sinθ = (x-1)
cosθ = (y+1)/2
sin^2 θ = (x-1)^2
cos^2 θ = (y+1)^2/4
(x-1)^2 + (y+1)^2/4 = 1
Looks like an ellipse with center (1,-1)
Convert the curve to an equation in rectangular coordinates:
x=sin theta + 1
y=2cos theta - 1
4 answers
Steve, thank you so very much!!! I am *trying* to wrap up my studying/homework and am having trouble with one more... It's asking the exact same question for the curve:
x=ln(t), y=ln sqrt(t)
A BIG THANK YOU AHEAD OF TIME!!!!! =)
x=ln(t), y=ln sqrt(t)
A BIG THANK YOU AHEAD OF TIME!!!!! =)
x=ln(t)
y = ln(sqrt(t)) = 1/2 ln(t)
so, y = 1/2 x
y = ln(sqrt(t)) = 1/2 ln(t)
so, y = 1/2 x
Find the arc length of the curve described by the parametric equation over the given interval:
x=t^(2) + 1
y=2t - 3
...0 < t < 1
x=t^(2) + 1
y=2t - 3
...0 < t < 1
1 answer