Answer
$11ft/sec$
Work Step by Step
Step 1. Using the figure given in the Exercise, after $3sec$, we have $x=3\times17=51ft$, $y=65+3\times1=68ft$
Step 2. Using the relation $s^2=x^2+y^2$, we have $2ss'+2xx'+2yy'$, which gives $s'=\frac{xx'+yy'}{s}$
Step 3. We are given $x'=17ft/sec, y'=1ft/sec, s=\sqrt {51^2+68^2}=85ft$
Step 4. We can get $\frac{ds}{dt}=s'=\frac{(51)(17)+(68)(1)}{85}=\frac{935}{85}=11ft/sec$
