(1 point) Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 6 cubic feet per minute. If the pool has radius 3 feet and height 6 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 2 feet

dh/dt = 2/3*π  ft/min

Step-by-step explanation:

We have a right circular cylinder, and water is pouring at a constant rate, we must expect the height  of the water will rise a constant rate, therefore it does not matter the height

V(c) = π*r²*h     ⇒  V(c) = π*(3)²*h  ⇒ V(c) =9*π*h

DV(c)/dh  =  9*π

DV(c)/dh = DV(c)/dt  *  dt/ dh    ⇒   9*π  =   DV(c)/dt  *  dt/ dh  

dh/dt  =   DV(c)/dt /  9*π

DV(c)/dt  = 6 ft³/m     ( from problem statement )

Then

dh/dt = 6/9*π ft/min

dh/dt = 2/3*π  ft/min

step-by-step explanation:

you didn't post any answer choices, so its not 100% clear what your teacher is looking for

however, log(15/7) is equivalent to log(15) - log(7)

i'm using the rule log(x/y) = log(x) - log(y)