Question

The top of a ladder 6 metres 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 the ladder is 4 metres from the wall, it is sliding away from the wall at the rate of 0.5 m/sec. How fast is the top-sliding downwards at this instance? How far is the foot from the wall when it and the top are moving at the same rate?

Answer 1

With the given details, we can draw a right angle triangle.
Please refer to video to see the traingle.
From the diagram,
`x^2+y^2 = 6^2`
Here, `x = 4`
`:. y = sqrt(6^2-4^2) = sqrt20 = 2sqrt5 m`
Now, we can create an equation using pythagoras theorem.
`x^2+y^2 = 6^2->(1)`
Now, differentiating (1),
`=>2xdx/dt +2ydy/dt = 0`
It is given that `dx/dt = 0.5` m/sec , `x = 4 m and y = 2 sqrt5 m`.
`:. 2(4)(0.5) + 2(2sqrt5)dy/dt = 0`
`=>dy/dt = -(4sqrt5)/4= -1/sqrt5` m/sec.
So, the rate at which the top sliding downwards is `-1/sqrt5` m/sec.