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.
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
1 answer