Answer:
The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is [tex]\sqrt{725}[/tex] ft/ sec
Explanation:
Given:
h(t) = 25 ft/sec
x(t) = 10 ft/ sec
h(5) = 25 ft/sec . 5 = 125 ft
x(5) = 10 ft/sec . 5 = 50 ft
Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem
[tex]D(t) = \sqrt{h^2 + x^2}[/tex]
Lets find the derivative of distance with respect to time
[tex]\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}[/tex]
Substituting the values of h(t) and x(t) and simplifying we get,
[tex]\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}[/tex]
[tex]\frac{dh}{dt} = 25ft/sec[/tex]
[tex]\frac{dx}{dt} = 10 ft/sec[/tex]
[tex]\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}[/tex] = [tex]\frac{725}{\sqrt{725}}[/tex] = [tex]\sqrt{725}[/tex] ft / sec
The rate of change of the distance between the helicopter and yourself is 25.72 ft/s.
The given parameters;
The position of the helicopter after 5 s;
a = v₁t = 25 ft/s x 5s = 625 ft
b = v₂t = 10 ft/s x 5s = 50 ft
The displacement from the initial position;
[tex]c^2 = a^2 + b^2\\\\c= \sqrt{a^2 + b^2} \\\\c= \sqrt{625^2 + 50^2}\\\\c = 627 \ ft[/tex]
The rate of change of the distance between the helicopter and yourself is calculated as follows;
[tex]c^2 = a^2 + b^2\\\\2c\frac{dc}{dt} = 2a\frac{da}{dt} + 2b\frac{db}{dt} \\\\c\frac{dc}{dt} = a\frac{da}{dt} + b\frac{db}{dt}\\\\627 \frac{dc}{dt}= 625(25) + 50(10)\\\\627 \frac{dc}{dt} = 16125\\\\\frac{dc}{dt} = \frac{16125}{627} \\\\\frac{dc}{dt} = 25.72 \ ft/s[/tex]
Thus, the rate of change of the distance between the helicopter and yourself is 25.72 ft/s.
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