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Claculus help!?
Calculus question!?
Water is leaking out of an inverted conical tank at a rate of 10,000 cm^3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height of 6m and the diameter at the top is 4m. If the water level is rising at a rate of 20cm/min when the height of the water is 2m (200cm), find the rate at which water is being pumped into the tank.
Volume of cone=(pi*r^2 *h)/3
Mike, I have already done 45 questions of my own homework! I'm stock at this one!
4 Answers
- gudspelingLv 71 decade agoFavorite Answer
V = (1/3)πr²h
r = (2/6)h = h/3
V = (1/27)πh³
dV/dh = (1/9)πh²
When h=200cm, dh/dt = 20cm/min
dV/dt = (dV/dh)(dh/dt) = (1/9)π*200² * 20
= 800,000π/9 cm³/min
Water is leaking at the rate of 10,000cm³/min
Rate at which water is being pumped into the tank
= 800,000π/9 cm³/min + 10,000cm³/min
≈ 289,250 cm³/min
- Stephen YLv 61 decade ago
he trick seems to be to find the radius of the cone at a height of 2 meters.
I do not have access to a trigonometric function table but if we take a cross section of the cone we will have an isosceles triangle. bisecting that triangle we get a right triangle with sides a and b = 6 meters and 2 meters.
That should let us find tangent of the angle. recalculate using a side of 2 meters instead of 6 and you should be able to find the radius of the cone at that height.
10,000 cm*3 + the change in volume should give you the rate at which water is being pumped in.
I hope this helps.
