Answer:
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
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Sorry!
This cannot be answered. We don't have weight, height, etc.
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