Related Rates: Kite String Angle Change

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SUMMARY

The discussion focuses on a related rates problem involving a kite flying 100 ft above the ground, moving horizontally at a speed of 8 ft/s. The angle between the string and the horizontal is decreasing at a rate of 0.055 rad/sec when 200 ft of string has been let out. The solution involves using the tangent function to relate the horizontal distance (x) and the height (y), and applying derivatives to find the rate of change of the angle (dφ/dt). Key equations include tan(φ) = y/x and the derivative relationship derived from the geometry of the situation.

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Homework Statement



A kite 100 ft above the ground moves horizontally at a speed
of 8 ft!s. At what rate is the angle between the string and the
horizontal decreasing when 200 ft of string has been let out?


Homework Equations



tan \o = x/y

The Attempt at a Solution



After solving for the sides of the triangle i got X = 173.21, Y = 100 and Z = 200 and \o = 1.047 rads

then i took the derivative of the equation that links \o with the opposite and adjacent of the triangle and plugged in the variables which gave me:

I got d\o/dt = .055 rad/sec
 
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String mαkes an angle φ with horizontal.
Hence tanφ = y/x, where y is constant and x is variable.
When the string in overhead, the length of the string is 100 ft. 200 ft of string is let out. So the final length of the string is 300 ft.
Taking the derivative on both the side, with respect to time, you get
(-y/x^2)(dx/dt) = sec^2(φ)(dφ/dt)...(1)
sec^2(φ) = 1 + tan^2(φ) = 1 + (y/x)^2
L = 300 ft, y = 100 ft. find x. dx/dt is given. Substitute these values in eq.(1) and find d(φ)/dt
 

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