Answer
(a) See explanations.
(b) See graph.
Work Step by Step
(a) Step 1. Use the figure given in the Exercise, as the dog runs along the curve, the rope will be unwrapped from the cylinder which means that the length of TD should equal to the arc length of $r\theta=\theta$ or $TD=\theta$
Step 2. As TD is tangent to the circle, we have $\angle OTD=90^{\circ}$ so that the slope of the line TD should be $-\frac{1}{tan\theta}$ where $tan\theta$ is the slope of line OT.
Step 3. Knowing the coordinates of point T as $(cos\theta, sin\theta)$, assume the coordinates of D as $(x,y)$, using the results of steps 1 and 2, we have slope $\frac{y-sin\theta}{x-cos\theta}=-\frac{1}{tan\theta}$ and distance $(x-cos\theta)^2+(y-sin\theta)^2=(\theta)^2$
Step 4. The first equation gives $x-cos\theta=-tan\theta (y-sin\theta)$, use this in the second equation we get
$(y-sin\theta)^2(1+tan^2\theta)=(\theta)^2$ with $(1+tan^2\theta)=sec^2\theta$ we have $y=sin\theta\pm\theta cos\theta$ thus $x=cos\theta\mp\theta sin\theta$
Step 5. As the x-value should increase initially, we choose the $+$ sign for $x$ (corresponding to $-$ in $y$) and the final equations are $x=cos\theta+\theta sin\theta$ and $y=sin\theta-\theta cos\theta$
(b) See graph.
