# Ladder + Related rates and expressing it in terms of time

There are two questions:

1. A ladder, L[m] long, is leaning against a vertical wall. If the bottom of the ladder is pulled horizontally away from the wall at $v_h[\frac{m}{s}]$, how fast is the top of the ladder sliding down the wall when the bottom is the distance $x_0[m]$ from the wall

2. Suppose a particular downward vertical speed is given as $v_d[\frac{m}{s}]$ . Determine how long it takes the top of the ladder to reach the speed by expressing the time in terms of $v_h, v_d, and L$

So for the first one I used related rates and set up my triangle such that

Given:

$L [m] \rightarrow$ Length of ladder

$V_h[\frac{m}{s}] \rightarrow$ Speed at which the ladder is sliding away from the wall

$x_0[m] \rightarrow$ Some distance the ladder is away from the wall.

Let us represent our rate at which the ladder is sliding down as

$\frac{\delta d}{\delta t}$ which we will later call $V_d[\frac{m}{s}]$

using the Pythagorean Theorem:

$$x_0^2 + d^2 = L^2\quad$$ where d is the point where the ladder is leaning against the wall (height of ladder)

$$2x_0 \frac{\delta x_0}{\delta t} + 2d\frac{\delta d}{\delta t} = 0\quad$$

$L^2$ goes to zero because it is a constant and never changes (ladder)

So solving for the rate at which the ladder is sliding down the wall when the bottom is a $x_0$ distance we isolate $\frac{\delta d}{\delta t}$ such that:

$$\frac{\delta d}{\delta t} = - \frac{x_0}{d}\frac{\delta x_0}{\delta t}$$

Here we can replace the rate at which $x_0$ changes with our $V_h[m]$ which gives us

$$\frac{\delta d}{\delta t} = - \frac{x_0}{d}V_h$$

We do not have a value for $d$, but can express it in terms of $x_0$ and $L$:

$$d = \sqrt{-x_0^2+L^2}$$

So replacing $d$ with this equation we have

$$\frac{\delta d}{\delta t} = - \frac{x_0}{\sqrt{-x_0^2+L^2}}V_h$$

Checking our units: $$\frac{m}{s} = \frac{m}{m} \frac{m}{s} \implies \frac{m}{s} = \frac{m}{s}$$

This solves (1) I believe. Moving onto (2), is where I am confused. How do I find the time it takes the ladder to reach a downward velocity when we are given $V_h, V_d$, and $L$.

I know that the first answer is the velocity at which the ladder slides down, and the integral of velocity is position. So would I have to integrate which would give me the right hand side of (1) * t for some equation $y(t)$ ?

You have the equation for a downward velocity in terms of $x$. Now the downward velocity is given. Knowing it, solve the equation for $x$. The time necessary for a bottom to reach this position would be $\frac{x}{V_h}$.
• Yes, here is what I got. $$- \frac{ (-x^2 + L^2)^{3/2}}{V_h} = \frac{x}{V_d}$$ Is this correct, doing the algebra I was not sure if the answer was supposed to be in this format as it looks rough to look at. – Hawaiian Rolls Apr 10 '17 at 2:44
$$t = \frac{x}{V_h} \rightarrow x = tV_h$$
$$t = \frac{V_dL}{V_h\sqrt{V_d^2 - V_h^2}}$$