How do you convert the following to a rectangular equation: $r = cos (θ) - sin (θ)$ $r = cos (theta) - sin (theta)$ ?

Question

How do you convert the following to a rectangular equation: $r = cos (θ) - sin (θ)$ $r = cos (theta) - sin (theta)$ ?

2 Answers

Impact of this question

29003 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-05T14:26:34

${(x - \frac{1}{2})}^{2} + {(y + \frac{1}{2})}^{2} = \frac{1}{2}$ $(x-1/2)^2+(y+1/2)^2=1/2$

Explanation:

When converting from polar to rectangular equations you have to switch the references to $r$ $r$ and $θ$ $theta$ to $x$ $x$ and $y$ $y$ .

$x = r cos (θ)$ $x=rcos(theta)$
$y = r sin (θ)$ $y=rsin(theta)$
$r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$

Multiply each term by $r$ $r$

$r \cdot r = r cos (θ) - r sin (θ)$ $r*r=rcos(theta)-rsin(theta)$

Simply

$r^{2} = r cos (θ) - r sin (θ)$ $r^2=rcos(theta)-rsin(theta)$

Make the necessary substitutions

$x^{2} + y^{2} = x - y$ $x^2+y^2=x-y$

Gather the $x$ $x$ and $y$ $y$ variables to the same side

$(x^{2} - x) + (y^{2} + y) = 0$ $(x^2-x)+(y^2+y)=0$

Now complete the square. Take the coefficient of both the $x$ $x$ and $y$ $y$ variable then divide by 2 and then square

${(- \frac{1}{2})}^{2} = \frac{1}{4}$ $(-1/2)^2=1/4$

${(\frac{1}{2})}^{2} = \frac{1}{4}$ $(1/2)^2=1/4$

Insert the above 2 quantities on both sides of the equation to keep it balanced.

$(x^{2} - x + \frac{1}{4}) + (y^{2} + y + \frac{1}{4}) = \frac{1}{4} + \frac{1}{4}$ $(x^2-x+1/4)+(y^2+y+1/4)=1/4+1/4$

Simplify

$(x^{2} - x + \frac{1}{4}) + (y^{2} + y + \frac{1}{4}) = \frac{1}{2}$ $(x^2-x+1/4)+(y^2+y+1/4)=1/2$

Convert the 2 perfect square trinomials

${(x - \frac{1}{2})}^{2} + {(y + \frac{1}{2})}^{2} = \frac{1}{2}$ $(x-1/2)^2+(y+1/2)^2=1/2$

Check out this tutorial on the conversion of equations from polar to rectangular.

Check out this tutorial on completing the square graphically

Check out this tutorial on completing the square analytically.

Answer 2 · 2016-10-05T14:40:56

${(x - \frac{1}{2})}^{2} + {(y + \frac{1}{2})}^{2} = \frac{1}{2}$ $(x-1/2)^2+(y+1/2)^2=1/2$ ,
representing the circle through the origin,
with center at $(\frac{1}{2}, - \frac{1}{2}) and r a d i u s \frac{1}{\sqrt{2}}$ $(1/2, -1/2) and radius 1/sqrt2$ .

Explanation:

Use the conversion equation $r (cos θ, sin θ) = (x, y)$ $r(cos theta, sin theta)=(x, y)$ that

gives $r = \sqrt{x^{2} + y^{2}}, cos θ = \frac{x}{r} and sin θ = \frac{y}{r}$ $r=sqrt(x^2+y^2), cos theta =x/r and sin theta = y/r$ .

Here, $r = \sqrt{x^{2} + y^{2}} = \frac{x - y}{r} = \frac{x - y}{\sqrt{x^{2} + y^{2}}}$ $r = sqrt(x^2+y^2)=(x-y)/r=(x-y)/sqrt(x^2+y^2)$ .

Cross multiplying, $x^{2} + y^{2} = x - y$ $x^2+y^2=x-y$ , or in the standard form,

${(x - \frac{1}{2})}^{2} + {(y + \frac{1}{2})}^{2} = \frac{1}{2}$ $(x-1/2)^2+(y+1/2)^2=1/2$ , representing the circle through the origin,

with center at $(\frac{1}{2}, - \frac{1}{2}) and r a d i u s \frac{1}{\sqrt{2}}$ $(1/2, -1/2) and radius 1/sqrt2$ .

Noe that the general polar equation of the circles through the pole r

= 0 is $r = d cos (θ + α)$ $r = d cos (theta+alpha)$ . Here, the diameter d = sqrt 2 and

$α = \frac{π}{4}$ $alpha= pi/4$ ..

Science

Math

Humanities

... and beyond

How do you convert the following to a rectangular equation: $r = cos (θ) - sin (θ)$ $r = cos (theta) - sin (theta)$ ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you convert the following to a rectangular equation: r = cos (theta) - sin (theta)r=cos(θ)−sin(θ)r = cos (theta) - sin (theta)?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you convert the following to a rectangular equation: $r = cos (θ) - sin (θ)$ $r = cos (theta) - sin (theta)$ ?