I dont even know if there is an equation for this, but what is the equation for a quarter circle that has a center at the origin?

I dont even know if there is an equation for this, but what is the equation for a quarter circle that has a center at the origin?

Do you want the area or the length of the arc?

im just wondering what the equation of it would be if it were plotted on a graph

Area = π × r² : 1/4 of that would be 1/4 × π × r²

or

Area of sector = 1/2 × r² × θ : θ = angle in radians :1/4 circle is 90º = 0.5π rad = 1.5707963268 rad

Area of 1/4 circle = 1/2 × r² × 1.5707963268

 

Circumference = 2 × π × r : 1/4 of that would be 1/4 × 2 × π × r = 1/2 × π × r

Area = π × r² : 1/4 of that would be 1/4 × π × r²

or

Area of sector = 1/2 × r² × θ : θ = angle in radians :1/4 circle is 90º = 0.5π rad = 1.5707963268 rad

Area of 1/4 circle = 1/2 × r² × 1.5707963268

 

Circumference = 2 × π × r : 1/4 of that would be 1/4 × 2 × π × r = 1/2 × π × r

oh sorry i don't think u understand what i mean im talking about the equation of a quarter circle on a graph which has an origin as its center

oh sorry i don't think u understand what i mean im talking about the equation of a quarter circle on a graph which has an origin as its center

??

x^2+y^2<=r^2 and 0<=tan^(-1)(x, y)<=pi/2

(assuming radius r, center (0, 0), and rotation angle 0°)

Well a circle is (x-a)^2+(y-b)^2=r^2

 

where a and b are coordinates of the centre and r is the radius

 

- you cannot get a simple function mapping as multiple x's give the same y

- for only positive x and y would give you a quarter circle (edge only) in the top right quadrant.

I dont even know if there is an equation for this, but what is the equation for a quarter circle that has a center at the origin?

Although you worded incorrectly, I believe I know what you're asking. However, there isn't an equation that is defined only for a quarter circle without restricting it to some interval such as

 

[latex]F\left(\theta\right)=\left \langle r\,\text{cos}\,\theta,\ r\,\text{sin}\,\theta \right \rangle \ \ \left \{r,\theta\in\mathbb{R}\,\vert \, 0 \le \theta \le 45^\circ \right \}[/latex]

 

Here we have the parametric equation for a circle which makes defining the [latex]x[/latex] and [latex]y[/latex] coordinates easier as you are wanting to restrict the equation according to a specific angle. The interval [[latex]0,45^\circ[/latex]] is an arbitrary choice that is dependent upon our parametrization of the circle or vice versa. We can achieve the same restriction using an interval [[latex]0,1[/latex]].

 

[latex]F\left(t\right)=\left \langle r\,\text{cos}\left(45 t\right),\ r\,\text{sin}\left(45 t\right) \right \rangle \ \ \left \{r,t\in\mathbb{R}\,\vert \, 0 \le t \le 1 \right \}[/latex]

 

Given the equation for a circle that Imatfaal has provided [math]x^2+y^2=r^2[/math], we can demonstrate that our parametric equation does indeed satisfy the conditions of the equation for a circle centered at the origin - (0,0).

 

[math]x=r\,\text{cos}\,\theta[/math]

[math]y=r\,\text{sin}\,\theta[/math]

 

[math]\left(r\,\text{cos}\,\theta\right)^2+\left(r\,\text{sin}\,\theta\right)^2=r^2[/math]

 

[math]r^2\,\text{cos}^2\theta+r^2\,\text{sin}^2\theta=r^2[/math]

 

[math]r^2\left(\text{cos}^2\theta+\text{sin}^2\theta\right)=r^2[/math]

 

Use the Pythagorean trigonometric identity to simplify the equation.

 

[math]\text{cos}^2\theta+\text{sin}^2\theta=1[/math]

 

[math]r^2\left(1\right)=r^2[/math]

 

[math]r=r[/math]

Edited July 14, 201411 yr by Daedalus