Asked by Joshua
I wanted to confirm that I solved these problems correctly (we had to convert the polar curves to Cartesian equations).
1.rcos(theta)=1
x=1
2.r=2*sin(theta)+2*cos(theta)
r^2=2rsin(theta)+2rcos(theta)
x^2+y^2=2y+2x (a little unsure what do next if this is correct)
3.r=tan(theta)sec(theta)
r=sin(theta)/(cos(theta))^2
r(cos(theta))^2=sin(theta)
r^2(cos(theta))^2=rsin(theta)
x^2=y
1.rcos(theta)=1
x=1
2.r=2*sin(theta)+2*cos(theta)
r^2=2rsin(theta)+2rcos(theta)
x^2+y^2=2y+2x (a little unsure what do next if this is correct)
3.r=tan(theta)sec(theta)
r=sin(theta)/(cos(theta))^2
r(cos(theta))^2=sin(theta)
r^2(cos(theta))^2=rsin(theta)
x^2=y
Answers
Answered by
Reiny
I agree with your answers
Nice work on #3
in #2, form
x^2 + y^2 = 2y + 2x you have a circle
you could complete the square, and find its centre and radius in this way ...
x^2 + y^2 - 2y - 2x = 0
x^2 - 2x + 1 + y^2 - 2y + 1 = 2
(x-1)^2 + (y-1)^2 = 2
so centre is (1,1) and radius is √2
Nice work on #3
in #2, form
x^2 + y^2 = 2y + 2x you have a circle
you could complete the square, and find its centre and radius in this way ...
x^2 + y^2 - 2y - 2x = 0
x^2 - 2x + 1 + y^2 - 2y + 1 = 2
(x-1)^2 + (y-1)^2 = 2
so centre is (1,1) and radius is √2
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