a ladder 6m long is leaning against a wall. if the top of th ladder is slipping down at th rate of 13 m/s, how fast is the bottom moving away from the wall when it is 5m from the wall?
(answer in m/s)
Please Help Before 230 pm central time
if top is at height h and base is at distance x,
x^2 + h^2 = 36
when x=5, h=√11
2x dx/dt + 2h dh/dt = 0
2(5) dx/dt + 2(√11)(-13) = 0
dx/dt = 13√11/5 = 8.6 m/s
To solve this problem, we can use related rates, since we have information about how fast the top of the ladder is moving and we want to find how fast the bottom is moving.
Let's denote the distance from the bottom of the ladder to the wall as x (in meters), and the distance from the top of the ladder to the ground as y (in meters). We are given that the ladder is 6 meters long, so we have the equation: x^2 + y^2 = 6^2.
Now, we want to find how fast the bottom is moving, dx/dt, when it is 5 meters from the wall, so we have x = 5.
To find dx/dt, we need to differentiate the equation x^2 + y^2 = 6^2 with respect to time (t), and then solve for dx/dt.
Differentiating both sides of the equation gives us: 2x(dx/dt) + 2y(dy/dt) = 0.
Since we are interested in finding dx/dt when x = 5, we can plug in these values into the equation, along with the given information that dy/dt (the rate at which the top of the ladder is moving down) is 13 m/s.
So we have: 2(5)(dx/dt) + 2y(13) = 0.
From the equation x^2 + y^2 = 6^2, we can solve for y using y = sqrt(6^2 - x^2).
Plugging this into our equation, we have: 2(5)(dx/dt) + 2(sqrt(6^2 - 5^2))(13) = 0.
Now we can solve for dx/dt:
10(dx/dt) = -26(sqrt(6^2 - 5^2)).
Simplifying further, we get: dx/dt = -26(sqrt(11)) / 10.
Therefore, when the bottom of the ladder is 5 meters from the wall, it is moving away from the wall at a rate of approximately -14.69 m/s (negative since it is moving away).