A ladder 10 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 = π/3?

Solution:

Given, a ladder of length 10ft rests against a vertical wall.

The angle between the top of the ladder and the wall is θ.

The distance from the bottom of the ladder to the wall is x.

We have to find the rate of change of x with respect to θ when θ = π/3.

Let L be the length of the ladder.

So, L = 10ft

The relation between the angle and distance x is given by

x = L sinθ

On differentiating,

dx/dt = L cosθ

Put θ = π/3,

dx/dt = (10) cos(π/3)

= (10)(1/2)

= 5

Therefore, dx/dt = 5ft/s

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A ladder 10 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 = π/3?

Summary:

A ladder 10 ft long rests against a vertical wall. Let theta be the angle between the top of the ladder and the wall and let be the distance from the bottom of the ladder to the wall. If the bottom of the ladder slides away from the wall, then the rate of change of x with respect to theta when theta = π/3 is 5ft/s.