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

**Problem 33**

From a car traveling east at 40 miles per hour, an airplane traveling horizontally north at 100 miles per hour is visible 1 mile east, 2 miles south, and 2 miles up. Find when this two will be nearest together.

**Solution 33**

$s = \sqrt{x^2 + 2^2} = \sqrt{x^2 + 4}$

where:

x^{2} = (1 - 40t)^{2} + (2 - 100t)^{2}

x^{2} = (1 - 80t + 1600t^{2}) + (4 - 400t + 10000t^{2})

x^{2} = 5 - 480t + 11600t^{2}

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*

**Problem 34**

In Problem 33, find how fast the two will be separating after along time.

**Solution 34**

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

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

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

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

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*

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