Question
Yan
water leaks from a hemispherical bowl of 8 cm diameter at the rate of 10 cm per hour .How fast is the surface
falling when the water is 2 cm deep in the bowl
water leaks from a hemispherical bowl of 8 cm diameter at the rate of 10 cm per hour .How fast is the surface
falling when the water is 2 cm deep in the bowl
Answers
Steve
The volume of a spherical cap of radius r and height h is
v = π/3 h^2(3r-h)
= πrh^2 - π/3 h^3
dv/dt = πh^2 dr/dt + 2πrh dh/dt - πh^2 dh/dt
= π(2rh-h^2) dh/dt
Now just plug in the numbers:
π(2*4*2-4) dh/dt = -10
dh/dt = -5/(6π) cm/hr
v = π/3 h^2(3r-h)
= πrh^2 - π/3 h^3
dv/dt = πh^2 dr/dt + 2πrh dh/dt - πh^2 dh/dt
= π(2rh-h^2) dh/dt
Now just plug in the numbers:
π(2*4*2-4) dh/dt = -10
dh/dt = -5/(6π) cm/hr