Question

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.)

Answer 1

Answer: The ladder is sliding down the wall at a rate of $5\dfrac{17}{50}\ ft/sec$

Step-by-step explanation:

Since we have given that

Length of ladder = 25 foot

Distance from the wall to the bottom of ladder = 15 feet

Let base be 'x'.

Let length of wall be 'y'.

So, by pythagorus theorem, we get that

$x^2+y^2=25^2\\\\15^2+y^2+625\\\\225+y^2=625\\\\y^2=625-225\\\\y^2=400\\\\y=\sqrt{400}\\\\y=20\ feet$

$\dfrac{dy}{dx}=-4\ ft/sec$

Now, the equation would be

$x^2+y^2=625$

Differentiating w.r.t x, we get that

$2x\dfrac{dx}{dt}+2y.\dfrac{dy}{dt}=0\\\\2\times 15\dfrac{dx}{dt}+2\times 20\times -4=0\\\\30\dfrac{dx}{dt}-160=0\\\\\dfrac{dx}{dt}=\dfrac{160}{30}=5.34\ ft/sec$

Hence, the ladder is sliding down the wall at a rate of $5\dfrac{17}{50}\ ft/sec$