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Homework Help: Ladder Rate of Change

  1. Jun 29, 2013 #1

    1. The problem statement, all variables and given/known data
    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.

    3. The attempt at a solution

    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 = -(16t+32)/(-16t²-64t+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?
    Last edited: Jun 29, 2013
  2. jcsd
  3. Jun 29, 2013 #2
    No, this is not given. The length of the base is some ##f(t)##, and you are given that ##f(0) = 8 \ \text{ft} ## and ##f'(0) = 4 \ \text{ft/s} ##.
  4. Jun 29, 2013 #3
    Not given, but can be derived. The base begins at a length of 8 and will expand as a function of time. I can define time, t, to be any unit I like. In this case, t represents seconds.
    Thus, (8 + 4t)

    t = 0 → 8ft
    t = 1 → 12ft

    If the base expands at a rate of 4 per second, this is valid. The problem statement is not saying that the base is expanding at a rate of [itex]\frac{4ft}{sec^2}[/itex]
  5. Jun 29, 2013 #4
    No, it cannot be derived. You cannot derive a function from the value of the function and its derivative at a single point - not without making assumptions, which you have no basis for in this case.

    Nor do you actually need to derive the entire function. All you really need is these two values.
  6. Jun 29, 2013 #5
    I wanted to derive a generalized equation representing the rate at which the height would be changing at any time(I wanted to verify that it would be constant) If you follow the link at the top, the video explains a way to derive the rate of change at the exact moment shown in the illustration.

    What if you wanted to derive an equation representing the generalized case?

    What type of assumptions would have to be made and what circumstances needed? I do not understand why, if given the rate of change, you cannot derive a function as I did. You know that the value is going to be changing at a given rate, and you also have the original values.

    The equation seems valid. After one second, the value had increased by the rate at which it changes in that time (4). So, it seems to be a valid case that (8+4t) would work.

    I am not a physics guy, but I study EE. I am just reviewing my calculus to ensure a firm foundation.
  7. Jun 29, 2013 #6
    You would need more information, such as why the ladder is sliding in the first place.

    In calculus, one learns that any (well-behaved) function can be represented via its own Taylor expansion. The latter requires the values of all (infinitely many!) its derivatives. Not just one as you have in this case. By only taking one derivative's value into account, you essentially assume that all the others are zero, which is why you end up with f(x) = 8 + 4t. But why do you think you can make such an assumption?
  8. Jun 29, 2013 #7
    I was thinking more in terms of a practical application of the fact that the base is growing at a given rate. I also considered that, after half a second, the base would be 10ft(its max), and of course mathematically could expend infinitely.

    So, I simply defined the base to be (8+4t) in order to capture and encapsulate this changing base in order to derive a function that represents the generalized case, given any t, of the rate of change at that instant.
  9. Jun 29, 2013 #8
    I may have misunderstood the intent, then. If you postulate that the base is changing at a constant rate, then 8 + 4t is valid. And you found that the rate of change of h is not constant, which is correct.
  10. Jun 29, 2013 #9
    What the problem says is that the base is changing at the rate given, that is all. However, the problem further states that the surfaces in which the ladder is sliding are friction-less. There is not mention of a gravitational affect(meaning an acceleration) on the rate at which the height is changing. So, from the mentioned absence, I assumed that the base would be changing at a constant rate. (Since the ladder was not being pulled downward at a changing rate)

    Now, the final result had me confused. I hypothesized before attempting this problem that the height should be changing at a constant rate. However, I derived an equation, with a very limited domain, that governed this problem. The problem is that the rate of change of the height was a function of time, and not constant. So, then I ask, which is incorrect, the resulting equation or my conjecture?

    Further, I also solved for the governing equation representing the rate at which the angle, with the terminal axis if you were to draw it how the picture depicts, changes. I resolved this governing equation to be a constant which is what I expected in the first place!

    What am I doing incorrectly in either of these attempts?
    Last edited: Jun 29, 2013
  11. Jun 29, 2013 #10
    Even without gravity, the ladder experiences reaction forces from the wall and the floor. And the ladder has a non-zero moment and angular moment, too. The motion is quite complicated here.

    On what grounds?

    Both are. But the conjecture is more wrong than the equation, which has just a couple of arithmetic mistakes.

    Your mistake in the final step was to set the angle constant.
  12. Jun 29, 2013 #11
    The only mistake I see is that t² should have a leading coefficient of 16, which I have on the paper I used to work this problem. I am not concerned about why the rate at which it is changing is strange(given an incorrect equation I derived), but why the rate at which the height changes in this problem is a function of t to begin with.

    This problem is one of a simple application of derivatives. Assuming that all forces on the ladder are negligible aside from the outward motion of the base of the ladder at a rate of 4ft/sec, what did I do wrong?

    If what I did was incorrect, how would you derive an explicit equation representing the derivative of θ(t) ?

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

    Okay, so θ(t) is a function of time and is equal to 36.87 at t = 0. I realize what I did here. Although, how could you derive an equation representing the changing angle? ( θ - ... ) ?
    Last edited: Jun 29, 2013
  13. Jun 29, 2013 #12
    I do not see anything wrong there. ## h^2(t) + f^2(t) = 100 \Rightarrow h(t) = \sqrt{100 - f^2(t)} \Rightarrow h'(t) = - (100 - f^2(t))^{-1/2} f(t) f'(t) ##. Assuming you calculate everything properly, you should get the correct result. But note, however, that even if ## f'(t) = \text{const} ##, ##h'(t) \ne \text{const} ## because of ## (100 - f^2(t))^{-1/2} f(t) ##.

    You have the equation for ## \cos \theta ##, so you can use that one to find ## \theta(t) ##. That is not the only way, though, you could use any other relevant trig identity.
  14. Jun 29, 2013 #13
    The math works out that way, I was just confused from a physical standpoint. I cannot understand why that ##f(t)## changing at a constant rate does not correspond to an ##h(t)## changing at a constant rate as well.

    To me, it would seem that as time passed if the the rate at which the height was changing increased, the base would be forced to increase at a faster rate to compensate since the ladder is a rigid structure.
  15. Jun 29, 2013 #14
    Wait. I understand. The answer is so obvious.

    Given the original diagram, h = 6ft, b = 8ft
    Considering that the ladder is 10ft, the base can only extend 2ft further.

    If the base were to have traveled 1ft, making it 9ft, the associated height would be h = 4.36ft

    So, during the first quarter-second, h changed by 1.64ft
    and during the second quarter-second, h would have to change 4.36ft

    Thank you for your posts. I must sit back and marvel at my ignorance for a moment.
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