the top of a 28 foor ladder, leaning against a vertical wall, is slipping down the wall at the rate of 3 feet per second. How fast is the bottom of the ladder sliding along the ground when the bottom of the ladder is 10 feet away from the base of the wall?

We get:

2(10) dx /dt + 2 (26.15) (-3) = 0

20 dx / dt  -  156.9 = 0

dx / dt = 156.9

Step-by-step explanation:

Solution:

Given:

Top of the ladder from floor= 28 feet

Ladder slipping down at the rate of = 3 feet per second

The bottom is 10 feet botton away from the base of the wall.

Firstly, a sketch triangle whose hypotenuses is the ladder.

Let y(t) be the  ladder height with a vertical wall, and

Let x(t) be the bottom length of the ladder with a vertical wall.

Then      x2(t) +y2(t) = 282

Differentiate with respect to time:

2x (t) dx / dt+ 2y (t)dy /dt = 0

To find the value of y, put x = 10 in eq x2(t) +y2(t) = 202

We get:

102 + y2(t) = 282

Y2(t) = 784 – 100 = 684

Y = 26.15

Put this value in

2x (t) dx / dt+ 2y (t)dy /dt = 0

If dy/dt = -3, then what is dx/dt?

We get:

2(10) dx /dt + 2 (26.15) (-3) = 0

20 dx / dt  -  156.9 = 0

dx / dt = 156.9

55? i think maybe

step-by-step explanation:

i have no clue at all sorry

step-by-step explanation: