[tex]7.2\:\text{ft/s}[/tex]
Step-by-step explanation:
We can apply the Pythagorean theorem here:
[tex]26^2 = x^2 + y^2\:\:\:\:\:\:\:\:\:(1)[/tex]
where x is the distance of the ladder base from the wall and y is the distance of the ladder top from the ground. Taking the time derivative of the expression above, we get
[tex]0 = 2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt}[/tex]
Solving for [tex]\frac{dx}{dt},[/tex] we get
[tex]\dfrac{dx}{dt} = -\dfrac{y}{x}\dfrac{dy}{dt}[/tex]
We can replace y by rearranging Eqn(1) such that
[tex]y = \sqrt{26^2 - x^2}[/tex]
Therefore,
[tex]\dfrac{dx}{dt} = - \dfrac{\sqrt{26^2 - x^2}}{x}\dfrac{dy}{dt}[/tex]
Since y is decreasing as the ladder is being lowered, we will assign a negative sign to [tex]\frac{dy}{dt}[/tex]. Hence,
[tex]\dfrac{dx}{dt} = - \dfrac{\sqrt{26^2 - (10)^2}}{10}(-3\:\text{ft/min})[/tex]
[tex]\:\:\:\:\:\:\:= 7.2\:\text{ft/min}[/tex]
