A ladder , 5 meter long , standing on a horizontal floor , leans against a vertical wall . If the top of the ladder slides downwards at the rate of 10 cm/sec . Find the rate at which the angle between the floor and the ladder is decreasing when the lower end of the ladder is 2 meter from the wall .
let the ladder reach h m high on the wall, let the base angle be Ø
sinØ = h/5
h = 5sinØ
dh/dt = 5cosØ dØ/dt
given: dh/dt = -10m/sec
when base = 2m, cosØ = 2/5
-10 = 5(2/5) dØ/dt
dØ/dt = -5 radians/sec
the angle is decreasing at a rate of 5 radians / sec
(by using the word "decreasing" I do not include the negative sign)
Change first 5m to 500cm and 2m to 200cm and then solve using above procedure of reiny answer is 1/20 radians/sec
To find the rate at which the angle between the floor and the ladder is decreasing, we can use trigonometry. Let's call the distance of the lower end of the ladder from the wall "x" meters.
We can form a right triangle with the ladder as the hypotenuse, the distance from the wall as the adjacent side, and the height of the ladder as the opposite side. Let's call the angle between the floor and the ladder "θ" radians.
Using Pythagoras theorem, we have:
x^2 + h^2 = 5^2
Differentiating both sides with respect to time, we get:
2x(dx/dt) + 2h(dh/dt) = 0
Since we need to find the rate at which θ is decreasing, we can differentiate the above equation with respect to time again. But first, we need to express h in terms of x.
From the right triangle, we have:
h = √(5^2 - x^2)
Substituting this into the derivative equation, we get:
2x(dx/dt) + 2(√(5^2 - x^2))(dh/dt) = 0
Now, we need to find the value of dx/dt when x = 2 meters. Also, from the problem statement, we know that dh/dt = -10 cm/sec (the top of the ladder slides downwards at the rate of 10 cm/sec).
Substituting these values, we get:
2(2)(dx/dt) + 2(√(5^2 - 2^2))(-10/100) = 0
Simplifying, we have:
4(dx/dt) - 2(√(21)) = 0
Solving for dx/dt, we get:
dx/dt = (√(21))/2
Therefore, the rate at which the angle between the floor and the ladder is decreasing when the lower end of the ladder is 2 meters from the wall is (√(21))/2 radians per second.