Respuesta :
Answer:
(a) h(t) = 300 - 30t
(b) θ = [tex]tan^{-1}[/tex]((200 - 30t)150)
[tex]-\frac{3}{35-3t}[/tex]
(c) t = [tex]6\frac{2}{3} sec[/tex]
h = 100m above ground
Step-by-step explanation:
(a)
change of distance of the elevator is governed by equation of speed
speed = distance/time
distance = speed x time
= 30 x t = 30t
since the elevator starts at t = 0 on top of the hotel at 300ft, and the distance between the elevator keep decreasing to the ground, the equation becomes
h(t) = 300 - 30t
(b)
θ = angle between the line of horizon of the observer to the line of sight to elevator
since the observer is 100 feet above the ground, the vertical distance between observer and the top of the hotel is 200ft.
hence, the opposite side of the angle θ is 300 - 30t - 100 = 200 - 30t
therefore, tan θ = (200 - 30t)/150
and θ = [tex]tan^{-1}[/tex]((200 - 30t)150)
Differentiate θ with respect to time, t.
Use chain rule: [tex]\frac{de}{dt}=\frac{de}{du}. \frac{du}{dt}[/tex]
let u = (200 - 30t)/150
u' = -1/5
θ = [tex]tan^{-1}[/tex]u
θ' = [tex]\frac{1}{1+u^{2} }[/tex]
[tex]\frac{de}{dt} = \frac{15}{35-3t}.(-\frac{1}{5})[/tex]
= [tex]-\frac{3}{35-3t}[/tex]
(c)
The elevator will seem to be moving the fastest at θ = 0 since that is the fastest change of θ as the elevator move downward
tan θ = (200 - 30t)/150
tan (0) = 0
(200 - 30t) = 0
200 = 30t
t = 200/30 = [tex]6\frac{2}{3} sec[/tex]
height = 100m above ground
The speed of the motion of the elevator relative to the observer depends on the position of the elevator.
The correct responses are;
(a) h(t) = 300 - 30·t
[tex](b) \ \theta(t)= arctan \left(\dfrac{h(t) - 100}{150}\right)[/tex]
(c) The rate of change of θ with time is [tex]\dfrac{d\theta(t)}{dt} = \dfrac{-45}{9 \cdot t^2-120 \cdot t + 625}[/tex]
(d) The time when the elevator will be moving fastest is after [tex]6\frac{2}{3}[/tex] seconds
(e) The elevator will appear to be moving fastest at height of 100 feet.
Reasons:
Height of the building = 300 ft.
Height of observer above ground = 100 feet
Distance of the hotel from the building = 150 feet.
Speed with which the elevator descends = 30 ft./sec
Starting time, t = 0
θ = Angle of elevation of line of sight of the observer
(a) The formula for h(t) is h(t) = 300 - 30·t
(b) Angle θ is given as follows;
[tex]tan (\theta) = \dfrac{Opposite \ distance}{Adjacent \ distance}[/tex]
The opposite distance = The elevators vertical height relative to the observer
∴ The opposite distance = h(t) - 100
The adjacent distance = The horizontal distance of the elevator from the observer
∴ The adjacent distance = 150
[tex]tan(\theta) = \dfrac{h(t) - 100}{150}[/tex]
Which gives;
[tex]\theta(t)= arctan \left(\dfrac{h(t) - 100}{150}\right)[/tex]
[tex]\theta(t) = arctan \left(\dfrac{300 - 30 \cdot t - 100}{150}\right) = arctan \left(\dfrac{200 - 30 \cdot t }{150}\right)[/tex]
[tex]\underline{\theta(t)= arctan \left(\dfrac{200 - 30 \cdot t }{150}\right)}[/tex]
(c) The rate of change of θ with time is given as follows;
[tex]\dfrac{d\theta(t)}{dt} = \dfrac{d}{dt} \left(arctan \left(\dfrac{200 - 30 \cdot t }{150}\right) \right) = -\dfrac{1}{5} \cdot \dfrac{1}{\left(\dfrac{200 - 30 \cdot t }{150}\right)^2 + 1}[/tex]
[tex]\dfrac{d\theta(t)}{dt} = -\dfrac{1}{5} \cdot \dfrac{1}{\left(\dfrac{200 - 30 \cdot t }{150}\right)^2 + 1} = \dfrac{-45}{9 \cdot t^2-120 \cdot t + 625}[/tex]
(d) The elevator will be moving fastest when the rate of change of angle of elevation with time is he fastest
Therefore, when the denominator of the derivative, [tex]\dfrac{d\theta(t)}{dt}[/tex] approaches zero, which is given as follows;
[tex]\dfrac{d\theta(t)}{dt} = -\dfrac{1}{5} \cdot \dfrac{1}{\left(\dfrac{200 - 30 \cdot t }{150}\right)^2 + 1}[/tex]
Therefore, when 200 - 30·t = 0, the elevator will be moving fastest, which gives;
[tex]t = \dfrac{200}{30} = 6\dfrac{2}{3}[/tex]
The time when the elevator will be moving fastest is t = [tex]6\frac{2}{3}[/tex] seconds.
(e) The rate of change of angle of elevation with time can be expressed as follows;
[tex]\dfrac{d\theta(t)}{dt} = -\dfrac{1}{5} \cdot \dfrac{1}{\left(\dfrac{200 - 30 \cdot t }{150}\right)^2 + 1} = -\dfrac{1}{5} \cdot \dfrac{1}{\left(\dfrac{h(t) -100 }{150}\right)^2 + 1}[/tex]
Therefore, the function is a minimum when the height of the elevator is h(t) = 100 feet, which is the height of the observer above the ground
Which gives;
When the height of the elevator is at the same level as the, height of the observer, and therefore when the angle of elevation is zero.
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