Ladder Sliding Down Question General Formula Proof

Question

I'm self teaching myself from a textbook for fun.

There's a question about a ladder sliding down at a constant rate. There's no specific length for the ladder and the question is in regards to the the vertical speed in relation to the top of the ladder to the wall.

The answers to the question are as follows:

$x=$ Horizontal component
$y=$ Vertical component
$L=$ Ladder length
$u=$ ladder's feet sliding down rate.

1.) $\dfrac {dx}{dt}= u$

2.) $\dfrac {dy}{dx} = -x\sqrt{L^2-x^2}$

3.) $\dfrac {dy}{dt} = -ux\sqrt{L^2-x^2}$

I'm confused as to how to get this answers.

Answer 1

One reason you may be confused is that the formulas you've given are not correct. The $\sqrt{L^2 - x^2}$ factor should be divided in, not multiplied. (I did double-check that I didn't change this in my format edit. You have it multiplied in your original post.)

1. is just the meaning of the definition of $u$ - except that since the feet are sliding horizontally on the floor, $u$ is the rate of sliding out, not down. $\dfrac{dx}{dt}$ is the rate at which $x$ is increasing with respect to $t$, and that is also what $u$ has been defined to be.
2. By the Pythagorean theorem $x^2 + y^2 = L^2$, so $y = \sqrt{L^2 - x^2}$.
   Let $z = L^2 - x^2$, so $y = \sqrt z$. Now $\dfrac {dy}{dz} = \dfrac 1{2\sqrt z}$ and $\dfrac {dz}{dx} = -2x$, so by the chain rule: $$\frac {dy}{dx} = \frac{dy}{dz}\frac{dz}{dx} = \left(\dfrac 1{2\sqrt z}\right)\left(-2x\right) = \dfrac {-x}{\sqrt z} = \dfrac {-x}{\sqrt{L^2 - x^2}}$$
3. Again by the chain rule, $$\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt} = \left(\dfrac {-x}{\sqrt{L^2 - x^2}}\right)(u) = \dfrac {-ux}{\sqrt{L^2 - x^2}}$$

You may note that if you pull the ladder feet out at a constant rate, the downward velocity of the top of the ladder increases to infinity. This unrealistic phenomenon is not a problem with the mathematics, but rather with the idea that the ladder will continue to lean on the wall for the entire trip. No matter how slowly you pull it, at some point the ladder will leave the wall because it can no longer fall fast enough to keep up.