What is the name of the curve where its radial vector ....

[yoyo]

what is the name of the curve where its radial vector drawn from the origin intersects the curves tangent line at a=3pi/4 or 135 degree at every point?

can anybody even show me how the curve looks like or what is the general eqaution of this curve?

thanks

 

[Gamma]

Take the genaral equation of the radial vector as y=mx and tangent line as y= m1x + c1. Let the point of intersection be (x1, y1). This means both the above equations should be satisefied by (x1,y1). So plug in x1, y1 and get m and m1 in terms of x1, y1 and c1. Now use the fact that the angle 'a' between two lines is given by tan (a) = m2 - m1 / (1 + m2m1). You will get an expression in terms of x1, y1 and c1. This is the equation of the curve. Should look familiar.

Note: This may be a long way. There may be short cuts. I can't think of any right now. :zzz:

 

[thepatientmental]

The curve you are describing is called a cardioid. It gets its name from the Latin word for heart, "cardia", because its shape resembles a heart with a cusp at the top. The general equation for a cardioid is r = a(1 + cosθ), where r is the radial distance from the origin and θ is the angle measured counterclockwise from the positive x-axis. When a = 3π/4 or 135 degrees, the radial vector intersects the tangent line at every point, creating the distinctive heart shape. You can visualize this curve by plotting points with different values of θ and r on a polar coordinate system.