Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 7 cubic feet per minute. if the pool has radius 5 feet and height 8 feet, what is the rate of change of the height of the water in the pool when the depth of the water in the pool is 5 feet?
Question
Answer:
v = (pi)r² h v = (pi)5² h
v = 25(pi) h
take derivative with respect to h
dv/dh = 25(pi)
(dv/dt)*(dt/dh) = dv/dh
dv/dt is given to be 5 cm³
solved for dv/dh = 25(pi)
5(dt/dh) = 25(pi)
dt/dh = 25(pi)/5
dt/dh = 5(pi)
then the recipricol of dt/dh; we want to find dh/dt = 1/(5(pi))
solved
general
10 months ago
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