How do you convert $r = 5 cos (t)$ $r = 5 cos (t)$ into cartesian form?

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How do you convert $r = 5 cos (t)$ $r = 5 cos (t)$ into cartesian form?

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Answer 1 · 2015-11-28T19:17:36

The equation relates $r and θ$ $r and theta$ , and describes a set of points $(r, θ)$ $(r,theta)$ which form a curve in the plane. We want to know the curve's equation in $(x, y)$ $(x,y)$ coordinates.

Explanation:

You used $t$ $t$ instead of $θ$ $theta$ . We can do that.

Start with knowing where we're going: the variables are related by things like
$x = r cos (t) and y = r sin (t) and r^{2} = x^{2} + y^{2} .$ $x = r cos(t) and y = r sin(t) and r^2 = x^2+y^2.$

You have $r = 5 cos (t),$ $r = 5 cos(t),$ let's create an $r cos (t$ $r cos(t$ ) by multiplying both sides by $r .$ $r.$

$r^{2} = 5 r cos (t)$ $r^2 = 5 r cos(t)$ . . . Now substitute and see that it turns into

$x^{2} + y^{2} = 5 x .$ $x^2+y^2=5x.$

This is the answer. If you complete the square you get the form

$x^{2} - 5 x + y^{2} = 0$ $x^2 - 5x + y^2 = 0$ . . . Half of 5 is 5/2, whose square is 25/4:

$x^{2} - 5 x + \frac{25}{4} + y^{2} = \frac{25}{4}$ $x^2 - 5x+25/4 + y^2 = 25/4$ , or in circle form:

${(x - \frac{5}{2})}^{2} + {(y - 0)}^{2} = {(\frac{5}{2})}^{2}$ $(x-5/2)^2+(y-0)^2=(5/2)^2$ ,

You answer these ?'s: "It's a circle of center $(?, ?)$ $(?,?)$ and radius $?$ $?$ ."

...// dansmath strikes again! \\...

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How do you convert $r = 5 cos (t)$ $r = 5 cos (t)$ into cartesian form?

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Explanation:

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Science

Math

Humanities

... and beyond

How do you convert r = 5 cos (t)r=5cos(t)r = 5 cos (t) into cartesian form?

1 Answer

Explanation:

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Impact of this question

How do you convert $r = 5 cos (t)$ $r = 5 cos (t)$ into cartesian form?