Finding the Rate of Change of a Falling Ladder

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SUMMARY

The discussion revolves around the mathematical analysis of a falling ladder scenario, specifically a 10ft ladder positioned against a wall with its base 8ft away, slipping at a rate of 4 ft/s on a frictionless surface. The participants explore the rate of change of the ladder's height and angle using calculus, particularly the Pythagorean theorem and derivatives. The calculations reveal that while the height's rate of change is complex and undefined for certain values, the angle's rate of change remains constant. The confusion arises from the interpretation of these rates in the context of the ladder's motion.

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sherrellbc
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https://www.khanacademy.org/math/calculus/derivative_applications/rates_of_change/v/falling-ladder-related-rates
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Essentially, the question states that a ladder, with length 10ft, is laid up against a wall with its bottom 8ft out. The ladder begins to slip, on a friction-less surface, at a rate of 4 ft/s. The problem asks to find the rate of change of height at that particular instant in time.

Now, my question to you is ..
Given that this ladder is on a friction-less surface, and the falling action is not affected by gravity, wouldn't the rate of change of the height be constant? Similarly, the rate of change of the angle should also be constant, given that the height is changing at a constant rate.

Correct me where I went wrong.

The base of the triangle, as a function of time, can be written as (8 + 4t)
The hypotenuse remains 10, and the height is, of course, h.
Solve for h using the pythagorean theorem =>
h² = 100 - (8+4t)²
h = ±√(100 - (8+4t)²)
dh/dt = -(t+32)/(-t²-64t+36)^½

If you evaluate at 0, the correct answer is derived. But this function is undefined for any integer number > 0, and any rational number less than 0 really doesn't make any sense for this problem. What kind of a result is that? Although, you can get real values for rational numbers approximately 0 < x < 1. That shows that the height is changing as a function of time?!

However, if one were to execute the necessary steps to determine how the angle(bottom right) was changing with respect to time, you would resolve a constant!

cos(Θ) = (8+4t)/10
-sin(Θ) * dΘ/dt = 4/10
dΘ/dt = -2/(5sin(36.87°))

This result is most certainly a constant. What am I doing wrong?
 
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h^2 = 100 - (8 + 4t)^2
2h dh/dt = -2(8 + 4t)(4)
dh/dt = -4(8 + 4t)/h
dh/dt = -(32 + 16t)/sqrt(36 - 64t - 16t^2)
 
rude man said:
h^2 = 100 - (8 + 4t)^2
2h dh/dt = -2(8 + 4t)(4)
dh/dt = -4(8 + 4t)/h
dh/dt = -(32 + 16t)/sqrt(36 - 64t - 16t^2)

Perhaps this is a better way to do it, but both solutions are the same. I realized later than I had mistakenly typed -t² instead of -16t².

Anyway, this thread was moved from another subforum and remained locked. I created a new thread and then this thread became unlocked.

https://www.physicsforums.com/showthread.php?t=699231
 

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