A 20m long ladder is leaning against a wall. The top of the ladder is sliding down the wall at a constant speed of 2m/s. How quickly is the bottom of the ladder moving away from the wall when the bottom of the ladder is 5m from the wall? Give your answer as an exact expression.
If the top of the ladder is at height y, then
x^2+y^2=400
when x=5, y=5√15
x dx/dt + y dy/dt = 0
Now just plug in your numbers.
To solve this problem, we can use the principles of related rates and trigonometry.
Let's define the following variables:
- y: The distance between the bottom of the ladder and the wall (we want to find dy/dt, how quickly y changes with respect to time).
- x: The distance between the top of the ladder and the wall.
- L: The length of the ladder (given as 20m).
- θ: The angle between the ladder and the ground.
Given that the top of the ladder is moving down the wall at a constant speed of 2m/s, we can find dx/dt, how quickly x changes with respect to time. Since the ladder is 20m long, we know that dx/dt = -2m/s (negative because it is moving down).
Using trigonometry, we can relate x, y, and θ in the following way:
cos(θ) = x / L
sin(θ) = y / L
Differentiating both equations with respect to time (t) using the chain rule, we get:
- sin(θ) * dθ/dt = dx/dt
- cos(θ) * dθ/dt = dy/dt
We know that dx/dt = -2m/s, so by substituting that into the first equation, we can solve for dθ/dt:
- sin(θ) * dθ/dt = -2m/s
dθ/dt = (-2m/s) / sin(θ)
Now we need to find the value of sin(θ) when y = 5m (when the bottom of the ladder is 5m from the wall). Using the equation sin(θ) = y / L, we can substitute the values to find sin(θ):
sin(θ) = 5m / 20m
sin(θ) = 1/4
Now we can substitute this value of sin(θ) into the dθ/dt equation to find dθ/dt:
dθ/dt = (-2m/s) / (1/4)
dθ/dt = -8m/s
Finally, we can substitute this value of dθ/dt and θ = 1/4 into the equation dy/dt = cos(θ) * dθ/dt to find dy/dt:
dy/dt = cos(1/4) * (-8m/s)
The exact expression for dy/dt is -8cos(1/4) m/s.