UPSC NDA Math Model Paper 3

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.

dx

dt

+2y.

dy

dt

=0
Now by substituting dx/dt = 2 m/sec in the above equation we get,
⇒2x.2+2y.

dy

dt

=0
⇒

dy

dt

=−

2x

y

Now when x=4⇒y=√52−42=3
By substituting the value of x and y in the equation ⇒

dy

dt

=−

2x

y

we get ⇒

dy

dt

=−

2×4

3

=−

8

3

m∕sec
Hence, the required rate of decrease in the height of the ladder on the wall is: 8/3 m/sec