The top of a 24 foot ladder, leaning against a vertical wall, is slipping down the wall at the rate of 4 feet per second. How fast is the bottom of the ladder sliding along the ground when the bottom of the ladder is 7 feet away from the base of the wall?
x^2+y^2=24^2
x dx/dt + y dy/dt = 0
Now you just solve for dx/dt, since you have x,y,dy/dt
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To find the rate at which the bottom of the ladder is sliding along the ground, we can use related rates and trigonometry.
Let's define the following variables:
- The height of the wall: h
- The distance from the wall to the bottom of the ladder: x
- The length of the ladder: L
We are given:
- The rate at which the top of the ladder is sliding down the wall: dh/dt = -4 ft/s (negative because the height is decreasing)
We need to find:
- The rate at which the bottom of the ladder is sliding along the ground: dx/dt
To relate the variables, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the ladder) is equal to the sum of the squares of the other two sides. In this case, we have:
L^2 = h^2 + x^2
Differentiating both sides of this equation with respect to time (t), we get:
(2L)(dL/dt) = (2h)(dh/dt) + (2x)(dx/dt)
Now, we substitute the given values and the known rate:
24^2 = h^2 + 7^2
576 = h^2 + 49
h^2 = 576 - 49
h^2 = 527
h ≈ 22.96 feet
We can now substitute all the values into the equation and solve for dx/dt:
(2(24))(dL/dt) = (2(22.96))(-4) + (2(7))(dx/dt)
48(dL/dt) = -91.84 - 14(dx/dt)
48(dL/dt) + 14(dx/dt) = -91.84
(48)(-4) + (14)(dx/dt) = -91.84
-192 + 14(dx/dt) = -91.84
14(dx/dt) = -91.84 + 192
14(dx/dt) = 100.16
dx/dt = 100.16/14
dx/dt ≈ 7.16 ft/s
Therefore, when the bottom of the ladder is 7 feet away from the base of the wall, the bottom of the ladder is sliding along the ground at a rate of approximately 7.16 ft/s.
To find the rate at which the bottom of the ladder is sliding along the ground, we can use related rates. Let's denote the distance from the base of the wall to the bottom of the ladder as x.
Given that the top of the ladder is slipping down the wall at a rate of 4 feet per second, we can say that the rate of change of the height of the ladder (y) with respect to time (t) is -4 ft/s (negative because it is sliding down).
We need to find the rate of change of the distance x with respect to time when x = 7 ft. Let's denote the rate of change of x with respect to t as dx/dt.
We can form a right triangle with the ladder as the hypotenuse, the vertical wall as one side, and the ground as the other side. The height (y) and the distance from the base to the ladder (x) are the other two sides of the right triangle.
Using the Pythagorean theorem, we have:
y^2 + x^2 = 24^2
Differentiating both sides with respect to time (t), we get:
2y(dy/dt) + 2x(dx/dt) = 0
Since we are interested in finding dx/dt when x = 7 ft, and we are given that dy/dt = -4 ft/s, we have:
2y(-4) + 2(7)(dx/dt) = 0
Simplifying, we obtain:
-8y + 14(dx/dt) = 0
Now, let's solve for dx/dt:
14(dx/dt) = 8y
(dx/dt) = (8y) / 14
To find the current value of y, we can use the Pythagorean theorem:
y^2 + x^2 = 24^2
Plugging in x = 7, we get:
y^2 + 7^2 = 24^2
y^2 + 49 = 576
y^2 = 527
y ≈ 22.977
Now that we have the value of y, we can substitute it into the equation for dx/dt:
(dx/dt) = 8(22.977) / 14
(dx/dt) ≈ 13.032
Therefore, when the bottom of the ladder is 7 feet away from the base of the wall, the bottom of the ladder is sliding along the ground at a rate of approximately 13.032 feet per second.