Aladder placed up against a wall is sliding down. the distance between the top of the ladder and the foot of the wall is decreasing at a rate of 9 inches per second. when this distance is 61 inches, how fast is the distance between the bottom of the ladder and the foot of the wall changing? the ladder is 152 inches long. (do not include units in your answer, and round to the nearest hundredth.)

distance changing at rate of 3.94 inches/sec

Explanation:

Given data

wall decreasing at a rate = 9 inches per second

ladder L = 152 inches

distance  h = 61 inches

to find out

how fast is the distance changing

solution

we know that

h² + b² = L²   1

h² + b² = 152²

Apply here derivative w.r.t. time

2h dh/dt + 2b db/dt = 0

h dh/dt + b db/dt = 0

db/dt = - h/b × dh/dt     2

and

we know

h = 61

so h² + b² = L²

61² + b² = 152²

b² = 19383

so b = 139.223

and we know dh/dt = -9 inch/sec

so from equation 2

db/dt = -61/139.223  (-9)

so

db/dt = 3.94 inches/sec

distance changing at rate of 3.94 inches/sec

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge

it is called kinetic energy