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

Well, well, well, it seems like we have a classic trigonometry problem here. Don't worry, I'm here to help you with a touch of humor!

Now, let's call the rate at which x changes with respect to θ as "dx/dθ" - but don't worry about that right now, I'll make it simple for you.

So, we have our ladder against the wall and we're sliding the bottom away. If we focus on that lovely angle θ = π/3 (which is 60 degrees for my non-math friend), we have a special relationship with the sides of a 30-60-90 triangle. The length of the ladder is 20 ft, which means we actually have a 10 ft base and a 10√3 ft height.

Now, to find the rate of change, we need to find dx/dθ. Since x represents the distance from the bottom of the ladder to the wall, we can use some trigonometry to relate x and θ.

In this case, we have x = 10√3 * tan(θ). Yes, I know it looks complicated, but trust me, it's quite simple. Just plug in θ = π/3 (or 60 degrees) into this equation, and you'll get the value of x.

Once you have that, just differentiate the equation with respect to θ to find dx/dθ, or how fast x changes relative to θ. To keep the jokes going, I'll leave the math part to you, my math-savvy friend. You got this!

To find how fast x changes with respect to θ, we need to differentiate the equation x = 20sinθ with respect to θ.

Differentiating both sides of the equation, we get:

dx/dθ = d/dθ(20sinθ)

To find the derivative of sinθ, we use the chain rule. Let's break it down step-by-step:

1. Differentiate the outer function (20sinθ) with respect to the inner function (θ):
d/dθ(20sinθ) = 20(d/dθ(sinθ))

2. Differentiate sinθ with respect to θ using the derivative of sine:
d/dθ(20sinθ) = 20(cosθ)

So, dx/dθ = 20cosθ.

We need to evaluate dx/dθ when θ = π/3. Plugging in θ = π/3 into the equation, we get:

dx/dθ = 20cos(π/3)

Since cos(π/3) = 1/2, we can substitute this value:

dx/dθ = 20 * (1/2) = 10.

Therefore, when θ = π/3, the rate of change of x with respect to θ is 10 ft/radian.

To find how fast x changes with respect to θ, we need to take the derivative of x with respect to θ. In this case, we can use trigonometry and related rates to solve the problem.

Let's draw a right triangle representing the situation. The vertical wall is one side, the ladder is the hypotenuse, and the distance x is the other side of the right triangle.

Since we have a right triangle, we can use the trigonometric relationship between the sides and angles:

sin(θ) = x / 20

To find dx/dθ (the rate of change of x with respect to θ), we need to take the derivative of both sides of this equation with respect to θ.

d/dθ (sin(θ)) = d/dθ (x / 20)

cos(θ) = (dx/dθ) / 20

Now, we have an equation that relates dx/dθ to θ:

(dx/dθ) = 20cos(θ)

To find the rate of change of x with respect to θ when θ = π/3, we substitute θ = π/3 into the equation:

(dx/dθ) = 20cos(π/3)
(dx/dθ) = 20 * 1/2 = 10

So, when θ = π/3, dx/dθ is equal to 10. This means that x is changing at a rate of 10 units per unit change in θ.

x -- as defined

y = 20cosØ

x^2 + y^2 = 400
x^2 + 400cos^2 Ø = 400
2x dx/dØ - 2sinØcosØ = 0
dx/dØ = sinØcosØ/x

when x = π/3 , sinπ/3 =√3/2, cosπ/3 = 1/2
x^2 + 400(3/4) = 400
x = 10
dx/dØ = (√3/2)(1/2)/10 = √3/40