A ladder 10 meter long is leaning against a wall. If the top of the ladder is sliding down the wall at 3m/s, how fast is the angle between the top of the ladder and the wall changing when the foot of the ladder is 6 meter away from the wall?
let the angle you mentioned be Ø
Let the foot of the ladder be x ft from the wall, and the height of the ladder be y ft
cosØ = y/10
y = 10cosØ
dy/dt = -10sinØ dØ/dt
when x = 6
x^2+y^2=100
y = 8 , and sinØ = 8/10
Also we know dy/dt = -3 m/s
-3 = -10(8/10)(dØ/dt)
dØ/dt = 3/8 radians/s
To find the rate at which the angle between the ladder and the wall is changing, we can use trigonometry. Let's call the length of the ladder (hypotenuse) 'x', the distance the foot of the ladder is from the wall (base) 'y', and the angle between the ladder and the wall 'θ'.
We are given the following information:
- The length of the ladder, x = 10 meters
- The rate at which the top of the ladder is sliding down the wall, dx/dt = -3 m/s (negative sign because the ladder is sliding down)
We need to find dθ/dt, the rate at which the angle θ is changing when the foot of the ladder is 6 meters away from the wall (y = 6 meters).
Using trigonometry, we know that:
sin(θ) = y / x
Differentiating both sides of the equation with respect to time (t):
d/dt(sin(θ)) = d/dt(y/x)
The derivative of sin(θ) with respect to time is:
cos(θ) * dθ/dt
Applying the chain rule to d/dt(y/x):
d/dt(y/x) = (dx/dt * y - x * dy/dt) / x^2
Substituting the known values into the equation:
-3 = (10 * 6 - 10 * dy/dt) / 10^2
Simplifying the equation:
-3 = (60 - 10 * dy/dt) / 100
-300 = 60 - 10 * dy/dt
10 * dy/dt = 60 + 300
10 * dy/dt = 360
Dividing both sides by 10:
dy/dt = 36 m/s
Therefore, the foot of the ladder is moving away from the wall at a rate of 36 meters per second when the foot is 6 meters away from the wall.
To solve this problem, we'll use trigonometry and differentiation.
Let's define the variables:
- Ladder length (hypotenuse): L = 10 meters
- Rate of change of the top of the ladder sliding down the wall: dx/dt = -3 m/s (negative because it's sliding down)
- Distance between the foot of the ladder and the wall: x = 6 meters
- Angle between the top of the ladder and the wall: θ (we need to find dθ/dt)
We can use the Pythagorean theorem to relate L, x, and the height of the ladder (h):
L^2 = x^2 + h^2
Differentiating both sides with respect to time (t), we get:
2L(dL/dt) = 2x(dx/dt) + 2h(dh/dt)
Substituting the given values and the unknowns we want to find:
2(10)(dL/dt) = 2(6)(-3) + 2h(dh/dt)
20(dL/dt) = -36 + 2h(dh/dt)
We know that h = L*sin(θ), so we substitute it into the equation:
20(dL/dt) = -36 + 2L*sin(θ)(dθ/dt)
Now we have an equation in terms of θ and dθ/dt. We need to find dθ/dt. Let's rearrange the equation to isolate it:
20(dL/dt) - 2L*sin(θ)(dθ/dt) = -36
2L*sin(θ)(dθ/dt) = 20(dL/dt) + 36
dθ/dt = (20(dL/dt) + 36) / (2L*sin(θ))
Plugging in the known values:
dθ/dt = (20(-3) + 36) / (2(10)*sin(θ))
dθ/dt = (-60 + 36) / (20*sin(θ))
dθ/dt = -24 / (20*sin(θ))
dθ/dt = -6 / (5*sin(θ))
Finally, when the foot of the ladder is 6 meters away from the wall, we can find the angle θ using the sine function:
sin(θ) = h / L
sin(θ) = (10^2 - 6^2) / 10
sin(θ) = sqrt(100 - 36) / 10
sin(θ) = sqrt(64) / 10
sin(θ) = 8 / 10
sin(θ) = 0.8
Now we can substitute this value back into our equation for dθ/dt:
dθ/dt = -6 / (5*0.8)
dθ/dt = -6 / 4
dθ/dt = -1.5 rad/s
Therefore, the angle between the top of the ladder and the wall is changing at a rate of -1.5 rad/s when the foot of the ladder is 6 meters away from the wall.