A 25-foot ladder is placed against a vertical wall. Suppose the bottom of the ladder slides away from the wall at a constant rate of 4 feet per second. How fast is the top of the ladder sliding down the wall when the bottom is 15 feet from the wall? The ladder is sliding down the wall at a rate of nothing ft/sec. (Type an integer or a simplified fraction.)
Accepted Solution
A:
Answer: The ladder is sliding down the wall at a rate of [tex]5\dfrac{17}{50}\ ft/sec[/tex]Step-by-step explanation:Since we have given that Length of ladder = 25 foot Distance from the wall to the bottom of ladder = 15 feetLet base be 'x'.Let length of wall be 'y'.So, by pythagorus theorem, we get that [tex]x^2+y^2=25^2\\\\15^2+y^2+625\\\\225+y^2=625\\\\y^2=625-225\\\\y^2=400\\\\y=\sqrt{400}\\\\y=20\ feet[/tex][tex]\dfrac{dy}{dx}=-4\ ft/sec[/tex]Now, the equation would be [tex]x^2+y^2=625\\[/tex]Differentiating w.r.t x, we get that [tex]2x\dfrac{dx}{dt}+2y.\dfrac{dy}{dt}=0\\\\2\times 15\dfrac{dx}{dt}+2\times 20\times -4=0\\\\30\dfrac{dx}{dt}-160=0\\\\\dfrac{dx}{dt}=\dfrac{160}{30}=5.34\ ft/sec[/tex]Hence, the ladder is sliding down the wall at a rate of [tex]5\dfrac{17}{50}\ ft/sec[/tex]