How do you graph #r = 1 + sin(t)#?

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Answer 1 · 2017-02-15T03:05:25

See graph and explanation.

The general equation to a cardioid having its dimple at r = 0 is

#r = a(1+cos(t-alpha))#,

a gives the size and #alpha# = inclination of its axis of symmetry to

#t = 0#.

Here, a = 1 and #alpha=pi/2#.

# r (t)# has a period #2pi#.

#sqrt(x^2+y^2)=r = 1+sint>=0# and r in (0, 2), when #t in (0, pi/2)i#.

The graph is symmetrical about #t = pi/2#.

Now, see Socratic graph, depicting all these features.

I have used Cartesian form of the equation.

graph{x^2+y^2-sqrt(x^2+y^2)-y=0 [-5, 5, -2.5, 2.5]}