33-34 Time Rates: A car traveling east and airplane traveling north

where:
x² = (1 - 40t)² + (2 - 100t)²
x² = (1 - 80t + 1600t²) + (4 - 400t + 10000t²)
x² = 5 - 480t + 11600t²
 

Thus,
$s = \sqrt{(5 - 480t + 11600t^2) + 4}$

$s = \sqrt{9 - 480t + 11600t^2}$

$\dfrac{ds}{dt} = \dfrac{-480 + 23200t}{2\sqrt{9 - 480t + 11600t^2}} = 0$

$-480 + 23200t = 0$

$t = \frac{3}{145} \,\, \text{ hr }$

$t = 1\frac{7}{29} \, \text{ min}$           answer

$\dfrac{ds}{dt} = \dfrac{-480 + 23200t}{2\sqrt{9 - 480t + 11600t^2}}$

$\dfrac{ds}{dt} = \dfrac{-240 + 11600t}{\sqrt{9 - 480t + 11600t^2}} \, \times \, \dfrac{1/t}{1/t}$

$\dfrac{ds}{dt} = \dfrac{-\dfrac{240}{t} + 11600}{\sqrt{\dfrac{9}{t^2} - \dfrac{480}{t} + 11600}}$
 

after a long time, $t \to \infty$
$\dfrac{ds}{dt} = \dfrac{-\dfrac{240}{\infty} + 11600}{\sqrt{\dfrac{9}{\infty^2} - \dfrac{480}{\infty} + 11600}}$

$\dfrac{ds}{dt} = \dfrac{11600}{\sqrt{11600}}$

$\dfrac{ds}{dt} = \sqrt{11600}$

$\dfrac{ds}{dt} = 107.7 \, \text{ mi/hr}$           answer