A bug is crawling along a parabolla y= x^2 (in quadrant 1) at a rate such that its distance from the origin is increasing at 6 cm/min. At what rate are the x- and y- coordinates of the bug increasing when the bug is at the point (3,9)?
We know that y(t) = (x(t))^{2}. Therefore:
$y'(t) = 2x(t)(x'(t))$
We also know that the distance $d(t)$ from the origin is:
$d(t) = \sqrt{(x(t))^2 + (y(t))^2}$.
Taking $d'(t)$ we get:
$6 = d'(t) = \dfrac{(2x(t))(x'(t)) + (2y(t))(y'(t))}{2\sqrt{(x(t))^2 + (y(t))^2}}$
$= \dfrac{(x'(t))(x(t))(1 + 2(x(t))^2)}{x(t)\sqrt{1 + (x(t))^2}}$
so that:
$x'(t) = \dfrac{6\sqrt{1 + (x(t))^2}}{(1 + 2(x(t))^2}$
Substituting in $x(t) = 3$, we obtain:
$x'(t) = \dfrac{6\sqrt{10}}{19}$ and:
$y'(t) = 2x(t)(x'(t)) = \dfrac{36\sqrt{10}}{19}$