Convert to Rectangular Form...1

[nycmathguy]

Can you show me how to do 14 and 16?

 

[MathLover1]

14.

r^2=1/(3+cos^2(theta))

r^2=x^2+y^2

x=r *cos(θ) =>cos(θ) =x/r =>cos(θ) =x/sqrt(x^2+y^2), then cos^2(θ) =x^2/(x^2+y^2)
y=r*sin(θ)=>sin(θ)=y/r

substitute in given formula

x^2+y^2=1/(3+x^2/(x^2+y^2))

(x^2+y^2) (3+x^2/(x^2+y^2))=1

3x^2+3y^2+x^2=1

4x^2+3y^2=1


16.

r*sin(theta+pi/4)=6

r=sqrt(x^2+y^2)
y=rsin(θ)=>sin(θ)=y/r

Apply the formula sin(α+β)=cos(α)sin(β)+sin(α)cos(β) with α=θ and β=π /4
sin(π /4)=1/sqrt(2)=sqrt(2)/2
cos(π /4)=sqrt(2)/2

sin(θ +π /4)
=cos(θ)(sqrt(2)/2)+sin(θ)(sqrt(2)/2)
=(sqrt(2)cos(θ)/2+sqrt(2)sin(θ)/2)

then

..............substitute sin(θ)=y/r , cos(θ) =x/r , and r=sqrt(x^2+y^2)

......simplify

x*sqrt(2)+y*sqrt(2)=12

y = 6sqrt(2) - x



.

 

[nycmathguy]

14.

r^2=1/(3+cos^2(theta))

r^2=x^2+y^2

x=r *cos(θ) =>cos(θ) =x/r =>cos(θ) =x/sqrt(x^2+y^2), then cos^2(θ) =x^2/(x^2+y^2)
y=r*sin(θ)=>sin(θ)=y/r

substitute in given formula

x^2+y^2=1/(3+x^2/(x^2+y^2))

(x^2+y^2) (3+x^2/(x^2+y^2))=1

3x^2+3y^2+x^2=1

4x^2+3y^2=1


16.

r*sin(theta+pi/4)=6

r=sqrt(x^2+y^2)
y=rsin(θ)=>sin(θ)=y/r

Apply the formula sin(α+β)=cos(α)sin(β)+sin(α)cos(β) with α=θ and β=π /4
sin(π /4)=1/sqrt(2)=sqrt(2)/2
cos(π /4)=sqrt(2)/2

sin(θ +π /4)
=cos(θ)(sqrt(2)/2)+sin(θ)(sqrt(2)/2)
=(sqrt(2)cos(θ)/2+sqrt(2)sin(θ)/2)

then

View attachment 2207

View attachment 2208..............substitute sin(θ)=y/r , cos(θ) =x/r , and r=sqrt(x^2+y^2)

View attachment 2209......simplify

x*sqrt(2)+y*sqrt(2)=12

y = 6sqrt(2) - x



.

Thank you very much. I will try a few on my own.