a 5 m long ladder rests against a vertical wall with its feet on horizontal ground. the feet on the ground slip, and at the instant when they are 3 m from the wall, they are moving at 10 ms^-1. at what speed is the other end of the ladder moving at this instant?
The classic introduction question to "rates of change"
At a given time of t seconds, let the foot of the ladder be x m from the wall
and let the top of the ladder be y m above the ground
we know:
x^2 + y^2 = 5^2
2x dx/dt + 2y dy/dt = 0
when x = 3, y = 4 , (using Pythagoras in my above equation), and dx/dt = 10 m/s
2(3) (10) + 2(4)(dy/dt) = 0
dy/dt = -60/8 m/s = -7.5 m/s
the negative tells me the top of the ladder is moving down, or the y value is decreasing.
To solve this problem, we can use the concept of related rates.
Let's consider the following variables:
- x: the distance from the base of the ladder to the wall (changing)
- y: the height of the ladder on the wall (changing)
- L: the length of the ladder (constant)
We are given that x is changing at a certain rate and want to find the rate of change of y.
Using the Pythagorean theorem, we have:
x^2 + y^2 = L^2
Differentiating both sides of the equation with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since the ladder is resting against a vertical wall, the height (y) is constant, so dy/dt = 0.
Substituting dx/dt = 10 m/s (given) and y = 3 m (given), we can solve for dy/dt:
2x(10) + 2(3)(0) = 0
20x = 0
x = 0
Since x = 0, the other end of the ladder is not moving at this instant. Therefore, the speed of the other end of the ladder is 0 m/s.
To find the speed at which the other end of the ladder is moving, we can use the concept of similar triangles.
Let's denote the distance between the wall and the slipping feet of the ladder as x. According to the problem, x = 3 m, and the rate of change of x with respect to time is given as 10 m/s.
The length of the ladder, which remains constant, is 5 m.
By applying similar triangles, we can relate the distances of x and the other end of the ladder, which we'll call y.
From the similar triangles, we can set up the following ratio:
x / y = 3 / 5
Now, we can differentiate both sides of the equation with respect to time, t, to find the rate of change of y:
d(x) / dt / d(y) / dt = 0 / d(y) / dt
To find d(y) / dt, the rate of change of y with respect to time, we can rearrange the equation and solve for it:
d(y) / dt = (d(x) / dt * y) / x
Plugging in the known values, we have:
d(y) / dt = (10 m/s * 5 m) / 3 m
Simplifying, we find:
d(y) / dt = 50 m/s / 3 m
d(y) / dt ≈ 16.67 m/s
Therefore, at the instant when the slipping feet are 3 m from the wall and moving at 10 m/s, the other end of the ladder is moving at approximately 16.67 m/s.