually do :P

Reply 6

EricPiphany

17

A vertical line touches the curve

r=a(1+\cos(\theta))

at two points. Show that the area enclosed between the line and the curve is equal to

Unparseable latex formula:

\diaplaystyle a^2\left(\frac{15\sqrt{3}}{16}-\frac{\pi}{2}\right)

.

(edited 8 years ago)

Reply 7

Zacken

22

Original post by EricPiphany

A vertical line touches the curve

r=a(1+\cos(\theta))

at two points. Show that the area enclosed between the line and the curve is equal to

Unparseable latex formula:

\diaplaystyle a^2\left(\frac{15\sqrt{3}}{16}-\frac{\pi}{2}\right)

.

OMG, I actually managed to do this. :ahee:

x = r\cos \theta = ar\cos \theta(1 + \cos\theta) \Rightarrow x' = 0 \iff \sin \theta (1 + 2\cos \theta) = 0

The two values that concern us are:

\theta = \frac{4\pi}{3}

and

\theta = \frac{2\pi}{3}

The are bounded in the polar curve (using symmetry) is then:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}a^2\int_{2\pi/3}^{\pi} (1 + \cos \theta)^2 \, \mathrm{d}\theta = \frac{a^2}{8}\left(4\pi - 7\sqrt{3}\right)\end{equation*}

Now draw the lines corresponding to the two arguments and we get a triangle that has area

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\frac{1}{2} \times \frac{a}{2} \times \frac{a}{2} \times \sin \frac{2\pi}{3} = \frac{a^2\sqrt{3}}{16}\end{equation*}

So the area required is given by:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{a^2}{8}\left(4\pi - 7\sqrt{3}\right) - \frac{a^2\sqrt{3}}{16} = \frac{a^2}{8}\left(\frac{15\sqrt{3}}{16} - \frac{\pi}{2}\right)\end{equation*}

as required.

(edited 8 years ago)

Reply 8

EricPiphany

17

Original post by Zacken

OMG, I actually managed to do this. :ahee:

x = r\cos \theta = ar\cos \theta(1 + \cos\theta) \Rightarrow x' = 0 \iff \sin \theta (1 + 2\cos \theta) = 0

The two values that concern us are:

\theta = \frac{4\pi}{3}

and

\theta = \frac{2\pi}{3}

The are bounded in the polar curve (using symmetry) is then:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}a^2\int_{2\pi/3}^{\pi} (1 + \cos \theta)^2 \, \mathrm{d}\theta = \frac{a^2}{8}\left(4\pi - 7\sqrt{3}\right)\end{equation*}

Now draw the lines corresponding to the two arguments and we get a triangle that has area

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\frac{1}{2} \times \frac{a}{2} \times \frac{a}{2} \times \sin \frac{2\pi}{3} = \frac{a^2\sqrt{3}}{16}\end{equation*}

So the area required is given by:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{a^2}{8}\left(4\pi - 7\sqrt{3}\right) - \frac{a^2\sqrt{3}}{16} = \frac{a^2}{8}\left(\frac{15\sqrt{3}}{16} - \frac{\pi}{2}\right)\end{equation*}

as required.