The rate at which water is being drained from the container is;
dV/dt = 24.4 in³/s
I've attached an image showing this inverted cone with the corresponding letters to represent the height and radius.
We are given;
Radius of Cone; R = 5 inches
Height of cone; H = 6 inches
h = 4 inches
Height rate of change; dh/dt = 0.7 in/s
Looking at the image attached, we can use similar triangle property to derive the relationship;
r/R = h/H
Where r is the radius at the instant where dh/dt = 0.7 in/s
Thus;
r/5 = 4/6
r = (5 × 4)/6
r = 3.33 inches
Now, from r/R = h/H
We can write with initial values of cone and differentiate;
r/5 = h/6
6r = 5h
6(dr/dt) = 5(dh/dt)
Since dh/dt = 0.7 in/s
Thus;
dr/dt = (5/6) × 0.7
dr/dt = 0.5833
Volume of cone is given by;
V = ⅓πr²h
We can find the rate at which the rate at which water is being drained from the container using partial differentiation on the volume equation.
Thus;
dV/dt = ⅓π((2rh × dr/dt) + (r² × dh/dt))
Plugging in the relevant values, we have;
dV/dt = ⅓π((2 × 3.33 × 4 × 0.5833) + (3.33² × 0.7))
dV/dt = 24.4 in³/s
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