A ladder 6m long rests against the vertical wall. How far is the top of the ladder sliding down the wall when the bottom of the ladder is 4m from the wall and sliding at a speed of 1m/s?
let the height along the wall be y m
let the distance from the wall along the ground be x
x^2 + y^2 = 6^2
2x dx/dt + 2y dy/dt = 0
x dx/dt + y dy/dt = 0
when x = 4
16 + y^2 = 36
y = √20 , and dx/dt = 1
4(1) + √20 dy/dt = 0
dy/dt = -4/√20 m/sec
the negative shows that y is decreasing, (the ladder sliding down)
To find how far the top of the ladder is sliding down the wall, we need to use the concept of similar triangles. Let's consider the ladder as one side of a right-angled triangle, the wall as another side, and the distance the top of the ladder slides down the wall as the third side.
We can use the Pythagorean theorem to relate the sides of the triangle:
a^2 + b^2 = c^2
where a is the distance the bottom of the ladder is from the wall, b is the distance the top of the ladder slides down the wall, and c is the length of the ladder.
Given that the length of the ladder is 6m and the distance the bottom of the ladder is from the wall is 4m, we can substitute these values into the equation:
4^2 + b^2 = 6^2
Simplifying, we get:
16 + b^2 = 36
Now, we can solve for b:
b^2 = 36 - 16
b^2 = 20
Taking the square root of both sides, we find:
b ≈ √20
b ≈ 4.47m
Therefore, the top of the ladder is sliding down the wall approximately 4.47m when the bottom of the ladder is 4m from the wall.
To find how fast the top of the ladder is sliding down the wall, we need to differentiate the equation we derived above with respect to time. Given that the bottom of the ladder is sliding at a speed of 1m/s, we can substitute this value into the derivative:
2b * db/dt = 0
2 * 4.47 * db/dt = 1
Simplifying, we find:
db/dt ≈ 1 / (2 * 4.47)
db/dt ≈ 0.112 m/s
Therefore, the top of the ladder is sliding down the wall at a speed of approximately 0.112 m/s.