Particle moving along an ellipse; related rates This is my very first post here, so sorry if I did anything wrong. This is a related rates problem for first semester calculus. I've been trying for some time and still have no idea how to solve it...
A particle is moving around the ellipse $4x^2 + 16y^2 = 64$. At any time t, its x- and y-coordinates are given by $x(t) = 4 \cos (t)$ and $y(t) = 2 \sin (t)$. At what rate is the particle's distance to the point $(2,0)$ changing at any time $t$? At what rate is the distance changing when $t = (\pi)/4$?
 A: For the first part, construct the distance function

$$ S = \sqrt{ (x-2)^2 + y^2 },$$

which we got by considering the distance between the point $(x,y)$ on the curve and the point $(2,0)$, and change it in terms of the parameter $t$ using the relations you have been given, then find $\frac{d S}{dt}$. For the second part, just substiyute $t=\frac{\pi}{4}$ in the last equation you get. 
A: Suppose for a second we forget about the assumption that the particle moves on the ellipse
$4x^2 + 16y^2 = 64, \tag{1}$
and that the $x$ and $y$ components of the motion are given by
$x(t) = 4 \cos t \tag{2}$
and
$y(t) = 2 \sin t, \tag{3}$
and just look at any (differentiable) curve $(x(t), y(t))$; furthermore, let us suppose we want to write down the distance between the point on the curve at time $t$,  $(x(t), y(t))$, and any particular point $(x_0, y_0)$, where the only assumption we will make is that
$(x_0, y_0)$ is not on the curve $(x(t), y(t))$; that is, $(x(t), y(t) \ne (x_0, y_0)$ for any $t$ under consideration.  If $r(t)$ denotes that distance at time $t$, then by the Euclidean distance formula (AKA The Pythagorean Theorem) we have
$r^2(t) = (x(t) - x_0)^2 + (y(t) - y_0)^2; \tag{4}$
then $r(t) \ne 0$ for any $t$.  We want to know $\dot r(t)$; if we differentiate (4) we obtain
$2r(t) \dot r(t) = 2(x(t) - x_0) \dot x(t) + 2(y(t) - y_0) \dot y(t), \tag{5}$
and since $r(t) \ne 0$ we can divide by $2r(t)$ to yield
$\dot r(t) = ((x(t) - x_0) / r(t)) \dot x(t) + ((y(t) - y_0) / r(t)) \dot y(t); \tag{6}$
this result is quite general and applies to any curve $(x(t), y(t))$ and point $(x_0, y_0)$ not on the curve.  We apply (6) to the case at hand; with $x(t), y(t)$ as in (2), (3) the derivatives are
$\dot x(t) = -4 \sin t \tag{7}$
and
$\dot y(t) = 2 \cos t. \tag{8}$
Substituting (2), (3), as well as $(x_0, y_0) = (2, 0)$ into (4) we see that the distance $r(t)$ satisfies
$r^2(t) = (4 \cos t -2)^2 + (2 \sin t)^2, \tag{9}$
which may be simplified using the identity $\sin^2 t + \cos^2 t = 1$:
$r^2(t) = (4 \cos t -2)^2 + (2 \sin t)^2 = 16 \cos^2 t - 8 \cos t + 4 + 4 \sin^2 t$
$= 12 \cos^2 t + 4(\cos^2 t + \sin^2 t) - 8 \cos t + 4 = 12 \cos^2 t - 8 \cos t + 8, \tag{10}$
so that, taking square roots,
$r(t) = \sqrt{12 \cos^2 t - 8 \cos t + 8}. \tag{11}$
Turning now to (6), we insert the formulas for $x(t), y(t), \dot x(t), \dot y(t)$ from (2)-(3), (7)-(8), as well as the values of $x_0, y_0$ into it to obtain
$\dot r(t) = -4 \sin t ((4 \cos t - 2) / r(t)) + 2 \cos t ((2 \sin t) / r(t))$
$= (8 \sin t -12 \sin t \cos t) / r(t), \tag{12}$
where we have left the expression $r(t)$ intact for the sake of simplicity.  (12) is the sought-for general formula for $r(t)$; if $t = \pi / 4$ we have, since $\sin (\pi / 4) = \cos (\pi / 4) = \sqrt 2 / 2$, first for $r(\pi / 4)$,
$r(\pi / 4) = \sqrt{12 (1 / 2) - 4 \sqrt 2 + 8} = \sqrt{14 - 4\sqrt 2}, \tag{13}$
and then for $\dot r(\pi / 4)$, using (13) as well,
$\dot r(\pi / 4) = (-6 + 4 \sqrt 2) /(\sqrt{14 - 4\sqrt 2}), \tag{14}$
assuming I did all that arithmetic correctly!
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
