A ladder 20 ft long rests against a vertical wall. Let \theta 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 \theta when \theta = \pi / 3
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5\9 divd 1\3
a 13 foot ladder is leaning against a vertical wall. when jack begins pulling the foot of the ladder away from the wall at a rate of 0.4 ft/s. how fast is the top of the ladder sliding down the wall when the foot of the ladder is 5 ft from the wall?
To find how fast x changes with respect to θ, we can use trigonometry and differentiation.
Let's first draw a diagram to visualize the problem:
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x |/__ θ|
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In this right triangle, the ladder is the hypotenuse of length 20 ft, x is the horizontal distance from the bottom of the ladder to the wall, and θ is the angle between the ladder and the wall. Our goal is to find dx/dθ, the rate of change of x with respect to θ.
We can use the trigonometric relationship of the right triangle to relate x and θ:
cos(θ) = x/20
To differentiate both sides of this equation with respect to θ, we apply the chain rule on the left side:
d(cos(θ))/dθ = d(x/20)/dθ
Now let's differentiate each term:
-sin(θ) dθ/dθ = 1/20 dx/dθ
Simplifying, we get:
-dx/dθ = (sin(θ))/20
Now to find dx/dθ, we can multiply both sides of the equation by -1:
dx/dθ = -(sin(θ))/20
Finally, substitute θ = π/3 into the equation to find the value of dx/dθ when θ = π/3:
dx/dθ = -(sin(π/3))/20 = -√3/40
Therefore, when θ = π/3, dx/dθ = -√3/40.