A ladder rests against a vertical pole. The foot of the ladder is sliding away from the pole along horizontal ground. Find the inclination of the ladder to the horizontal at the instant when the top of the ladder is moving 3 times as
fast as the foot of the ladder.
If the ladder has length z, then
x^2+y^2 = z^2
x dx/dt + y dy/dt = 0
when dy/dt = -3 dx/dt,
x dx/dt - 3y dx/dt = 0
x = 3y
Thus, the angle θ is
tanθ = 1/3
To solve this problem, we can use a property of similar triangles. Let's assume that the distance between the foot of the ladder and the pole is represented by "x", and the height at the top of the ladder is represented by "y".
Since the top of the ladder is moving 3 times as fast as the foot of the ladder, we can express the relationship between the rates of change:
(dy / dt) = 3(dx / dt)
This means that the rate of change of the height (dy / dt) is 3 times the rate of change of the distance (dx / dt).
We also know that the height and distance are related by the Pythagorean theorem:
x^2 + y^2 = L^2
where L represents the length of the ladder.
Taking the derivative of this equation with respect to time t, we get:
2x(dx / dt) + 2y(dy / dt) = 0
Substituting the given relationship between the rates of change, we can rewrite the equation as:
2x(dx / dt) + 2y(3(dx / dt)) = 0
Simplifying this equation, we have:
2x(dx / dt) + 6y(dx / dt) = 0
Dividing both sides of the equation by 2(dx / dt), we get:
x + 3y = 0
Solving this equation for y, we have:
y = (-1/3)x
Now, we can find the inclination of the ladder to the horizontal. The inclination is the angle between the ladder and the horizontal ground. We can use the tangent function to find this angle:
tan(θ) = (y / x)
Substituting the value of y, we get:
tan(θ) = (-1/3)x / x
Simplifying this expression, we have:
tan(θ) = -1/3
To find the angle θ, we take the inverse tangent (arctan) of both sides:
θ = arctan(-1/3)
Therefore, the inclination of the ladder to the horizontal at the instant when the top of the ladder is moving 3 times as fast as the foot of the ladder is arctan(-1/3).
To find the inclination of the ladder to the horizontal, we can use trigonometry.
Let's denote the length of the ladder as "l", the distance between the foot of the ladder and the pole as "d", and the height of the pole as "h". We can also denote the speed of the top of the ladder as "v_top" and the speed of the foot of the ladder as "v_foot".
We are given that the top of the ladder is moving 3 times as fast as the foot of the ladder, so we can write the equation:
v_top = 3 * v_foot
Now, let's analyze the situation using trigonometry. We have a right-angled triangle formed by the ladder, the ground, and the pole. The angle between the ladder and the horizontal is the inclination we want to find.
Using the Pythagorean theorem, we can write:
d^2 + h^2 = l^2
Differentiating this equation with respect to time, we get:
2d * (d/dt) + 2h * (d/dt) = 2l * (d/dt)
Since we are given that the foot of the ladder is sliding away from the pole along the horizontal ground, we know that (d/dt) is the speed of the foot, which we denoted as v_foot. Therefore, we can rewrite the equation as:
2d * v_foot + 2h * v_foot = 2l * (d/dt)
Now, substitute the given relationship between the top and the foot speed:
2d * v_foot + 2h * v_foot = 2l * 3 * v_foot
Simplifying this equation, we get:
2(d + h) * v_foot = 6l * v_foot
Divide both sides of the equation by v_foot:
2(d + h) = 6l
Simplify further:
d + h = 3l
Substituting the equation we obtained above into the Pythagorean theorem equation:
(3l - d)^2 + d^2 = l^2
Expanding and simplifying:
9l^2 - 6dl + d^2 + d^2 = l^2
Simplifying further:
9l^2 - 6dl + 2d^2 - l^2 = 0
Rearranging terms:
8l^2 - 6dl + 2d^2 = 0
Now, we can differentiate this equation with respect to time to find the relationship between the variation of l and d:
16l * (d/dt) - 6d * (d/dt) + 4d * (d/dt) = 0
Since we are looking for the inclination at the instant when the top of the ladder is moving 3 times as fast as the foot, we know that (d/dt) is 3 times the speed of the foot, which is v_foot. Therefore:
16l * 3v_foot - 6d * 3v_foot + 4d * 3v_foot = 0
Simplifying:
48lv_foot - 18dv_foot + 12dv_foot = 0
Combine like terms:
48lv_foot - 6dv_foot = 0
Divide both sides by 6v_foot:
8l - d = 0
Rearrange terms:
d = 8l
Now, substitute this result into the equation we obtained earlier:
8l + h = 3l
Subtract 8l from both sides:
h = -5l
Since h is the height of the pole, it cannot be negative. Therefore, we need to consider the negative value and positive value of h:
1. h = -5l
2. h = 5l
Taking the positive value, we have h = 5l.
Now, we can use this information to find the inclination of the ladder to the horizontal. The inclination is defined as the arctangent of the height divided by the distance:
θ = arctan(h/d)
Substituting the values we found:
θ = arctan(5l/d)
Therefore, the inclination of the ladder to the horizontal at the instant when the top of the ladder is moving 3 times as fast as the foot of the ladder is given by the equation:
θ = arctan(5l/d)