Minimum Distance between a kitten and a puppy moving on a coordinate plane At time $t= 0$, the kitten is at $(1,0)$ and the puppy is at $(-1,0)$. The kitten starts walking counterclockwise along the unit circle, and the puppy runs right along the $x$-axis. If the puppy's speed is always twice the kitten's, what will the shortest distance ever between the two be?
I've been trying to create parametric equations for the two, and see if I can use the pythagorean theorem with their $x$-coordinates, but wasn't able to get anywhere with finding the minimum distance.
Puppy: $x= t$, $y = 0$
Kitten: $x = \cos(\frac{-t}{2})$, $y = \sin(\frac{-t}{2})$ [since counterclockwise movement]
All help is appreciated.
 A: Let's first make 2 little equations for the position of the puppy and kitten with respect to time. Let's do the kitten first (it's a bit harder). If imagine the kitten is moving a distance of 1 unit per second (or whatever time unit) over the circle surface, then the formula for it's angle relative to the origin point will be $\frac{t}{2\pi}$ in radians. So, to find the $x$ value at some time you take $\cos(\frac{t}{2\pi})$ and to find the $y$ value you take $\sin(\frac{t}{2\pi})$. Now the puppy is easier, the y value will always equal $0$, but the $x$ value is simply given with: $f(t) = 2t-1$ (moving right at a rate of 2 units per second to the right from a starting location of $-1$). Now to find the distance between them we use the Pythagorean theorem:
$$d(t) = \sqrt{(puppy\text{ }x - kitten\text{ }x)^2 + (puppy\text{ }y - kitten\text{ }y)^2}$$
Plugging in the formulas we just found:
$$d(t) = \sqrt{\left (2t-1- \cos\frac{t}{2\pi}\right )^2 + \left (0 - \sin \frac{t}{2\pi} \right )^2}$$
Now you have a formula for the distance between the two as a function of time. Plug this into your calculator and find the time when the distance between the two is minimized. If you can't use your calculator find this point by taking the derivative of this formula. Plug this time into the formula again to find the distance between the two.
