Convert to a rectangular equation? #r + rsintheta = 1#

Answer is #x^2 + 2y = 1#

Thanks in advance

3 Answers
May 1, 2018

# r + r sin theta = 1 #

becomes

# x^2 + 2y = 1 #

Explanation:

We know

#r^2 = x^2 + y^2#

#x = r cos theta #

#y = r sin theta #

so

# r + r sin theta = 1 #

becomes

# \sqrt{x^2 + y^2 }+ y = 1 #

# \sqrt{x^2 + y^2} = 1-y #

# x^2 + y^2 = 1 - 2y + y^2 #

# x^2 + 2y = 1 #

The only iffy step is the squaring of the square root. Usually for polar equations we allow negative #r#, and if so the squaring doesn't introduce a new part.

May 1, 2018

Procedure in explanation.

Explanation:

To convert from polar to rectangular, we may use the following substitutions: #x=rcosθ#
#y=rsinθ#
#r^2 =x^2+y^2#
#tanθ=y/x#
Using 1 and 3,
#sqrt(x^2 + y^2) + y = 1#
Square the equation. Using the expansion of #(a + b)^2#
#x^2 + y^2 +y^2 +2ysqrt(x^2 + y^2) = 1#
#implies x^2 + 2y^2 + 2ysqrt(x^2 + y^2) = 1#
#implies x^2 + 2y(y + sqrt(x^2 + y^2)) = 1#

Notice that the coefficient of 2y is 1.(See the first equation I wrote using 1 and 3)
So #x^2 + 2y = 1#

Hope this helps!

May 1, 2018

#x^2 - 2y = 1#

Explanation:

#r + rsintheta = 1#

We need to convert from polar to rectangular form.

We know that:
#x = rcostheta#

#y = rsintheta#

and

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

#------------------#

We can substitute in these values for #color(red)r# and #color(red)(rsintheta)#:
#color(red)(sqrt(x^2 + y^2) + y) = 1#

Subtract #color(red)y# from both sides of the equation:
#sqrt(x^2 + y^2) +y quadcolor(red)(-quady)= 1 quadcolor(red)(-quady)#

#sqrt(x^2 + y^2) = 1-y#

Square both sides of the equation:
#(sqrt(x^2 + y^2))^color(red)(2) = (1-y)^color(red)(2)#

#x^2 + y^2 = 1 - 2y + y^2#

Subtract #color(red)(y^2)# from both sides of the equation so they cancel:

#x^2 + cancel(y^2 quadcolor(red)(-quady^2)) = 1 - 2y + cancel(y^2 quadcolor(red)(-quady^2))#

#x^2 = 1 - 2y#

Add #color(red)(2y)# to both sides of the equation to get the final answer in rectangular form:
#x^2 - 2y = 1#

Hope this helps!