Question

Calculus Application

2. A ladder 20 feet long leans a vertical building. If the bottom of the ladder slides away from the building
horizontally at a rate of 2 ft/sec, how fast is the ladder sliding down the building when the top of the
ladder is 12 feet above the ground?

Answer 1

Answer:

-8/3 ft/s

Step-by-step explanation:

We are given:

distance of the top of the ladder from the ground (h) = 12 ft

height of the ladder = 20 ft

rate of change of the distance of the base of ladder from the wall (dx/dt):

2 ft/s

Finding the distance of the base of the ladder from the wall:

From the Pythagoras's Theorem, we know that:

hypotenuse² = height² + base²

replacing the given values

20² = 12² + x²

400 = 144 + x²

x² = 256                            [subtracting 144 from both sides]

x = 16 ft                            [taking the square root of both sides]

The rate of change of the height of the Ladder from the ground:

We know that:

h = 12 ft

($\frac{dh}{dt}$) = ?

x = 16 ft

($\frac{dx}{dt}$) = 2 ft/s

According to the Pythagoras's Theorem:

20² = x² + h²

differentiating both sides with respect to time

$\frac{d(400)}{dt} = \frac{d(x^{2} + h^{2})}{dt}$

$0 = \frac{d(x^{2})}{dt} + \frac{d(h^{2})}{dt}$

$0 = \frac{d(x^{2})}{dx}(\frac{dx}{dt}) + \frac{d(h^{2})}{dh}(\frac{dh}{dt})$

$0 = 2x(\frac{dx}{dt}) + 2h(\frac{dh}{dt})$

replacing the variables

$0 = 2(16)(2) + 2(12)(\frac{dh}{dt})$

$0 = 64 + 32(\frac{dh}{dt})$

$-64 =32(\frac{dh}{dt})$                                [subtracting 64 from both sides]

$\frac{-64}{32} =(\frac{dh}{dt})$                                    [dividing both sides by 32]

$\frac{dh}{dt} = \frac{-8}{3} ft/s$

Hence, the ladder will slide down at a speed of 8/3 feet per second