This question involves the concepts of Pythagora's Theorem and derivatives.
The plane's distance is increasing at a rate of "105.6 ft/" (OR) "72 mph".
First, we will write an expression for the distance between the person and the plane in terms of time. For that purpose, we will use Pythagora's Theorem. Here, the altitude will act as perpendicular, the horizontal distance on the ground will be the base, and the straight line distance between the plane and the person will be the hypotenuse.
[tex]Hyp^2 = Base^2+Perp^2\\s^2 = (vt)^2+(32000\ ft)^2\\s=\sqrt{(vt)^2+(32000)^2}[/tex]
where,
s = plane's distance from the person
v = horizontal speedof the plane = [tex](120\ mph)(\frac{1.47\ ft/s}{1\ mph})[/tex] = 176 ft/s
t = time
Therefore,
[tex]s=\sqrt{(176t)^2+(32000)^2}[/tex]
Taking the derivative with respect to time:
[tex]\\\\\frac{ds}{dt} = \frac{1}{2}\frac{(2t)(176)^2}{\sqrt{(176t)^2+(32000)^2}}\\\\\frac{ds}{dt} = \frac{61952\ t}{2s}[/tex]
where,
[tex]\frac{ds}{dt}[/tex] = rate at which plane's distance is increasing from the person = ?
s = distance = [tex]\sqrt{(24000\ ft)^2+(32000\ ft)^2} = 40000\ ft[/tex]
t = time = [tex]\frac{horizontal\ distance}{v} = \frac{24000\ ft}{176\ ft/s} = 136.36\ s[/tex]
Substituting these values in the equation, we get:
[tex]\frac{ds}{dt}=\frac{(61952)(136.36)}{(2)(40000)}[/tex]
[tex]\frac{ds}{dt}=105.6\ ft/s = 72\ mph[/tex]
Learn more about Pythagora's Theorem here:
https://brainly.com/question/343682?referrer=searchResults
The attahced picture illustrates Pythagora's Theorem.
