A bag is tied to the top of a 5m ladder resting against a vertical wall. Suppose the ladder begins sliding down the wall in such a way that the foot of the ladder is moving away from the wall. How fast is the bag descending at the instant the foot of the ladder is 4m from the wall and foot is moving away at the rate 2 m/sec.
when the foot of the ladder is x m from the wall, and the top of the ladder is y m up the wall,
x^2+y^2 = 25
x dx/dt + y dy/dt = 0
So now plug in your numbers to find dy/dt
To find the rate at which the bag is descending, we need to use related rates. We can start by drawing a diagram of the situation.
Let's denote the distance between the foot of the ladder and the wall as x, and the distance from the top of the ladder to the ground as y, which is the height the bag is above the ground.
From the diagram, we can observe that a right triangle is formed by the ladder, the wall, and the ground. The ladder acts as the hypotenuse of the triangle, while the wall and the ground represent the legs of the triangle.
Using the Pythagorean theorem, we have x^2 + y^2 = 5^2 since the ladder has a length of 5m.
Differentiating both sides of the equation with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0
We are given that dx/dt = 2 m/sec (the rate at which the foot of the ladder is moving away from the wall). By substituting this value into the equation above, we can solve for dy/dt, which represents the rate at which the bag is descending.
Plugging in the given values:
2(4)(2) + 2y(dy/dt) = 0
Simplifying the equation:
8 + 2y(dy/dt) = 0
2y(dy/dt) = -8
(dy/dt) = -8 / (2y)
To find the rate at which the bag is descending, we need to evaluate (dy/dt) when x = 4m. From the Pythagorean theorem, we know that when the foot of the ladder is 4m from the wall, y can be found by substituting x = 4 into the equation x^2 + y^2 = 5^2:
4^2 + y^2 = 5^2
16 + y^2 = 25
y^2 = 25 - 16
y^2 = 9
y = 3
Now we can substitute y = 3 into (dy/dt) = -8 / (2y) to find the rate at which the bag is descending:
(dy/dt) = -8 / (2 * 3)
(dy/dt) = -8 / 6
(dy/dt) = -4/3 m/sec
Therefore, the bag is descending at a rate of 4/3 m/sec at the instant the foot of the ladder is 4m from the wall and is moving away at a rate of 2 m/sec.