Ah, the classic case of the perpendicular sides of a right-angled △ changing length but with the hypotenuse unchanged.

We can think of the 3 contact points (the top of the ladder touching the wall, the bottom of the ladder touching the ground and the intersection of the wall and the ground) as the vertices of a right-angled triangle.

We can assume the ladder has a perfectly rectangular cross section (i.e its ends are not rounded or odd shaped or of some trapezoidal shape)

We can treat it as if the ladder will slide down smoothly. When the ladder is fully on the ground,its top will be flush with the ground and be in contact the intersection of the wall and the ground.

Edit : to add on, the most important assumption is that the ground and the wall are perfectly smooth i.e no bumps or undulations.

Let a m be the vertical distance from the top of the ladder to the intersection point and b m be the horizontal distance from the intersection point to the bottom of the ladder, in metres.

These are two unknowns, but we know that we have a right-angled triangle, so we can apply Pythagoras's Theorem.

a² + b² = 4.00²
a² + b² = 16

When b = 2.00,

a² + 2.00² = 16
a² + 4 = 16
a² = 16 - 4 = 12
a = √12 = √(4 × 3) = √4 √3 = 2√3

We can also say that : da/dt = -3.00 s-¹
(The ladding is sliding down so the rate of change of the vertical distance/height with respect to time is negative)

We are asked to find db/dt when b = 2.00


From the initial equation,

a² + b² = 4.00²

Since we have da/dt already and need to find db/dt, we are expected to perform implicit differentiation.

Differentiate both sides with respect to a,

2a + 2b db/da = 0

Substitute a = 2√3, b = 2.00,

2(2√3) + 2(2.00) db/da = 0

4√3 + 4 db/da = 0

4 db/da = -4√3

db/da = -√3


Lastly,


db/dt = db/da × da/dt

Substitute db/da = -√3 and da/dt = -3.00,

db/dt = -√3 × (-3.00)

db/dt = 3√3


The bottom of the ladder is moving along the ground at 3√3 m s-¹ when it is 2.00 m away from the wall.