Chapter 3: Derivatives - Section 3.8 - Related Rates - Exercises 3.8 - Page 164: 37

$-5m/sec$

Step 1. We are given $x'=\frac{dx}{dt}=-1m/sec$, $y'=\frac{dy}{dt}=-5m/sec$, point $(5,12)$ Step 2. The distance from point $P(x,y)$ to the origin is $s^2=x^2+y^2$; thus we have $2ss'=2xx'+2yy'$ or $s'=\frac{xx'+yy'}{s}$ Step 3. As $s=\sqrt {x^2+y^2}=\sqrt {5^2+12^2}=13$, we have: $\frac{ds}{dt}=s'=\frac{5(-1)+12(-5)}{13}=-\frac{65}{13}=-5m/sec$