# Calculus II

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### 2 Answers

`r=1+sin(theta/2)=>theta=2arcsin(r-1)`

` ` `r=1+cos(theta/2)=>theta=2arccos(r-1)`

Now we have system of two equations

`theta=2arcsin(r-1)`

`theta=2arccos(r-1)`

From this we get

`arcsin(r-1)=arccos(r-1)`

Take `sin` of both sides of equation and apply formula `sin(arcsosx)=sqrt(1-x^2)`

`r-1=sqrt(1-(r-1)^2)`

`r-1=sqrt(1-r^2+2r-1)`

Square whole equation.

`r^2-2r+1=-r^2+2r`

`2r^2-4r+1=0`

Apply quadratic formula.

`r_(1,2)=(4pmsqrt(16-8))/4`

`r_1=1-sqrt2/2`

`r_2=1+sqrt2/2`

Now we return to our first equation `r=1+sin(theta/2)`

For `r_1` we have

`1-sqrt2/2=1+sin(theta/2)`

`sqrt2/2=sin(theta/2)` ` `

Here we have two cases

**(i) case**

`theta/2=-pi/4+2kpi,kinZZ`

`theta=-pi/2+4kpi,kinZZ`

Notice that, since we are working in polar coordinates period `4pi` doesn't matter because everything is defined only for `theta in[0,2pi).` Thus we have

`(1-sqrt2/2,-pi/2)`** <--First intersection**

**(ii) case**

`theta/2=-pi+pi/4`

`theta=-(3pi)/2`

`(1-sqrt2/2,-(3pi)/2)` **<-- Second intersection**

For `x_2` we have

`1+sqrt2/2=1+sin(theta/2)`

`sqrt2/2=sin(theta/2)`

Again we have two cases

**(i) case**

`theta/2=pi/4`

`theta=pi/2`

`(1+sqrt2/2,pi/2)` **<-- Third intersection**

**(ii) case**

`theta/2=pi-pi/4`

`theta=(3pi)/2`

`(1+sqrt2/2,(3pi)/2)` **<-- Fourth intersection**

Those are all intersection points. Of course you could say that point of origin is also intersection point (as you can see from the image below) but that's rather matter of perspective.

In the image below blue is curve `r=1+sin(theta/2)` and red is `r=1+cos(theta/2).`

Given

`r=1+sin(theta/2)` (i)

`r=1+cos(theta/2)` (ii)

(i) and (ii) will intersect each other if

`1+sin(theta/2)=1+cos(theta/2)`

`tan(theta/2)=1`

`theta/2=npi+pi/4`

`theta=(npi)/2+pi/2`

When `theta=pi/2 =>r=1+1/sqrt(2)`

Thus point of intersection is

`(r,theta)=(1+1/sqrt(2),(npi)/2+pi/2), n=0,1,2,3`