Rough Cubic
Rough Cubic
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Rough Cubic

Calculus BC question?
You have a wine glass, approximated by the parabola y = -.8(x^2) +9.5. If water is being poured in at a constant rate of 2 cubic centimeters per second, at what rate is the water level rising when the object contains half its volume?
I determined the volume to be 177.2055 cubic centimeters (rotating half of the parabola around the y-axis), so half that would be 88.60275 cubic centimeters. A very rough approximation gives that the height at that time would be about 5.75 cm. I don't have a formula for the volume (I used integrals to find it originally), so I don't know how to do the related rates for this. Anyone know?
I'm sure you didn't mean the minus sign in the parabola equation. Is there a given maximum value for y?
You are given dV/dt and are asked the value of dy/dt at some particular y.
dV/dt = (dV/dy)(dy/dt)
so
dy/dt = (dV/dt)/(dV/dy) = 2/(dV/dy)
If you used the method of disks to compute the volume, then
dV = π[x(y)]² dy so
dV/dy = π[x(y)]² so
dy/dt = 2/(π[x(y)]²)
If y = 0.8x² + 9.5, then
x² = (y - 9.5)/0.8
so
dy/dt = 2/(π[(y - 9.5)/0.8] = (1.6/π)[1/(y - 9.5)]
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