A 20 foot ladder is leaning against a wall. The base of the ladder is placed 6 feet from the bottom of the wall. The base of the ladder slips away from the wall at a constant rate of 2 feet/sec. Find the velocity of the top of the ladder as it slider down the wall 3 seconds after it begins to fall.
At a time of t seconds, let the distance the foot is away from the wall be x ft, and let the top of the ladder be y ft above the ground
x^2 + y^2 = 20^2
2x dx/dt + 2y dy/dt = 0
given dx/dt = 2
when t = 3, x = 2(3) = 6ft
when x = 6, 36 + y^2 = 400
y = √(400-36) = 2√91
in 2x dx/dt + 2y dy/dt = 0
12(2) + 4√91 dy/dt = 0
dy/dt = .... , notice the negative indicates y is decreasing, thus the ladder is sliding down
6^2 + y^2 = 20^2,
Y = 19 Ft. = ht. of ladder against wall.
r * T = 19.
r * 3 = 19,
r = 6.36 Ft/s.
To find the velocity of the top of the ladder as it slides down the wall, we can use the concept of similar triangles.
Let's denote the distance between the base of the ladder and the wall as x, and the height of the ladder on the wall as y.
According to the problem, the base of the ladder slips away from the wall at a constant rate of 2 feet/sec. This means that the rate of change of x, denoted as dx/dt, is equal to 2 ft/sec.
We can use the Pythagorean theorem to relate x and y:
x^2 + y^2 = 20^2
Differentiating both sides of the equation with respect to time t, we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since we know dx/dt = 2 ft/sec, we can substitute this value into the equation:
2x(2) + 2y(dy/dt) = 0
Simplifying the equation, we have:
4x + 2y(dy/dt) = 0
We can solve this equation for dy/dt, which represents the rate of change of y (the height of the ladder on the wall):
2y(dy/dt) = -4x
(dy/dt) = -4x / (2y)
Since we're interested in finding the velocity of the top of the ladder as it slides down the wall, we need to find dy/dt when t = 3 seconds.
To obtain x and y when t = 3 seconds, we can use the given information that the base of the ladder is placed 6 feet from the bottom of the wall (x = 6) and the ladder is initially 20 feet long (y = 20).
Substituting these values into the equation, we have:
(dy/dt) = -4(6) / (2(20))
(dy/dt) = -24 / 40
(dy/dt) = -0.6 ft/sec
Therefore, the velocity of the top of the ladder as it slides down the wall 3 seconds after it begins to fall is -0.6 ft/sec.
To find the velocity of the top of the ladder after 3 seconds, we need to use the concept of related rates.
Let's start by visualizing the situation. We have a right-angled triangle formed by the wall, the ground, and the ladder. The ladder is the hypotenuse of the triangle, with a length of 20 feet, and the base of the ladder is placed 6 feet from the bottom of the wall.
Now, let's define some variables:
- Let t represent time (measured in seconds).
- Let x represent the distance from the base of the ladder to the wall at time t (measured in feet).
- Let y represent the distance from the top of the ladder to the ground at time t (measured in feet).
Given that the base of the ladder slips away from the wall at a constant rate of 2 feet/sec, we can express this relationship as: dx/dt = 2.
We are asked to find the velocity of the top of the ladder, which is dy/dt, when t = 3 seconds.
Using the Pythagorean Theorem, we can relate the variables x, y, and the hypotenuse (ladder length):
x^2 + y^2 = 20^2
Differentiating both sides of this equation with respect to time t, we get:
2x * dx/dt + 2y * dy/dt = 0
Now, plug in the known values:
x = 6 (since the base of the ladder is placed 6 feet from the bottom of the wall)
dx/dt = 2 (since the base of the ladder slips away from the wall at a constant rate of 2 feet/sec)
y = ? (since we want to find the distance from the top of the ladder to the ground)
dy/dt = ? (since we want to find the velocity of the top of the ladder)
We have a system of equations:
2(6)(2) + 2y * dy/dt = 0
Now, solve for y * dy/dt:
12 + 2y * dy/dt = 0
2y * dy/dt = -12
dy/dt = -12 / (2y)
Since we are interested in finding the velocity of the top of the ladder after 3 seconds, we can substitute t = 3 into the equation:
dy/dt = -12 / (2y)
dy/dt = -12 / (2y) = -6 / y
Now, we need to find the value of y when t = 3. From the Pythagorean Theorem, when t = 0, x = 6 and y = 18. Using similar triangles, we can determine that when t = 3, x = 12 and y = 18 - (2 * 3) = 18 - 6 = 12.
Plug in the values:
dy/dt = -6 / y = -6 / 12 = -1/2
Therefore, the velocity of the top of the ladder as it slides down the wall 3 seconds after it begins to fall is -1/2 feet per second.