Doctors have to attend to a
patient on top of a building
using a 50 m ladder before
s/he is evacuated to a hospital.
The ladder is placed against
the building. The top of the
ladder is sliding down the
building at the rate of 2
m/min. Determine the rate at
which the base of the ladder is
moving away from the building
at the instant that the base is
30 m from the building.
Differentiation
Let the distance between the base of the ladder and the building be x meters.
Using the Pythagorean theorem, we have:
x^2 + 50^2 = (distance between top of ladder and ground)^2
Differentiating both sides with respect to time t:
2x(dx/dt) = 2(distance between top of ladder and ground)(d(distance between top of ladder and ground)/dt)
Since the ladder is sliding down the building at a rate of 2 m/min, we know:
d(distance between top of ladder and ground)/dt = -2
Plugging in x=30 and distance between top of ladder and ground = sqrt(50^2 - 30^2) = 40, we get:
60(dx/dt) = -2(40)
Therefore, dx/dt = -4/3 m/min.
So the base of the ladder is moving away from the building at a rate of 4/3 m/min when it is 30 m from the building.
To determine the rate at which the base of the ladder is moving away from the building, we can use the concept of related rates and differentiate the given equation.
Let's denote the height of the building as "h" and the distance of the base of the ladder from the building as "x."
Given:
h = 50 m
The top of the ladder is sliding down the building at a rate of 2 m/min (this represents the rate at which the height of the ladder is changing).
We want to find dx/dt, the rate at which the base of the ladder is moving away from the building when x = 30 m.
We can set up the following equation using the Pythagorean theorem:
x^2 + h^2 = 50^2
Differentiate both sides of the equation with respect to t (time) using implicit differentiation:
d/dt (x^2 + h^2) = d/dt (50^2)
2x * (dx/dt) + 2h * (dh/dt) = 0
Since we're looking for dx/dt, isolate it in the equation:
2x * (dx/dt) = -2h * (dh/dt)
Now we substitute the given values into the equation:
x = 30 m
h = 50 m
dh/dt = -2 m/min
We find the value of dh/dt negative because the height is decreasing.
2(30) * (dx/dt) = -2(50) * (-2)
60 * (dx/dt) = 100
dx/dt = 100/60
dx/dt = 5/3 m/min
Therefore, the rate at which the base of the ladder is moving away from the building at the instant when the base is 30 m from the building is 5/3 m/min.