Let OC be the wall as shown in the figure below. At a certain instant t, let AB be the position of the ladder such that OA = x meter and OB = y meter.
Given: dx/dt = 2 m/sec
From the figure given above we can see that:
In right angle triangle Δ AOB,
⇒x2+y2=52=25 .......-(By pythagoras theorem)
Now by differentiating the above equation with respect t we get
⇒2x.+2y.=0 Now by substituting dx/dt = 2 m/sec in the above equation we get,
⇒2x.2+2y.=0 ⇒=− Now when
x=4⇒y=√52−42=3 By substituting the value of
x and
y in the equation
⇒=− we get
⇒=−=−m∕sec Hence, the required rate of decrease in the height of the ladder on the wall is: 8/3 m/sec