Asked by RK
A cylindrical jar of radius 5 cm contains water to a depth of 8 cm. the water from jar is poured at a constant rate into a hemispherical bowl. After t seconds, the depth of the water in the bowl is x cm and the volume, V of water is given as 1/3πh^2(18-h). Given that all the water is transferred in π seconds, find
I) dV/do
I I) the rate of change of x when x=8, giving your answer in terms of 1/π
I) dV/do
I I) the rate of change of x when x=8, giving your answer in terms of 1/π
Answers
Answered by
Steve
Assuming you meant x instead of h,
V = 1/3πx^2(18-x)
= 6πx^2 - 1/3πx^3
dV/dx = 12πx - πx^2
the jar's volume is 400π cm^3. If it drains in π seconds, that's 400cm^3/s. That is dV/dt
dV/dt = (12π - 2πx) dx/dt
-400 = (12π-16π) dx/dt
dx/dt = 100(1/π)
V = 1/3πx^2(18-x)
= 6πx^2 - 1/3πx^3
dV/dx = 12πx - πx^2
the jar's volume is 400π cm^3. If it drains in π seconds, that's 400cm^3/s. That is dV/dt
dV/dt = (12π - 2πx) dx/dt
-400 = (12π-16π) dx/dt
dx/dt = 100(1/π)
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