Abuilding that is 245 feet tall casts a shadow of various lengths x as the day goes by. an angle of elevation θ is formed by lines from the top and bottom of the building to the tip of the shadow. 245 ftθx find the rate of change (in radians per foot) of the angle of elevation dθ dx when x = 300 feet. (round your answer to five decimal places.)

Explanation:

It is given that,

Height of the building, h = 245 ft

The base of the building and shadow is x.

Let is the angle of elevation that is formed by lines from the top and bottom of the building to the tip of the shadow. Using the trigonometric ratio as :

When x = 300 feet

So, the rate of change of the angle of elevation is -0.001633 rad/ft. Hence, this is the required solution.

The angle of elevation decreases as the length of the shadow which forms the angle increases

The rate of change of the angle of elevation, when x = 300 feet, is approximately -0.00163 radians per foot

Reasons:

Given parameter;

Height of the building = 245 ft.

Lengths of the shadow cast by the building = x

Angle of elevation of the top of the building to the tip of the shadow = θ

Required:

The rate of change of the angle of elevation, , when, x = 300 feet

Solution:

Using chain rule of differentiation, we get;

Therefore;

When x = 300 feet, we get;

Therefore,  when x = 300 feet, ≈ -0.00163 radians/ft.

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Explanation:

v = ω R

v is linear speed and ω is angular speed

ω = v / R

a ) Inner radius = 25 x 10⁻³ m

speed v = 1.25 m/s

ω = 1.25 / (25 x 10⁻³ )

= .05 x 10⁻³

= 5 x 10⁻⁵ rad / s

outer  radius = 58 x 10⁻³ m

speed v = 1.25 m/s

ω = 1.25 / (58 x 10⁻³ )

= .0215 x 10⁻³

= 2.15 x 10⁻⁵ rad / s

b )

linear constant speed v = 1.25 m /s

time = 74 min = 74 x 60 s

distance tracked = speed x time

= 1.25 x 74 x 60

= 5550 m

c ) time given

= 74 min = 74 x 60 s

angular acceleration

= (  2.15 - 5 ) x 10⁻⁵ /  (74 x 60 )

= -  6.42 x 10⁻⁹ rad / s²