A ladder 10 ft long rests against a vertical wall. let θ be the angle between the top of the ladder and the wall and let x be the distance from the bottom of the ladder to the wall. If the bottom of the ladder slides away from the wall, how fast does x change with respect to θ when θ = π/3
We know that x=Lsin(θ), where L=length of ladder = 10'.
x(θ)=L sin(θ)
x'(θ)=dx/dθ = L cos(θ)
Calculate x'(π/3).
To find the rate at which x changes with respect to θ, we need to find the derivative dx/dθ.
We know that the ladder is 10 ft long, so its length can be expressed as:
L^2 = x^2 + h^2,
where L is the length of the ladder, x is the distance from the bottom of the ladder to the wall, and h is the height from the bottom to the top of the ladder.
Differentiating the above equation implicitly with respect to θ, we get:
2L(dL/dθ) = 2x(dx/dθ) + 2h(dh/dθ).
Since h = x * tan(θ), we can substitute this into the equation:
2L(dL/dθ) = 2x(dx/dθ) + 2x * tan(θ)(dx/dθ).
Simplifying the equation, we have:
L(dL/dθ) = x(dx/dθ) + x * tan(θ)(dx/dθ).
Rearranging the equation, we get:
x(dx/dθ) = L(dL/dθ) - x * tan(θ)(dx/dθ).
Now let's substitute the given values into the equation. When θ = π/3, we have:
x(dx/dθ) = 10(dL/dθ) - x * tan(π/3)(dx/dθ).
Using the value of tan(π/3) = √3, we have:
x(dx/dθ) = 10(dL/dθ) - x * √3(dx/dθ).
Now we can solve for dx/dθ by isolating it on one side of the equation:
x(dx/dθ) + x * √3(dx/dθ) = 10(dL/dθ),
Factor out dx/dθ:
dx/dθ(x + x√3) = 10(dL/dθ),
Divide both sides by (x + x√3):
dx/dθ = 10(dL/dθ) / (x + x√3).
Since dL/dθ represents the change in the ladder length with respect to θ, it is given as dL/dθ = 0. (The length of the ladder is fixed, so it does not vary with θ.)
Substituting this in the equation, we have:
dx/dθ = 10(0) / (x + x√3),
dx/dθ = 0.
Therefore, when θ = π/3, the distance x from the bottom of the ladder to the wall does not change with respect to θ (dx/dθ = 0).
To find how fast x changes with respect to θ, we can use related rates. We will use the information given in the problem, which is that a ladder 10 ft long rests against a vertical wall.
Let's draw a diagram to better understand the problem:
|
\ | /
\|/
|
| _
|10 ft
|
-------
From the diagram, we can see that the ladder acts as the hypotenuse of a right triangle, with the wall forming one side and the distance x from the bottom of the ladder to the wall forming the other side. The angle between the top of the ladder and the wall is θ.
Using trigonometry, we can write the equation relating x, θ, and the length of the ladder:
x = 10 * cos(θ)
Now, to find how fast x changes with respect to θ, we need to differentiate both sides of the equation with respect to θ:
d(x)/d(θ) = d(10 * cos(θ))/d(θ)
Now, we can use the chain rule to differentiate the right side of the equation:
d(x)/d(θ) = -10 * sin(θ)
Since we are given that θ = π/3, we can substitute that into the equation:
d(x)/d(θ) = -10 * sin(π/3)
Simplifying further:
d(x)/d(θ) = -10 * (√3/2)
Therefore, when θ = π/3, the rate at which x changes with respect to θ is -10 * (√3/2) ft/radian.