Given as a ladder 13 m long leans against a wall. The foot of the ladder is pulled along the ground away from the wall, at the rate of 1.5m/sec
As to find how fast is the angle θ between the ladder and the ground is changing when the foot of the ladder is 12 m away from the wall
Suppose AC be the position of the ladder initially, then AC = 13m.
DE be the position of the ladder after being pulled at the rate of 1.5m/sec, then DE = 13m as shown in the below figure.
Therefore it is given that foot of the ladder is pulled along the ground away from the wall, at the rate of 1.5 m/sec
So, dx/dt = 1.5 m/sec ...(i)
Considering the ΔABC, it is fight angled triangle, therefore on applying pythagoras theorem
And in same for triangle
secθ = AC/BC = 13/12 ...(iii)
Differetiating equation (ii) with respect to time
Substitute the value of x,y,h and dx/dt,
Value of h is always constant as the ladder is not increasing in size, thus the equation becomes
And consider the same triangle
tanθ = AB/BC = y/x
Differentiate the above equation with respect to time
Substitute the value of secθ, x,y,h and dx/dt, the equation becomes
Thus the angle θ between the ladder and the ground is changing at the rate of 0.3 rad/sec (because angle cannot be negative) when the foot of the ladder is 12 m away from the wall.
