A kite 100 ft above the ground moves horizontally at a speed of 6 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out? rad/s g

Respuesta :

Answer:

0.015 radians per second.

Step-by-step explanation:

They tell us that at the moment the speed would be 6 ft / s, that is, dx / dt = 6 and those who ask us is dθ / dt.

Which we can calculate in the following way:

θ = arc sin 100/200 = pi / 6

Then we have the following equation of the attached image:

x / 100 = cot θ

we derive and we are left:

(1/100) * dx / dt = - (csc ^ 2) * θ * dθ / dt

dθ / dt = 0.01 * dx / dt / (- csc ^ 2 θ)

dθ / dt = 0.01 * 6 / (- csc ^ 2 pi / 6)

dθ / dt = 0.06 / (-2) ^ 2

dθ / dt = -0.015

So there is a decreasing at 0.015 radians per second.

Ver imagen jmonterrozar

The horizontal distance and the height of the kite are illustration of rates.

The angle is decreasing at a rate of 0.24 radian per second

The given parameters are:

[tex]\mathbf{Height =y= 100ft}[/tex]

[tex]\mathbf{Speed =\frac{dx}{dt}= 6fts^{-1}}[/tex]

[tex]\mathbf{Length = 200}[/tex]

See attachment for illustration

Calculate the angle using the following sine ratio

[tex]\mathbf{sin(\theta) = \frac{100}{200}}[/tex]

[tex]\mathbf{sin(\theta) = \frac{1}{2}}[/tex]

The horizontal displacement (x) is calculated using the following tangent ratio:

[tex]\mathbf{tan(\theta) = \frac{100}{x}}[/tex]

Take inverse of both sides

[tex]\mathbf{cot(\theta) = \frac{x}{100}}[/tex]

[tex]\mathbf{cot(\theta) = \frac{1}{100}x}[/tex]

Differentiate both sides with respect to time (t)

[tex]\mathbf{-csc^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dx}{dt}}[/tex]

Substitute known values

[tex]\mathbf{-csc^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 6}[/tex]

[tex]\mathbf{-csc^2(\theta) \cdot \frac{d\theta}{dt} = \frac{6}{100}}[/tex]

Recall that:

[tex]\mathbf{sin(\theta) = \frac{1}{2}}[/tex]

Take inverse of both sides

[tex]\mathbf{csc(\theta) = 2}[/tex]

Square both sides

[tex]\mathbf{csc^2(\theta) = 4}[/tex]

Substitute [tex]\mathbf{csc^2(\theta) = 4}[/tex] in [tex]\mathbf{-csc^2(\theta) \cdot \frac{d\theta}{dt} = \frac{6}{100}}[/tex]

[tex]\mathbf{-4 \cdot \frac{d\theta}{dt} = \frac{6}{100}}[/tex]

Divide both sides by -4

[tex]\mathbf{\frac{d\theta}{dt} = -\frac{24}{100}}[/tex]

[tex]\mathbf{\frac{d\theta}{dt} = -0.24}[/tex]

Hence, the angle is decreasing at a rate of 0.24 radian per second

Read more about rates at:

https://brainly.com/question/6672465