A rod of length 'l' is inclined at an angle `'theta'` with the floor against a smooth vertical wall. If the end A moves instantaneously with velocity `v_(1)`, what is the velocity of end B at the instant when rod makes `'theta'` angle with the horizontal.

Explanation

Let at any instant, end B and A are at a distance x and y respectively from the point .O..

Thus we have, `x^(2)+y^(2)=l^(2)" …………..(1)"`
Here l is the length of the rod, which is constant. Differentiating eq (1) with respect to time, we get
`(d)/(dt)(x^(2)+y^(2))=(d)/(dt)(l^(2)), 2x(dx)/(dt)+2y(dy)/(dt)=0`
If `(dx)/(dt)=v_(2) and (dy)/(dt)=-v_(1)`
`x(v_(2))+y(-v_(1))=0`
`rArr v_(2)=((y)/(x))v_(1)=v_(1)tan theta`