Angle Decrease Rate for Kite String Related-Rates Problem

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SUMMARY

The discussion focuses on calculating the rate at which the angle between a kite string and the horizontal decreases, specifically when the kite is 100 feet above the ground and 200 feet of string has been released. The correct rate of angle decrease is determined to be -0.08 rad/s, indicating that the angle is indeed decreasing. The relationship used for the calculation involves the cotangent function, where cot(theta) = x/100, and the differentiation of this relationship leads to the final result. The initial misunderstanding regarding the angle's rate being positive was clarified, emphasizing that a decreasing angle results in a negative rate.

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Homework Statement


A kite 100ft 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? Answer: 0.02 rad/s

Homework Equations

The Attempt at a Solution


costheta = x/200
taking the derivative and rearranging
theta prime= x'/-200sintheta

substituting x' = 8 and theta = pi/6
theta prime = -0.08 rad/s
 
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OmniNewton said:

Homework Statement


A kite 100ft 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? Answer: 0.02 rad/s

Homework Equations

The Attempt at a Solution


costheta = x/200
No, this is incorrect. The height of the kite is constant, but the length of the string is not constant.
I used ##\cot \theta = \frac x {100}## for my relationship between x and ##\theta##, and differentiated with respect to t to get the relationship between the rates.

My work agrees with the answer you posted, except that the answer should be negative -- the angle is decreasing, which means that ##\frac{d\theta}{dt}## is negative.
OmniNewton said:
taking the derivative and rearranging
theta prime= x'/-200sintheta

substituting x' = 8 and theta = pi/6
theta prime = -0.08 rad/s
 
Alright makes sense thank you! I solved it
 

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