Question

A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Answer 1

It is given that a 5m long ladder is leaning against a wall and its bottom is pulled, away from the wall along the ground at a rate of 2 cm/s.

Figure (1)

Figure (1) shows that AB is the wall and AC is the ladder leaning against the wall AB.

Let BC be x cm and AB be y cm.

It is given that dx dt =2 cm/s and AC=5 cm.

From Pythagoras theorem,

H 2 = B 2 + P 2

In triangle ABC,

AC 2 = AB 2 + BC 2

Substitute AC=5 cm, AB=y and BC=x in the above formula.

25= y 2 + x 2 y 2 =25− x 2 y= 25− x 2

Differentiate with respect to t,

dy dt = 1 2 25− x 2 d( − x 2 ) dt = −2x 2 25− x 2 dx dt = −x 25− x 2 dx dt

Substitute dx dt =2 cm/s.

dy dt = −x 25− x 2 ( 2 ) = −2x 25− x 2

Substitute x=4 cm.

dy dt = −2( 4 ) 25− ( 4 ) 2 = −8 25−16 = −8 9 = −8 3  cm/s

Thus, the height is decreasing at a rate of 8 3  cm/s.