# 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. - Mathematics

Sum

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.

#### Solution

Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x m away from the wall.

Then, by Pythagoras theorem, we have:

x2 + y2 = 169 ......................[Length of the ladder = 13 m]

⇒ "y" = sqrt(169-"x"^2)

Then, the rate of change of height (y) with respect to time (t) is given by,

"dy"/"dx" = (-"x")/sqrt(169-"x"^2) "dx"/"dt"

It is given that "dx"/"dt" = 2 "cm"//"s".

"dy"/"dt" = (-2"x")/sqrt(169-"x"^2)

Now, when x = 5 m, we have:

"dy"/"dt" = (-2xx5)/sqrt(169-5^2) = (-10)/sqrt144 = -5/6

Hence, the height of the ladder on the wall is decreasing at the rate of 5/6cm/sec.

Concept: Rate of Change of Bodies Or Quantities
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