The top of ladder 6m long is resting against a vertical wall on a level pavement ,when the ladder begins to slide outwards .At the moment when the foot of ladder is 4m from the wall ,it is sliding away from the wall at the rate of 0.5m/sec. how fast is the top-sliding downwards at instanse? How far is the foot the wall when it and the top are moving at the same rate

Dear rohit

x^2+y^2= l^2

x is the foot distance and y is the height at which top point of ladder touched wall

differentin you can get

2xdx/dt = -2ydy/dt

x=4

l=6

dx/dt = 0.5

so y = 4.472

dy/dt=0.447

In the same way for the second part dy/dt = dx/dt

so y=x

so y=x =3*2^0.5

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