How do you convert r=2+cosθ into rectangular form?

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

1
Feb 9, 2018

Answer:

#color(blue)(x^4+2x^2y^2-2x^3+y^4-2y^2x+x^2-4x^2-4y^2=0)#

Explanation:

enter image source here

From the diagram we can see that point P has polar coordinates
#( r , theta )# and Cartesian coordinates #(x,y)#.

And #color(white)(88)x=rcos(theta)# , #y = rsin(theta)#

#(x,y) -> (rcos(theta), rsin(theta))#

Also:

By Pythagoras' Theorem :

#r^2=(rcostheta)^2+(rsintheta)^2#

Since:

#x=rcos(theta)# and #y = rsin(theta)#

Then:

#r^2=x^2+y^2# #:. r=sqrt(x^2+y^2)#

Using these ideas:

#r=2+cos(theta)#

Substitute: #color(white)(88)r=sqrt(x^2+y^2)#

#sqrt(x^2+y^2)=2+cos(theta)#

Substitute: #color(white)(88)cos(theta)=x/r#

#sqrt(x^2+y^2)=2+x/r#

Substitute fro #bbr# again:

#sqrt(x^2+y^2)=2+x/sqrt(x^2+y^2)#

Multiply by: #sqrt(x^2+y^2)#

#sqrt(x^2+y^2)*sqrt(x^2+y^2)=2sqrt(x^2+y^2)+(xsqrt(x^2+y^2))/sqrt(x^2+y^2)#

#x^2+y^2=2sqrt(x^2+y^2)+(xcancelsqrt(x^2+y^2))/cancelsqrt(x^2+y^2)#

#x^2+y^2=2sqrt(x^2+y^2)+x#

#x^2+y^2-x=2sqrt(x^2+y^2)#

Squaring:

#x^4+2x^2y^2-2x^3+y^4-2y^2x+x^2=4x^2+4y^2#

#color(blue)(x^4+2x^2y^2-2x^3+y^4-2y^2x+x^2-4x^2-4y^2=0)#

I did not include the steps in expanding #(x^2+y^2-x)^2#. This would make the answer too long.

Was this helpful? Let the contributor know!
1500