A 15-ft ladder rests against a vertical wall. If the top of the ladder slides down the wall at a rate of 0.33 ft/sec, how fast, in ft/sec, is the bottom of the ladder sliding away from the wall, at the instant when the bottom of the ladder is 9 ft from the wall? Answer with 2 decimal places.
we have
x^2+y^2 = 225
2x dx/dt + 2y dy/dt = 0
Now plug in x=9, y=12, dy/dt=-1/3
0.44
To solve this problem, we will use related rates. Let's label the height of the wall "h" and the distance of the bottom of the ladder from the wall "x".
We are given that the ladder is sliding down the wall at a rate of 0.33 ft/sec. We need to find how fast the bottom of the ladder is sliding away from the wall, which is the rate of change of x with respect to time.
Using the Pythagorean theorem, we have the equation:
x^2 + h^2 = 15^2
Differentiating both sides of the equation with respect to time, we get:
2x(dx/dt) + 2h(dh/dt) = 0
We are given that dx/dt = 0.33 ft/sec and we need to find dh/dt when x = 9 ft.
Substituting the given values into the equation, we have:
2(9)(0.33) + 2h(dh/dt) = 0
Simplifying the equation, we get:
5.94 + 2h(dh/dt) = 0
To find dh/dt, we can solve for it by isolating the variable:
2h(dh/dt) = -5.94
dh/dt = -5.94 / (2h)
When x = 9 ft, we can use the Pythagorean theorem to find h:
9^2 + h^2 = 15^2
81 + h^2 = 225
h^2 = 144
h = 12 ft
Substituting h = 12 ft into the equation, we have:
dh/dt = -5.94 / (2 * 12)
dh/dt = -0.495 ft/sec (rounded to 2 decimal places)
Therefore, the bottom of the ladder is sliding away from the wall at a rate of -0.495 ft/sec (or approximately -0.50 ft/sec) when the bottom of the ladder is 9 ft from the wall.
To find the rate at which the bottom of the ladder is sliding away from the wall, we can use related rates. Let's assign variables to the relevant quantities:
Let y represent the distance from the bottom of the ladder to the wall.
Let x represent the distance from the top of the ladder to the ground.
We know that the ladder is 15 ft long, so we have the equation: x^2 + y^2 = 15^2.
We are given that dx/dt (the rate at which the distance from the top of the ladder to the ground is changing) is 0.33 ft/sec. We want to find dy/dt (the rate at which the distance from the bottom of the ladder to the wall is changing) when y = 9 ft.
To solve this, we first differentiate both sides of the equation x^2 + y^2 = 15^2 with respect to time:
2x(dx/dt) + 2y(dy/dt) = 0.
Next, we substitute the given values: x = 15 - y (since the ladder is 15 ft) and dx/dt = 0.33 ft/sec:
2(15 - y)(0.33) + 2y(dy/dt) = 0.
Simplifying the equation further:
(30 - 2y)(0.33) + 2y(dy/dt) = 0.
Now we can substitute y = 9 into the equation and solve for dy/dt:
(30 - 2(9))(0.33) + 18(dy/dt) = 0.
(30 - 18)(0.33) + 18(dy/dt) = 0.
12(0.33) + 18(dy/dt) = 0.
3.96 + 18(dy/dt) = 0.
18(dy/dt) = -3.96.
dy/dt = -3.96/18.
dy/dt ≈ -0.22 ft/sec.
Therefore, at the instant when the bottom of the ladder is 9 ft from the wall, the bottom of the ladder is sliding away from the wall at a rate of approximately -0.22 ft/sec.