Question

A ladder 13 m long is leaning against a vertical wall. The bottom of the ladder is dragged away from the wall along the ground at the rate of 2 cm/sec. How fast is the height on the wall decreasing when the foot of the ladder is 5 m away from the wall?

Answer 1

Let: 

x = distance from the wall to the foot of the ladder (in meters)

y = height of the ladder on the wall (in meters)

The length of the ladder is constant: L = 13 m

Apply the Pythagorean theorem

x² + y² = L²

Substituting L = 13:

x² + y² = 132 = 169

Differentiate with respect to time

Differentiating both sides of the equation with respect to time t:

`d/(dt) (x^2) + d/(dt)(y^2) = d/(dt) (169)`

This gives us:

`2x(dx)/(dt) + 2y(dy)/(dt) = 0`

Solve for `(dy)/(dt)`

Rearranging the equation:

`2x(dx)/(dt) + 2y(dy)/(dt) = 0`

⟹ `x(dx)/(dt) + y(dy)/(dt) = 0`

Thus: `(dy)/(dt) = -x/y (dx)/(dt)`

Substitute known values

We know: `(dx)/(dt)` = 2 m/sec (the rate at which the foot of the ladder is moving away from the wall)

When x = 5 m, we need to find y.

Using the Pythagorean theorem to find y when x = 5:

5² + y² = 169

⟹ 25 + y² = 169

⟹ y² = 144

⟹ y = 12 m

Now substitute x = 5, y = 12, and `(dx)/(dt) = 2` into the equation for `(dy)/(dt)`

`(dy)/(dt) = -5/12 * 2 = -10/12 = -5/6` m/sec

The negative sign indicates that the height is decreasing.

Therefore, the height on the wall is decreasing at a rate of `5/6` m/sec