Asked by Francois
Water is flowing into a hemispherical bowl having a radius of 10cm at constant rate of 3cm^3/min. When the water is x cm deep, The water level rises at a rate of 0.0149 cm/min. What is the value of x?
Answers
Answered by
bobpursley
The first thing is to develpe a relation between depth and volume
In the SPHERICAL CAP (which you can google),
Volume = (pi/3)x2(3R - x)
R=10cm
dV/dt=(PI/3)(6Rx-3x^2) dx/dt
you know dv/dt, and dx/dt. Solve for x. Looks like something of a quadratic equation, use the quadratic formula.
In the SPHERICAL CAP (which you can google),
Volume = (pi/3)x2(3R - x)
R=10cm
dV/dt=(PI/3)(6Rx-3x^2) dx/dt
you know dv/dt, and dx/dt. Solve for x. Looks like something of a quadratic equation, use the quadratic formula.
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