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?

-8/3 ft/s

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

() = ?

x = 16 ft

() = 2 ft/s

According to the Pythagoras's Theorem:

20² = x² + h²

differentiating both sides with respect to time

replacing the variables

                               [subtracting 64 from both sides]

                                   [dividing both sides by 32]

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

rectangle: 9 1/8 x 4 1/9 = 37 37/72in

parallelogram: b times h= 10.3 x 6.2 = 63.86

trapezoid: a+b/2 times h= 42.5

volume of prism: l times w times h= 36

alright well the answers to your question is

2xy/9 hope this have a nice day : )

step-by-step explanation: