A ladder 14 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?

Question

A ladder 14 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?

Answer 1

sinØ = x/14

cosØ = (1/4)dx/dØ
dx/dØ = 4cosØ

so when Ø = π/3
dx/dØ = 4cosπ/3 = 2

Answer 2

somehow my 14 changed to 4,

(need new specks)

so all the 4's are 14's
and last line is
dx/dØ = 14cos π/3 = 7

Answer 3

just realized that drwls had already answered the question correctly,

http://www.jiskha.com/display.cgi?id=1272944003

Answer 4

Well, looks like someone's dealing with some ladder math! Let's see if we can add a little humor to the equation.

So, we have a ladder resting against a wall, and we want to know how fast the distance from the bottom of the ladder to the wall changes as the ladder slides away.

Now, if I were that ladder, I'd be sliding away from the wall as fast as I could too! Who wants to be stuck against a wall all day?

But hey, let's get serious for a second. We know that the length of the ladder is 14 ft, and we're looking at a specific angle theta.

To find out how fast the distance x changes with respect to theta, we need to take a little derivative action.

Now, if I were a derivative, I'd be a little confused. It's like being a clown in a library – a square peg in a round hole. But hey, we're here to do the math, so let's do it!

We'll take the derivative of x with respect to theta, which gives us dx/dtheta.

Using trigonometry, we can relate x and theta through the equation x = 14*cos(theta).

If we take the derivative of both sides with respect to theta, we get dx/dtheta = -14*sin(theta).

So there you have it! The rate of change of x with respect to theta is -14*sin(theta).

Now, when theta is pi/3, we just plug in that value and find out how fast x is changing at that point. Get your sin(pi/3) calculator ready and crunch those numbers!

But remember, even though this ladder is sliding away from the wall, it still needs a hug from time to time. So go ahead, give it a little pat on the rungs and let it know it's doing a great job in the world of geometry!

Answer 5

To find how fast x changes with respect to θ, we need to use calculus. We will differentiate the equation relating x and θ.

Let's start by drawing a diagram of the situation.

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Here, the ladder is represented by the line connecting the wall and the top of the ladder. The angle θ is the angle between the ladder and the wall, and x is the distance from the bottom of the ladder to the wall.

Now, since we have a right triangle formed by the ladder, the wall, and the ground, we can use trigonometry to find the relation between x and θ.

We have: sin(θ) = x / 14 (using the definition of sine in a right triangle)

Rearranging, we get: x = 14sin(θ)

Now, we differentiate both sides of this equation with respect to θ to find how x changes with respect to θ:

d/dθ (x) = d/dθ (14sin(θ))

Using the chain rule on the right side, we have:

d/dθ (x) = 14cos(θ) * d/dθ (θ)

Since d/dθ (θ) = 1, we have:

d/dθ (x) = 14cos(θ)

So, the rate at which x changes with respect to θ is given by 14cos(θ).

To find how fast x changes with respect to θ when θ = π/3, we substitute θ = π/3 into the expression:

d/dθ (x) = 14cos(π/3) = 14 * 1/2 = 7

Therefore, when θ = π/3, x is changing at a rate of 7 units per radian.