Why does the top of a sliding ladder speed up?

1. 1. Guess first

   A $10$ m ladder leans on a wall. Its foot, $2$ m out, is pulled away along the floor at a steady $1$ m/s, and the top slides down the wall.

   Question: How does the top move?

   Slowly at first, then faster and faster, until near the floor it is far faster than the foot. The next step plays it.

   The figure: A 10 m ladder against a wall, its foot 2.00 m from the wall and its top 9.80 m up. The foot moves away at 1 m/s and the top falls at 0.20 m/s.

2. 2. Watch it slide

   The faint ladders mark where it stood at each whole second. The foot’s marks are evenly spaced. Watch how the top’s are spaced, and the two arrows, each one unit long per metre a second.

   The figure: A 10 m ladder against a wall, its foot 9.90 m from the wall and its top 1.41 m up. The foot moves away at 1 m/s and the top falls at 7.02 m/s. Faint copies mark the ladder at each of the first 8 seconds: the foot's marks are evenly spaced and the top's are further apart each second.

3. 3. True at every instant

   The ladder never changes length, so $x^2 + y^2 = 100$ at every instant. Differentiate both sides with respect to time and solve for $y'$: the top's rate depends on where the ladder is, through $x/y$.

   Drag the foot to where the top falls exactly as fast as the foot moves.

   There, $x = y \approx 7.07$ m, the ladder is at $45^\circ$ and $y' = -1$ m/s. Nearer the wall the top is slower than the foot; further out it is faster. The minus sign says $y$ is shrinking.

   $$2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} = 0 \quad\Longrightarrow\quad \frac{dy}{dt} = -\frac{x}{y}\,\frac{dx}{dt}$$$2x\frac{dx}{dt}+2y\frac{dy}{dt}$ $=0$$⟹$$\frac{dy}{dt}$ $=-\frac{x}{y}\frac{dx}{dt}$

   The figure: A 10 m ladder against a wall, its foot 7.07 m from the wall and its top 7.07 m up. The foot moves away at 1 m/s and the top falls at 1.00 m/s.

4. 4. Differentiate first

   A tempting shortcut puts the instant's numbers in first. At $x = 6$ the equation becomes $36 + y^2 = 100$, and differentiating that gives $2y\,y' = 0$: it says the top is not moving at all.

   Question: At $x = 6$, how fast is the top really moving?

   Here $y = 8$, so $y' = -\tfrac{6}{8}(1) = -0.75$ m/s. $36 + y^2 = 100$ is true at one instant only, and a fixed number has no rate of change: putting $x = 6$ in first throws away the fact that $x$ is moving. Differentiate first, substitute last.

   The figure: A 10 m ladder against a wall, its foot 6.00 m from the wall and its top 8.00 m up. The foot moves away at 1 m/s and the top falls at 0.75 m/s.

5. 5. Near the floor

   At $x = 9.9$ m the top is $1.41$ m up and falling at about $7$ m/s, seven times the foot's speed. As $y$ shrinks to $0$, $y' = -x/y$ grows without bound: the model only holds while the top stays on the wall.

   Question: Pull the foot at $2$ m/s instead. At $x = 6$, what is the top's rate?

   $y' = -\tfrac{x}{y}\,x' = -\tfrac{6}{8}(2) = -1.5$ m/s. The geometry sets the factor $x/y$; the given rate is multiplied by it. Double one rate and, at the same instant, the other doubles.

   The figure: A 10 m ladder against a wall, its foot 9.90 m from the wall and its top 1.41 m up. The foot moves away at 1 m/s and the top falls at 7.02 m/s. Faint copies mark the ladder at each of the first 8 seconds: the foot's marks are evenly spaced and the top's are further apart each second.

The film, as said, 0:59

1. 0:00

   A $10$ m ladder leans on a wall. Its foot, $2$ m out, is pulled away at a steady $1$ m/s.

2. 0:07

   Watch the top. A faint ladder marks each second.

3. 0:10

   The foot’s marks are evenly spaced. The top’s spread out: it falls faster and faster.

4. 0:21

   Why? The ladder never changes length, so $x^2 + y^2 = 100$ at every instant.

5. 0:26

   Differentiate in time: $2x\,x' + 2y\,y' = 0$, so $y' = -\frac{x}{y}\,x'$.

6. 0:28

   Where $x = y$, at $45^\circ$, the top falls exactly as fast as the foot moves.

7. 0:36

   At $x = 6$ the top is $8$ m up, and falls at $\frac{6}{8}$ of the foot's speed: $0.75$ m/s.

8. 0:44

   Put $x = 6$ in first, and $36 + y^2 = 100$ says the top is still. A fixed number has no rate: differentiate first.

9. 0:50

   Near the floor $y$ is small and $x/y$ is large. At $x = 9.9$ the top falls at about $7$ m/s.