Eliminate the parameter to find the Cartesian equation for the function:

Question

parametric equation: x = sin t, y = 1 + cos t.

Answer 1

sin²t + cos²t = 1

solve for cos t and substitute the appropriate squares of the trig values in the above equation; your result will contain x & y and no t.

Answer 2

Another way to solve this:

 

x = sint

 

y = 1 + cost = 1 + √(1 - sin²t) = 1 + √(1 - x²)

 

y =  1 + √(1 - x²)

 

Note this gives

 

1 - y = - √(1 - x²)

 

Square both sides and get

 

(1 - y)² = 1 - x² 

 

x² + (y - 1)² = 1

 

which is the equation of a circle of radius 1 centered at (0,1) as in Gregg's solution

 

 

 

 

Answer 3

This is a circle centered at (0,1) with a radus of 1

 

(x-0)^2 +(y-1)^2 = 1^2

 

x^2 + (y-1)^2 = 1  is the cartesian form

 

 

 

Here is why

 

sin^2(t) + cos^2(t) = 1   (identity)

 

if we square your original x and y equations, we obtain:

 

x^2 = sin^2(t)

y^2 = 1 +2cost + cos^2(t)

 

if we add the equations from above we get

 

x^2 + y^2 = sin^2(t) + cos^2(t) + 1 + 2cost apply the identity

 

x^2 + y^2 = 1 + 1 +2cos(t)    rearrange the equation by leaving one of the 1s on the right and moving one of the 1s and 2cost to the left side

 

x^2 + y^2-2cost - 1 = 1

 

cost can be written as y -1 because  the original equation states that y = 1 + cost

 

x^2 + y^2 - 2(y-1)-1 =1

 

x^2 + y^2 -2y +2 -1 = 1

 

x^2 y^2-2y+1 = 1        our underlined equation here is just (y-1)^2

 

Therefore we have our answer:

 

x^2 + (y-1)^2  + 1