a ladder of 10 ft long leans against a vertical wall. the upper end slips down the wall at 5 ft/sec. how fast is the ladder turning when it takes an angle of 30 degree with the ground?
if the ladder reaches up h feet,
sinθ = h/10
cosθ dθ/dt = 1/10 dh/dt
You know θ and dh/dt, so solve for dθ/dt.
dθ/dt = sqrt(3)/3
To find the rate at which the ladder is turning when it takes an angle of 30 degrees with the ground, we can use trigonometry and related rates.
Let's define some variables:
- Let y be the height of the ladder against the wall.
- Let x be the distance from the base of the ladder along the ground.
We are given:
- The length of the ladder is 10 ft, so at any time, sqrt(x^2 + y^2) = 10.
Differentiating both sides with respect to time t, we get:
d/dt(sqrt(x^2 + y^2)) = d/dt(10).
Now we need to find the rates of change of x and y with respect to time.
Given that the upper end of the ladder slips down the wall at a rate of 5 ft/sec, we know that dy/dt = -5.
To find dx/dt, we need to express x in terms of y:
y = 10 - x^2 (using the relationship of the ladder with the wall).
Differentiating both sides with respect to t, we get:
dy/dt = d/dt(10 - x^2).
Since we know that dy/dt = -5, we can substitute this value:
-5 = d/dt(10 - x^2).
Now, we can solve for dx/dt:
d/dt(10 - x^2) = -5.
-2x(dx/dt) = -5.
dx/dt = 5/(2x).
To find the angle θ between the ladder and the ground, we can use trigonometry:
tan(θ) = y/x.
Differentiating both sides with respect to t, we get:
sec^2(θ) * dθ/dt = (dy/dt * x - y * dx/dt) / x^2.
Since we know that dy/dt = -5 and dx/dt = 5/(2x), and we want to find dθ/dt when θ = 30 degrees, we have:
sec^2(30) * dθ/dt = (-5 * x - y * (5/(2x))) / x^2.
Recall that the ladder length is fixed, so x and y are related by the equation sqrt(x^2 + y^2) = 10. Substituting this relationship into the equation, we get:
sec^2(30) * dθ/dt = (-5 * x - y * (5/(2x))) / x^2.
sec^2(30) * dθ/dt = (-5 * x - y * (5/(2x))) / x^2.
sec^2(30) * dθ/dt = (-5 * x - y * (5/(2x))) / x^2.
At θ = 30 degrees, we have:
sec^2(30) = 4/3.
Using the relationship y = 10 - x^2, we can substitute y = 10 - x^2 into the equation:
4/3 * dθ/dt = (-5x - (10 - x^2) * (5/(2x))) / x^2.
Simplifying the equation further, we get:
4/3 * dθ/dt = (-5x - (10 - x^2) * 5/(2x)) / x^2.
Now, substitute x and y back into the equation:
4/3 * dθ/dt = (-5 * x - (10 - x^2) * 5/(2x)) / x^2.
4/3 * dθ/dt = (-5 * x - (10 - x^2) * 5/(2x)) / x^2.
Finally, we can plug in the value of x when θ = 30 degrees into the equation and solve for dθ/dt:
4/3 * dθ/dt = (-5 * (10 - x^2) - (10 - x^2) * 5/(2 * (10 - x^2)))) / (10 - x^2)^2.
Simplifying the expression further, we get:
4/3 * dθ/dt = (-5(10 - x^2)(2 + 1))/2(10 - x^2)^2.
Substituting x = 10 - y = 10 - (10 - x^2):
4/3 * dθ/dt = (-5(10 - (10 - x^2))(3))/2(10 - (10 - x^2))^2.
The expression simplifies to:
4/3 * dθ/dt = (-5(10 - (10 - x^2))(3))/2(10 - (10 - x^2))^2.
To find the value of dθ/dt, divide both sides by 4/3:
dθ/dt = (-5(10 - (10 - x^2))(3))/((2(10 - (10 - x^2))^2) * (4/3)).
Once we know the current value of x, we can substitute it into the equation to find the rate at which the ladder is turning when it takes an angle of 30 degrees with the ground.
To find how fast the ladder is turning when it forms a 30-degree angle with the ground, we can apply the formula for related rates from calculus.
Let's denote the length of the ladder as "L" and the distance along the ground between the base of the ladder and the wall as "x". Note that the ladder is forming a right triangle with the wall and the ground.
We are given that the ladder is 10 ft long, so L = 10 ft.
The problem states that the upper end of the ladder is slipping down the wall at a rate of 5 ft/sec. Let's call the variable representing the distance along the wall as "y".
We want to find dy/dt, the rate at which y is changing. We are given dx/dt, which represents the rate at which x is changing.
Based on the right triangle formed by the ladder, we can create the following equation using the Pythagorean theorem:
x^2 + y^2 = L^2
Differentiating both sides of the equation implicitly with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since we are interested in finding dy/dt when the ladder forms a 30-degree angle with the ground, we know that the angle formed between the ladder and the wall is 90 degrees - 30 degrees = 60 degrees.
Using trigonometry, we can determine the values of x and y when the ladder forms a 30-degree angle:
sin(60 degrees) = y / L
sin(60 degrees) = y / 10
sqrt(3) / 2 = y / 10
y = 5 sqrt(3)
From this, we can find x:
cos(60 degrees) = x / L
cos(60 degrees) = x / 10
1/2 = x / 10
x = 5
Substituting these values into our equation, we have:
2(5)(dx/dt) + 2(5 sqrt(3))(dy/dt) = 0
Simplifying the equation, we get:
10(dx/dt) + 10 sqrt(3)(dy/dt) = 0
Now, we need to find dx/dt when x = 5 ft. We can differentiate both sides of the equation x = 5 with respect to time to obtain:
dx/dt = 0
Substituting dx/dt = 0 into our equation, we have:
10(0) + 10 sqrt(3)(dy/dt) = 0
Simplifying further, we find:
10 sqrt(3)(dy/dt) = 0
To find dy/dt, we divide both sides of the equation by 10 sqrt(3) and get:
dy/dt = 0
Therefore, when the ladder takes an angle of 30 degrees with the ground, it is not turning (dy/dt = 0).