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.
Let's first draw a diagram to visualize the situation:
```
.
/|
/ |
/ |
50 / | y
/ |
/ |
/ |
/θ |
/________|
x
```
We can see that we have a right triangle with the ladder as the hypotenuse, and we can use trigonometry to relate the various sides and angles. Let:
- x = the distance from the building to the foot of the ladder
- y = the height of the ladder on the building
- θ = the angle between the ladder and the ground
Then, from the right triangle, we have:
sinθ = y/50
cosθ = x/50
Differentiating both equations with respect to time t, we get:
cosθ dθ/dt = -y/2500 (from sinθ)
-sinθ dθ/dt = x/2500 (from cosθ)
We want to find dx/dt, the rate at which the base of the ladder is moving away from the building. To do so, we can differentiate x/50 = cosθ with respect to time t:
dx/dt = -sinθ dθ/dt
Substituting from the earlier equations, we get:
dx/dt = -(x/2500)(y/50)
Now, we need to find x and y when x = 30 m. From the right triangle, we can use Pythagoras' theorem:
x^2 + y^2 = 50^2
Substituting x = 30, we get:
y = sqrt(50^2 - 30^2) = 40
We can also find θ using the earlier equation:
sinθ = y/50 = 0.8
θ = arcsin(0.8) = 53.13°
Now, we just need to substitute x = 30, y = 40, and dθ/dt = -2 (since the ladder is sliding down the building at a rate of 2 m/min) into the equation for dx/dt:
dx/dt = -(30/2500)(40/50)(-2) = 0.24 m/min
Therefore, the base of the ladder is moving away from the building at a rate of 0.24 m/min when it is 30 m from the building.
To solve this problem, we can use related rates. Let's denote the distance between the base of the ladder and the building as x at time t (in minutes). We are given that dx/dt (the rate at which x is changing) is what we need to find when x = 30 m.
We are also given that the top of the ladder is sliding down the building at a rate of 2 m/min. This means that the height of the ladder, h, is changing at a rate of -2 m/min (negative sign because the ladder is sliding down).
We can use the Pythagorean theorem to relate x and h:
x^2 + h^2 = 50^2
Differentiating both sides of the equation with respect to time t, we get:
2x(dx/dt) + 2h(dh/dt) = 0
Substituting the given values, we have:
2x(dx/dt) + 2(-2) = 0
2x(dx/dt) - 4 = 0
2x(dx/dt) = 4
dx/dt = 4/(2x)
When x = 30, we can substitute this value into the equation to find the rate at which the base of the ladder is moving away from the building:
dx/dt = 4/(2(30))
dx/dt = 4/60
dx/dt = 1/15
Therefore, the rate at which the base of the ladder is moving away from the building when the base is 30 m from the building is 1/15 m/min.