Answer:
[tex]\dfrac{dz}{dt} =181.11\ ft/s [/tex]
Explanation:
given,
constant altitude of aircraft (y)= 1000 ft
steady speed of aircraft = 232 ft/sec
distance between aircraft and trawler(z) = 1600 ft
using Pythagoras theorem
z² = x² + y²
[tex]x = \sqrt{1600^2-1000^2}[/tex]
x = 1249 m
[tex]\dfrac{dy}{dt} = 0[/tex]
[tex]\dfrac{dx}{dt}= 232[/tex]
differentiating both side w.r.t to t
[tex]2z \dfrac{dz}{dt} = 2x \dfrac{dx}{dt} + 2y\dfrac{dy}{dt}[/tex]
[tex]z \dfrac{dz}{dt} = x \dfrac{dx}{dt} [/tex]
[tex]\dfrac{dz}{dt} =\dfrac{x}{z} \dfrac{dx}{dt} [/tex]
[tex]\dfrac{dz}{dt} =\dfrac{1249}{1600} \times 232 [/tex]
[tex]\dfrac{dz}{dt} =181.11\ ft/s [/tex]
so the speed at which trawler is receding is 181.11 ft/s
This question involves the concepts of Pythagora's Theorem and derivatives.
The plane is receeding at a rate of "ft/s".
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+(1000\ ft)^2\\\\s=\sqrt{(vt)^2+(1000\ ft)^2}[/tex]
where,
s = plane's distance from the person
v = horizontal speedvof the plane = = 232 ft/s
t = time
Therefore,
[tex]s = \sqrt{(232)^2t^2+(1000)^2}[/tex]
Taking the derivative with respect to time:
[tex]\frac{ds}{dt}=\frac{1}{2}\frac{(2t)(232)^2}{\sqrt{(232)^2t^2+(1000)^2}}\\\\\frac{ds}{dt}=\frac{53824\ t}{s}[/tex]
where,
[tex]\frac{ds}{dt}[/tex] = rate at which plane is receding from the person = ?
s = distance = [tex]\sqrt{(1600\ ft)^2+(1000\ ft)^2}[/tex] = 1886.8 ft
t = time = [tex]\frac{horizontal\ distance}{v} = \frac{1600\ ft}{232\ ft/s} = 6.9\ s[/tex]
Substituting these values in the equation, we get:
[tex]\frac{ds}{dt}= \frac{(53824)(6.9)}{1886.8}\\\\\frac{ds}{dt}=196.8\ ft/s[/tex]
Learn more about Pythagora's Theorem here:
brainly.com/question/343682?referrer=searchResults
The attached picture illustrates Pythagora's Theorem.
