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

Answer:

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

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: $\alpha =Arctan(\frac{y}{x})$ 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:$\frac{d\alpha }{dx} =\frac{-1}{\sqrt{100-x^{2} } }$ evaluating at x=6 ft we get: -0.125(degree/feet). The negative signal means that the angle is decreasing.

Answer 2

Answer:

$\frac{d\theta}{dt} = -0.125 rad/s$

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

$\frac{x}{L} = cos\theta$

now we have

$x = L cos\theta$

now differentiate the above equation with time

$\frac{dx}{dt} = -L sin\theta\frac{d\theta}{dt}$

so we have

$\frac{d\theta}{dt} = \frac{-v_x}{L sin\theta}$

$\frac{d\theta}{dt} = \frac{-1 ft/s}{10 sin\theta}$

as we know that

$cos\theta = \frac{6}{10} = 0.6$

$\theta = 53 degree$

$\frac{d\theta}{dt} = \frac{-1 ft/s}{10 sin53}$

$\frac{d\theta}{dt} = \frac{-1 ft/s}{10 sin\theta}$

$\frac{d\theta}{dt} = -0.125 rad/s$