Equation for Point P's Path in Parametric Problem

[Mauserketi]

Circle A is fixed at center (1,0) with a radius 1. Circle B, also with radius 1, rotates at one revolution per (2*PI) seconds. Circle B is always connected to circle A at a single point. If at t=0, circle B is centered at (3,0) and point P (point p is on the edge of circle B) is at (4,0), what is the equation for P's path? (It should be a heart).
X=_________
Y=_________

 

[z-component]

Welcome to PF! Since this is a HW problem, you'll need to show what you've attempted at this problem in order for you to receive help. :)

 

[Architeuthis Dux]

The equation for point P's path in this parametric problem can be derived by first defining the position of circle B at any given time t. Since it rotates at one revolution per (2*PI) seconds, we can use the parametric equations of a circle to determine its position at any time t:

Xb = 1*cos(t) + 3
Yb = 1*sin(t)

Next, we need to define the position of point P on the edge of circle B. Since it is always connected to circle A at a single point, we can use the fact that the distance between the centers of the two circles is always 2 units (since both have a radius of 1). This means that the coordinates of point P can be defined as:

Xp = Xb + 2*cos(t)
Yp = Yb + 2*sin(t)

Substituting the equations for Xb and Yb from before, we get:

Xp = cos(t) + 3 + 2*cos(t)
Yp = sin(t) + 2*sin(t)

Simplifying and combining like terms, we get:

Xp = 3*cos(t) + 3
Yp = 3*sin(t)

This is the equation for point P's path in terms of t. To graph it, we can plot the points (Xp, Yp) for various values of t. This will result in a heart-shaped curve, as requested.