Answer
$(x-1)^{2}+(y+2)^{2}=1, \quad 1\leq x\leq 2$
Work Step by Step
Rewriting the parametric equations, isolating $\sin t$ and $\cos t,$
$\left\{\begin{array}{l}
x-1=\sin t\\
y+2=\cos t
\end{array}\right.$
Square both equations
$\left\{\begin{array}{l}
(x-1)^{2}=\sin^{2}t\\
(y+2)^{2}=\cos^{2}t
\end{array}\right.$
Add the two equations$,\qquad\cos^{2}t +\sin^{2}t=1$
$(x-1)^{2}+(y+2)^{2}=1$
- a circle of radius 1, centered at $(1,-2)$
From the parametric equation for x, we see that $x\in[1,2].$
so we restrict the Cartesian equation:
$(x-1)^{2}+(y+2)^{2}=1\quad 1\leq x\leq 2$
To graph, create a function value table using values for t, in ascending order of t, calculating the x- and y-coordinates of points on the graph.
Plot and join the points obtained with a smooth curve, noting the direction in which t increases.
