Question

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.0 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall? (That is, find the angle's rate of change when the bottom of the ladder is 6 ft from the wall.)

Answer 1

The angle's rate of change is: -0.125 (degree/feet).

Step-by-step explanation:

In this case of problem we need to find the angle's rate (α) of change when x=6 ft. First we need to relate (α) with x and y and the expression that do it is:
`\alpha =Arctan((y)/(x))` where (α) is the angle between the ladder and the ground, x is the horizontal distance and y the vertical distance, now we need to have the variable y at function of x, so we can do it using the Pythagorean theorem and gets:
`y^(2) +x^(2) =10^(2)` solving for y(x) we get:
`y(x)=\sqrt{100-x^(2)}`. Replacing all we have got in the first equation:
`\alpha =Arctan(\frac{\sqrt{100-x^(2) } }{x})`. Finally we derivate this equation at function of variable x and gets this result:
`(d\alpha )/(dx) =\frac{-1}{\sqrt{100-x^(2) } }` evaluating at x=6 ft we get: -0.125(degree/feet). The negative signal means that the angle is decreasing.

Answer 2

`(d\theta)/(dt) = -0.125 rad/s`

Step-by-step explanation:

Let at any moment of time the foot of the ladder is at distance "x" from the wall and if the length of the ladder is L = 10 ft

so we have

`(x)/(L) = cos\theta`

now we have

`x = L cos\theta`

now differentiate the above equation with time

`(dx)/(dt) = -L sin\theta(d\theta)/(dt)`

so we have

`(d\theta)/(dt) = (-v_x)/(L sin\theta)`

`(d\theta)/(dt) = (-1 ft/s)/(10 sin\theta)`

as we know that

`cos\theta = (6)/(10) = 0.6`

`\theta = 53 degree`

`(d\theta)/(dt) = (-1 ft/s)/(10 sin53)`

`(d\theta)/(dt) = (-1 ft/s)/(10 sin\theta)`

`(d\theta)/(dt) = -0.125 rad/s`