11-12 Two trains; one going to east, and the other is heading north

Problem 11
A train starting at noon, travels north at 40 miles per hour. Another train starting from the same point at 2 PM travels east at 50 miles per hour. Find, to the nearest mile per hour, how fast the two trains are separating at 3 PM.

Solution 11

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Two trains, one going to north, and one to east

$s^2 = 40^2t^2 + 50^2(t - 2)^2$

$s = \sqrt{1600t^2 + 2500(t - 2)^2}$

$\dfrac{ds}{dt} = \dfrac{3200t + 5000(t - 2)}{2\sqrt{1600t^2 + 2500(t - 2)^2}}$

$\dfrac{ds}{dt} = \dfrac{4100t - 5000}{\sqrt{1600t^2 + 2500(t - 2)^2}}$

at 3 PM, t = 3
$\dfrac{ds}{dt} = \dfrac{4100(3) - 5000}{\sqrt{1600(3^2) + 2500(3 - 2)^2}}$

$\dfrac{ds}{dt} = 56.15 \, \text{ mi/hr}$ answer

Problem 12
In Problem 11, how fast the trains are separating after along time?

Solution 12

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After a long time, t = ∞
$\dfrac{ds}{dt} = \dfrac{4100t - 5000}{\sqrt{1600t^2 + 2500(t - 2)^2}}$

$\dfrac{ds}{dt} = \dfrac{4100t - 5000}{\sqrt{1600t^2 + 2500t^2 - 10\,000t + 10\,000}}$

$\dfrac{ds}{dt} = \dfrac{4100t - 5000}{\sqrt{4100t^2 - 10\,000t + 10\,000}} \times \dfrac{1/t}{1/t}$

$\dfrac{ds}{dt} = \dfrac{4100 - 5000/t}{\sqrt{4100 - 10\,000/t + 10\,000/t^2}}$

$\dfrac{ds}{dt} = \dfrac{4100 - 5000/\infty}{\sqrt{4100 - 10\,000/\infty + 10\,000/\infty^2}}$

$\dfrac{ds}{dt} = 64.03 \, \text{ mi/hr}$ answer

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